# NCERT Solutions Class 12 Maths Chapter 13

## NCERT Solutions Class 12 Mathematics Chapter 13 – Probability

The Class 12 Mathematics textbook has a total of 13 essential chapters. The last chapter 13 is about ‘Probability’ which continues after the ‘Linear Programming’ section. With NCERT Solutions Class 12 Mathematics Chapter 13, students will learn about Probability in detail. Extramarks notes are easy to understand as the subject experts ensure that solutions are given in the simplest form. These solutions include all the exercise questions covered in the book and harmonise with the new update on the CBSE syllabus.

Students can also utilise the solutions for Class 12 Mathematics chapter 13 to practise at their liberty. Students will get to learn the basic principles of Probability and get a strong base when they refer to our NCERT Solutions Class 12 Mathematics Chapter 13 for solutions. Our chapters are prepared by Extramarks experts who have tried to explain this topic in as simple terms as possible.

Probability is a core subject of Mathematics that deals with the occurrence of a random incident. Probability has been presented in Mathematics to foresee how likely events are to happen. This is the basic Probability theory, which is also applied in the Probability distribution, where you will learn the possibilities of the outcomes for a random observation.

NCERT Solutions Class 12 help students judge the accuracy of an event by analysing seemingly persuasive data. This gives students a solid introduction to Probability Mathematics while studying both advantages and limitations of applying Probability theory to different situations under varied conditions.

### Key Topics Covered In NCERT Class 12 Mathematics Chapter 13

Probability is the measure of the chance of an event to occur. Most of the events cannot be predicted with total certainty, but we can expect the likelihood of an incident that is about to happen. Probability can range from 0 to 1, where 0 means that the situation would be impossible, and 1 specifies a particular event.

The important topics discussed in NCERT Solutions for Class 12 Mathematics Chapter 13 – Probability are:

 Exercise Topic 13.1 Introduction 13.2 Conditional Probability 13.3 Multiplication Theorem on Probability 13.4 Independent Events 13.5 Bayes’ Theorem 13.6 Random Variables and its Probability Distributions 13.7 Bernoulli Trials and Binomial Distribution

#### 13.1 Introduction

The Probability formula has been defined as the possibility of an event to occur. The formula for an incident’s possibility is P(E) = Number of favourable outcomes / Total number of outcomes. This is equal to the ratio of promising results and the total number of outcomes. NCERT Solutions Class 12 Mathematics Chapter 13 provides detailed theory and practical solutions to Probability Mathematics.

#### 13.2 Conditional Probability

Conditional Probability is the feasibility of an incident or outcome occurring based on the happening of the previous event or outcome. This can consider any event that has already taken place in the past. NCERT Solutions Class 12 Mathematics Chapter 13 takes you through all the calculations necessary in conditional Probability. Also, the properties of conditional Probability are one or two current independent events and a set of prior incidents.

#### 13.3 Multiplication Theorem on Probability

According to the Probability multiplication rule, the likelihood of two occurrences of both the events A and B is equal to the product of the Probability of B occurring and the conditional Probability that event A happens while simultaneously event B occurs. The outcome of their distinct possibilities represents the Probability of simultaneous occurrences of two self-determining incidents. NCERT Solutions Class 12 Mathematics Chapter 13 clarifies the Theorem with practical examples.

#### 13.4 Independent Events

The events are independent if the occurrence is not dependent on any other event, or the change in any one event doesn’t result in any change in the occurrence of another event. The events are mutually independent of each other. NCERT Solutions Class 12 Mathematics Chapter 13 lets you learn the definition of separate events in Probability and different diagrams and examples. There are various types of events such as mutually exclusive events, dependent events and independent events.

#### 13.5 Bayes’ Theorem

Bayes’ Theorem describes the Probability of occurrence of an incident related to any condition. The formula for the possibility of causes is known as Bayes’ Theorem. You can learn about Bayes’ Theorem, its formula, derivation and solved examples in NCERT Solutions Class 12 Mathematics Chapter 13 by Extramarks. The Probability of occurrences of an incident calculated depending on other conditions is also called conditional Probability. Bayes’ Theorem can be obtained for events and random variables separately, implementing the definition of conditional Probability and density.

#### 13.6 Random Variables and their Probability Distributions

Random Variables and its Probability Distributions is a real-valued function whose domain is the sample space of the random experiment. That means each outcome of a random experiment is related to a single real number, and a single real number may vary with the varied results of a random experiment. You can go through the example experiments provided in NCERT Solutions Class 12 Mathematics Chapter 13 to understand these distributions better. There are mean, variance and standard deviation of random variables.

#### 13.7 Bernoulli Trials and Binomial Distribution

Bernoulli distribution is the Probability distribution for a series of Bernoulli trials with only two possible results. Bernoulli Trials are also called a binomial distribution. In a binomial distribution, we obtain a number of successes in a series of independent experiments, but in Bernoulli trials, we have only two possible outcomes. With NCERT Solutions Class 12 Mathematics Chapter 13, you can learn about Bernoulli distribution, its properties, Bernoulli trial formula, and the relationship between Bernoulli trials and binomial distribution.

### Access NCERT Solutions Class 12 Mathematics Chapter 13 Exercises & Answer Solutions

The exercise and solutions paper are available on the Extramarks website. These questions will help students clear all doubts and understand the concept better to solve the Mathematical problems.  NCERT Solutions Class 12 Mathematics Chapter 13 includes questions according to the latest CBSE syllabus of 2022-23.

Exercise 13.1 Introduction Questions & Solutions refer to the below link

Exercise 13.1 Solutions – 3 Questions and Solutions

Exercise 13.2 Conditional Probability – Questions & Solutions refer to the below link

Exercise 13.2 Solutions – 18 Questions and Solutions

Exercise 13.3 Multiplication Theorem on Probability – Questions & Solutions click below.

Exercise 13.3 Solutions – 14 Questions and Solutions

Exercise 13.4 Independent Events – Questions & Solutions click here to know more.

Exercise 13.4 Solutions – 17 Questions and Solutions

Exercise 13.5 Bayes’ Theorem – Questions & Solutions refer to the link below for further details.

Exercise 13.5 Solutions – 15 Questions and Solutions

Exercise 13.6 Random Variables and its Probability Distributions – Questions & Solutions follow the links.

Exercise 13.7 Bernoulli Trials and Binomial Distribution – Questions & Solutions Click on this link

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### NCERT Exemplar for Class 12 Mathematics

NCERT Exemplars are practice books that include additional higher-level questions and are meant to aid meticulous learning. They are mainly used for competitive exams. These topics are up to date, consisting of different topics explained in every chapter of the Class 12 Mathematics book. Students can click on the links for these Exemplar questions and answers for Class 12 Mathematics chapters from the Extramarks website.

Considering the standard and variety, practising Exemplar problems is essential for school boards and competitive examinations. Students can perfectly evaluate themselves by working on these problems and improving their problem-solving skills for future studies. NCERT Solutions Class 12 Mathematics Chapter 13 provides a series of Exemplar Solutions for Class 12 for detailed clarification on all the questions given in the NCERT Exemplar Class 12 books.

### Key Features of NCERT Solutions for Class 12 Mathematics Chapter 13

To score well in competitive exams, one should have a solid base in Mathematics. The NCERT books present you with demanding questions that better your analytical skills and give you sufficient exposure to all kinds of questions in any of these exams. Our application is highly beneficial for your preparation for the following reasons.

• NCERT Solutions Class 12 Mathematics Chapter 13 answers to all numerical problems given in the textbook and are given with a step-by-step simplification.
• Free unlimited access.
• Highly detailed answers to all the theoretical as well as logical reasoning questions from the NCERT Solutions Class 12 Mathematics Chapter 13 that are verified by Extramarks subject experts.
• The reputed teachers’ accurate and straightforward answers cover all the complex and specific topics.
• The material provided by NCERT guarantees students to score remarkably in the exams.

Q.1

$\text{Compute P(A|B), if P(B) = 0.5and P(A∩B) =0.32.}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{P}\left(\mathrm{B}\right) = 0.5\mathrm{and}\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right) = 0.32\\ \mathrm{So},\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{0.32}{0.5}\\ =\frac{16}{25}\end{array}$

Q.2

$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{P}\left(\mathrm{A}\right)=0.8,\mathrm{P}\left(\mathrm{B}\right)=0.5\mathrm{and}\mathrm{}\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=0.4,\mathrm{}\mathrm{find}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{ItisgiventhatP}\left(\mathrm{A}\right)=0.8,\mathrm{P}\left(\mathrm{B}\right)=0.5,\mathrm{andP}\left(\mathrm{B}|\mathrm{A}\right)=0.4\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=0.4\\ \therefore \frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}=0.4\\ ⇒\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{0.8}=0.4\\ ⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=0.32\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \mathrm{ }=\frac{0.32}{0.5}\\ \mathrm{ }=0.64\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \mathrm{ }=0.8+0.5-0.32\\ \mathrm{ }=0.98\end{array}$

Q.3

$\text{E}\text{v}\text{a}\text{l}\text{u}\text{a}\text{t}\text{e}\text{}\text{P}\left(\text{A}\cup \text{B}\right),\text{}\text{i}\text{f}\text{}2\text{P}\left(\text{A}\right)=\text{P}\left(\text{B}\right)=\frac{5}{13}\text{}\text{a}\text{n}\text{d}\text{}\text{}\text{P}\left(\text{A}|\text{B}\right)=\frac{2}{5}.$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}2\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)=\frac{5}{13}⇒ \mathrm{P}\left(\mathrm{A}\right)=\frac{5}{26},\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)=\frac{5}{13}\\ \mathrm{and} \mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{2}{5}.\\ \therefore \mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \frac{2}{5}=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\left(\frac{5}{13}\right)}\\ ⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{ }\frac{2}{5}×\frac{5}{13}\\ \mathrm{ }=\frac{2}{13}\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \mathrm{ }=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}\\ \mathrm{ }=\frac{15}{26}-\frac{2}{13}\\ \mathrm{ }=\frac{15-4}{26}\mathrm{ }=\frac{11}{26}\end{array}$

Q.4

$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{P}\left(\mathrm{A}\right)=\frac{6}{11},\mathrm{P}\left(\mathrm{B}\right)=\frac{5}{11}\mathrm{and}\mathrm{}\left(\mathrm{A}\cup \mathrm{B}=\frac{7}{11},\mathrm{find}\left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\\ \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{6}}{11},\mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{5}}{11}\mathrm{and}\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\frac{\mathrm{7}}{11}\\ \left(\mathrm{i}\right)\mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \frac{\mathrm{7}}{11}\mathrm{ }=\frac{\mathrm{6}}{11}+\frac{\mathrm{5}}{11}-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ =\frac{11}{11}-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\mathrm{ }=\frac{11}{11}-\frac{7}{11}\mathrm{ }=\frac{4}{11}\\ \left(\mathrm{ii}\right)\mathrm{Since},\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)} =\frac{\left(\frac{4}{11}\right)}{\left(\frac{\mathrm{5}}{11}\right)}\\ =\frac{4}{5}\\ \left(\mathrm{iii}\right)\mathrm{Since},\\ \mathrm{ }\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\mathrm{ }=\frac{\left(\frac{4}{11}\right)}{\left(\frac{\mathrm{6}}{11}\right)}\\ =\frac{4}{6}=\frac{2}{3}\end{array}$

Q.5 A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(iii) E: at most two tails, F: at least one tail
Determine P (E|F).

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{a}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{three}\mathrm{times},\mathrm{then}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{is}\\ \mathrm{S}= \left\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THH},\mathrm{THT},\mathrm{TTH},\mathrm{TTT}\right\}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{has}8\mathrm{elements}.\\ \left(\mathrm{i}\right) \mathrm{ }\mathrm{E}=\left\{\mathrm{HHH},\mathrm{HTH},\mathrm{THH},\mathrm{TTH}\right\}\\ \mathrm{F}=\left\{\mathrm{HHH},\mathrm{HHT}\right\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HHH}\right\}\\ \mathrm{So},\mathrm{P}\left(\mathrm{E}\right)=\frac{4}{8}=\frac{1}{2}, \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{8}=\frac{1}{4}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{8}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{1}{8}\right)}{\left(\frac{1}{4}\right)}=\frac{1}{2}\\ \left(\mathrm{ii}\right)\right) \mathrm{E}=\left\{\mathrm{HHH},\mathrm{}\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH}\right\}\mathrm{}\\ \mathrm{ }\mathrm{F}=\left\{\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH},\mathrm{}\mathrm{TTT}\right\}\mathrm{}\\ \therefore \mathrm{ }\mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HHT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH}\right\}\\ \mathrm{So}, \mathrm{P}\left(\mathrm{F}\right)=\frac{7}{8}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{3}{8}\mathrm{}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{3}{8}\right)}{\left(\frac{7}{8}\right)}=\frac{3}{7}\\ \left(\mathrm{iii}\right) \mathrm{E}=\left\{\mathrm{HHH},\mathrm{}\mathrm{HHT},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH}\right\}\mathrm{}\\ \mathrm{ }\mathrm{F}=\left\{\mathrm{HHT},\mathrm{}\mathrm{HTT},\mathrm{}\mathrm{HTH},\mathrm{}\mathrm{THH},\mathrm{}\mathrm{THT},\mathrm{}\mathrm{TTH},\mathrm{}\mathrm{TTT}\right\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HHT},\mathrm{HTT},\mathrm{HTH},\mathrm{THH},\mathrm{THT},\mathrm{TTH}\right\}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{7}{8}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{6}{8}\\ \mathrm{So},\mathrm{}\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{6}{8}\right)}{\left(\frac{7}{8}\right)}=\frac{6}{7}\end{array}$

Q.6 Two coins are tossed once, where
(i) E: tail appears on one coin,
(ii) E: no tail appears, F : no head appears
Determine P (E|F).

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{two}\mathrm{coins}\mathrm{are}\mathrm{tossed}\mathrm{once},\mathrm{then}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{is}\\ \mathrm{ }\mathrm{S}= \left\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\right\}\\ \left(\mathrm{i}\right)\mathrm{E}= \left\{\mathrm{HT},\mathrm{TH}\right\}\\ \mathrm{F}= \left\{\mathrm{HT},\mathrm{TH}\right\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\mathrm{HT},\mathrm{TH}\right\}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{4}=\frac{1}{2}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{2}{4}=\frac{1}{2}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}=1\\ \left(\mathrm{ii}\right) \mathrm{ }\mathrm{E}=\left\{\mathrm{HH}\right\},\mathrm{F}=\left\{\mathrm{TT}\right\}\\ \mathrm{and}\mathrm{E}\cap \mathrm{F}=\mathrm{\varphi }\\ \therefore \mathrm{P}\left(\mathrm{F}\right)=\frac{1}{4},\mathrm{ }\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=0\\ \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{ }\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{0}{\left(\frac{1}{4}\right)}=0\end{array}$

Q.7 A die is thrown three times,
E: 4 appears on the third toss,
F: 6 and 5 appears respectively on first two tosses
Determine P (E|F).

Ans.

If a die is thrown three times, then the number of elements in the sample space will be 6× 6 × 6 = 216

$\begin{array}{l} \mathrm{ }\mathrm{E}=\left\{\begin{array}{l}\left(1,1,4\right),\left(1,2,4\right),\left(1,3,4\right),... \left(1,6,4\right)\\ \left(2,1,4\right),\left(2,2,4\right),\left(2,3,4\right),...\left(2,6,4\right)\\ \left(3,1,4\right),\left(3,2,4\right),\left(3,3,4\right),...\left(3,6,4\right)\\ \left(4,1,4\right),\left(4,2,4\right),\left(4,3,4\right),...\left(4,6,4\right)\\ \left(5,1,4\right),\left(5,2,4\right),\left(5,3,4\right),...\left(5,6,4\right)\\ \left(6,1,4\right),\left(6,2,4\right),\left(6,3,4\right),...\left(6,6,4\right)\end{array}\right\}\\ \mathrm{F}=\left\{\left(6,5,1\right),\left(6,5,2\right),\left(6,5,3\right),\left(6,5,4\right),\left(6,5,5\right),\left(6,5,6\right)\right\}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\left(6,5,4\right)\right\}\\ \mathrm{So},\mathrm{P}\left(\mathrm{F}\right)=\frac{6}{216}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{216}\\ \mathrm{Then}, \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ \mathrm{ }=\frac{\left(\frac{1}{216}\right)}{\left(\frac{6}{216}\right)}=\frac{1}{6}\end{array}$

Q.8 Mother, father and son line up at random for a family picture
E: son on one end,
F: father in middle
Determine P (E|F).

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{mother}\left(\mathrm{M}\right),\mathrm{father}\left(\mathrm{F}\right),\mathrm{and}\mathrm{son}\left(\mathrm{S}\right)\mathrm{line}\mathrm{up}\mathrm{for}\mathrm{the}\mathrm{family}\\ \mathrm{picture},\mathrm{then}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{will}\mathrm{be}\\ \mathrm{S}=\left\{\mathrm{MFS},\mathrm{}\mathrm{MSF},\mathrm{}\mathrm{FMS},\mathrm{}\mathrm{FSM},\mathrm{}\mathrm{SMF},\mathrm{}\mathrm{SFM}\right\}\mathrm{}\\ \mathrm{E}=\left\{\mathrm{MFS},\mathrm{}\mathrm{FMS},\mathrm{}\mathrm{SMF},\mathrm{}\mathrm{SFM}\right\}\\ \mathrm{F}=\left\{\mathrm{MFS},\mathrm{SFM}\right\}\\ \mathrm{So},\mathrm{E}\cap \mathrm{F}=\left\{\mathrm{MFS},\mathrm{SFM}\right\}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{6}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{2}{6}\\ \therefore \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{2}{6}\right)}{\left(\frac{2}{6}\right)}=1\mathrm{}\end{array}$

Q.9 A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{observation}\mathrm{be}\mathrm{from}\mathrm{the}\mathrm{black}\mathrm{die}\mathrm{and}\mathrm{second}\mathrm{from}\\ \mathrm{the}\mathrm{red}\mathrm{die}.\mathrm{When}\mathrm{two}\mathrm{dice}\left(\mathrm{one}\mathrm{black}\mathrm{and}\mathrm{another}\mathrm{red}\right)\mathrm{are}\mathrm{rolled},\\ \mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{has}6 × 6=36\mathrm{number}\mathrm{of}\mathrm{elements}.\\ \left(\mathrm{a}\right)\mathrm{Let}\\ \mathrm{A}:\mathrm{Obtaining}\mathrm{a}\mathrm{sum}\mathrm{greater}\mathrm{than}9\\ =\left\{\left(4, 6\right), \left(5, 5\right), \left(5, 6\right), \left(6, 4\right), \left(6, 5\right), \left(6, 6\right)\right\}\\ \mathrm{B}:\mathrm{Black}\mathrm{die}\mathrm{results}\mathrm{in}\mathrm{a}5.\\ =\left\{\left(5, 1\right), \left(5, 2\right), \left(5, 3\right), \left(5, 4\right), \left(5, 5\right), \left(5, 6\right)\right\}\\ \therefore \mathrm{ }\mathrm{A}\cap \mathrm{B}=\left\{\left(5, 5\right), \left(5, 6\right)\right\}\\ \mathrm{P}\left(\mathrm{B}\right)=\frac{6}{36}\mathrm{and}\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{2}{36}\mathrm{}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{of}\mathrm{obtaining}\mathrm{a}\mathrm{sum}\mathrm{greater}\mathrm{than}9,\\ \mathrm{given}\mathrm{that}\mathrm{the}\mathrm{black}\mathrm{die}\mathrm{resulted}\mathrm{in}\mathrm{a}5,\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right).\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ =\frac{\left(\frac{2}{36}\right)}{\left(\frac{6}{36}\right)}\\ =\frac{2}{6}=\frac{1}{3}\\ \left(\mathrm{b}\right)\right)\mathrm{E}:\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{observations}\mathrm{is}8.\\ =\left\{\left(2, 6\right), \left(3, 5\right), \left(4, 4\right), \left(5, 3\right), \left(6, 2\right)\right\}\\ \mathrm{F}:\mathrm{Red}\mathrm{die}\mathrm{resulted}\mathrm{in}\mathrm{a}\mathrm{number}\mathrm{less}\mathrm{than}4.\\ =\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(2,1\right),\left(2,2\right),\left(2,3\right)\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(4,1\right),\left(4,2\right),\left(4,3\right)\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(6,1\right),\left(6,2\right),\left(6,3\right)\end{array}\right\}\\ \mathrm{and}\mathrm{ }\mathrm{E}\cap \mathrm{F}=\left\{\left(5,3\right),\left(6,2\right)\right\}\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{18}{36}\mathrm{and}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{2}{36}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{of}\mathrm{obtaining}\mathrm{the}\mathrm{sum}\mathrm{equal}\mathrm{to}8,\\ \mathrm{given}\mathrm{that}\mathrm{the}\mathrm{red}\mathrm{die}\mathrm{resulted}\mathrm{in}\mathrm{a}\mathrm{number}\mathrm{less}\mathrm{than}4,\\ \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{1}{9}\end{array}$

Q.10 A fair die is rolled. Consider events
E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}
Find
(i) P(E|F) and P (F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)

Ans.

$\begin{array}{l}\mathrm{When}\mathrm{a}\mathrm{fair}\mathrm{die}\mathrm{is}\mathrm{rolled},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{S}\mathrm{will}\mathrm{be}\\ \mathrm{S}= \left\{1, 2, 3, 4, 5, 6\right\}\\ \mathrm{then}\mathrm{given}\mathrm{events}\mathrm{are}\mathrm{E}=\left\{1,3,5\right\},\mathrm{}\mathrm{F}=\left\{2,3\right\}\mathrm{}\mathrm{and}\mathrm{}\mathrm{G}=\left\{2,3,4,5\right\}\\ \mathrm{So},\mathrm{E}\cap \mathrm{F}=\left\{3\right\},\mathrm{then}\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{6}, \mathrm{P}\left(\mathrm{E}\right)=\frac{3}{6},\mathrm{}\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{2}{6}\mathrm{and}\mathrm{P}\left(\mathrm{G}\right)=\frac{4}{6}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ \mathrm{ }=\frac{\left(\frac{1}{6}\right)}{\left(\frac{2}{6}\right)}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{F}|\mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ \mathrm{ }=\frac{\left(\frac{1}{6}\right)}{\left(\frac{3}{6}\right)}=\frac{1}{3}\\ \left(\mathrm{ii}\right) \mathrm{ }\mathrm{E}\cap \mathrm{G}=\left\{5\right\}\\ \mathrm{P}\left(\mathrm{E}\cap \mathrm{G}\right)=\frac{1}{6}\\ \mathrm{ }\mathrm{P}\left(\mathrm{E}|\mathrm{G}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{G}\right)}\\ =\frac{\left(\frac{1}{6}\right)}{\left(\frac{4}{6}\right)}=\frac{1}{4}\\ \therefore \mathrm{P}\left(\mathrm{G}|\mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ =\frac{\left(\frac{1}{6}\right)}{\left(\frac{3}{6}\right)}=\frac{1}{3}\\ \left(\mathrm{iii}\right) \mathrm{ }\left(\mathrm{E}\cup \mathrm{F}\right)=\left\{1,2,3,5\right\}\\ \left(\mathrm{E}\cup \mathrm{F}\right)\cap \mathrm{G}=\left\{2,3,5\right\}\\ \mathrm{P}\left(\left(\mathrm{E}\cup \mathrm{F}\right)\cap \mathrm{G}\right)=\frac{3}{6}=\frac{1}{2}\\ \mathrm{ }\mathrm{P}\left\{\left(\mathrm{E}\cup \mathrm{F}\right)|\mathrm{G}\right\}=\frac{\mathrm{P}\left(\left(\mathrm{E}\cup \mathrm{F}\right)\cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{G}\right)}\\ =\frac{\left(\frac{1}{2}\right)}{\left(\frac{4}{6}\right)}=\frac{3}{4}\\ \left(\mathrm{E}\cap \mathrm{F}\right)\cap \mathrm{G}=\left\{3\right\}\\ \mathrm{ }\mathrm{P}\left(\left(\mathrm{E}\cap \mathrm{F}\right)\cap \mathrm{G}\right)=\frac{1}{6}\\ \therefore \mathrm{ }\mathrm{P}\left(\left(\mathrm{E}\cap \mathrm{F}\right)|\mathrm{G}\right)=\frac{\mathrm{P}\left(\left(\mathrm{E}\cap \mathrm{F}\right)\cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{G}\right)}\\ =\frac{\left(\frac{1}{6}\right)}{\left(\frac{4}{6}\right)}=\frac{1}{4}\end{array}$

Q.11 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{B}\mathrm{and}\mathrm{G}\mathrm{represent}\mathrm{the}\mathrm{boy}\mathrm{and}\mathrm{the}\mathrm{girl}\mathrm{child}\mathrm{respectively}.\mathrm{If}\\ \mathrm{a}\mathrm{family}\mathrm{has}\mathrm{two}\mathrm{children},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{will}\mathrm{be}\\ \mathrm{S}= \left\{\left(\mathrm{B},\mathrm{B}\right), \left(\mathrm{B},\mathrm{G}\right), \left(\mathrm{G},\mathrm{B}\right), \left(\mathrm{G},\mathrm{G}\right)\right\}\\ \mathrm{Let}\mathrm{A}=\mathrm{both}\mathrm{children}\mathrm{are}\mathrm{girls}=\left\{\left(\mathrm{G},\mathrm{G}\right)\right\}\\ \left(\mathrm{i}\right)\mathrm{Let}\mathrm{B}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{youngest}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl},\mathrm{then}\\ \mathrm{ }\mathrm{B}=\left\{\left(\mathrm{B},\mathrm{G}\right),\left(\mathrm{G},\mathrm{G}\right)\right\}\\ ⇒\mathrm{A}\cap \mathrm{B}=\left\{\left(\mathrm{G},\mathrm{G}\right)\right\}\\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{2}{4}=\frac{1}{2}\mathrm{and}\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{1}{4}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{that}\mathrm{both}\mathrm{are}\mathrm{girls},\mathrm{given}\mathrm{that}\mathrm{the}\\ \mathrm{youngest}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl}=\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\\ =\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ =\frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{2}\right)}=\frac{1}{2}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{1}{2}.\\ \left(\mathrm{ii}\right)\mathrm{Let}\mathrm{E}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl}.\\ \mathrm{E}=\left\{\left(\mathrm{B},\mathrm{G}\right),\left(\mathrm{G},\mathrm{B}\right),\left(\mathrm{G},\mathrm{G}\right)\right\}\\ \mathrm{Then}, \mathrm{A}\cap \mathrm{E}=\left\{\left(\mathrm{G},\mathrm{G}\right)\right\}\\ \therefore \mathrm{P}\left(\mathrm{A}\cap \mathrm{E}\right)=\frac{1}{4}\mathrm{and}\mathrm{P}\left(\mathrm{E}\right)=\frac{3}{4}\\ \mathrm{The}\mathrm{conditional}\mathrm{probability}\mathrm{that}\mathrm{both}\mathrm{are}\mathrm{girls},\mathrm{given}\mathrm{that}\mathrm{at}\mathrm{least}\\ \mathrm{one}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{girl}=\mathrm{P}\left(\mathrm{A}|\mathrm{E}\right)\\ =\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{E}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ =\frac{\left(\frac{1}{4}\right)}{\left(\frac{3}{4}\right)}=\frac{1}{3}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{1}{3}.\end{array}$

Q.12 An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Ans.

Number of easy True/False questions = 300
Number of easy multiple choice questions = 500
Number of difficult True/False questions = 200
Number of difficult multiple choice questions = 400
Total number of multiple choice questions = 900
Total questions = 1400
Let the notations of questions are as follows:
E = Easy questions, D = Difficult questions, M = Multiple choice questions and T = True/False questions.

$\begin{array}{l}\mathrm{So},\mathrm{Probability}\mathrm{of}\mathrm{easy}\mathrm{and}\mathrm{multiple}\mathrm{choice}\mathrm{questions}\\ \mathrm{P}\left(\mathrm{E}\cap \mathrm{M}\right)\mathrm{ }=\frac{500}{1400}\\ =\frac{5}{14}\\ \mathrm{Probability}\mathrm{of}\mathrm{multiple}\mathrm{choice}\mathrm{question},\mathrm{P}\left(\mathrm{M}\right)=\frac{900}{1400}\\ =\frac{9}{14}\\ \mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{randomly}\mathrm{selected}\mathrm{an}\mathrm{easy}\mathrm{question},\mathrm{given}\mathrm{that}\\ \mathrm{it}\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{choice}\mathrm{question}=\mathrm{P}\left(\mathrm{E}|\mathrm{M}\right)\\ =\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{M}\right)}{\mathrm{P}\left(\mathrm{M}\right)}\\ =\frac{\left(\frac{5}{14}\right)}{\left(\frac{9}{14}\right)}=\frac{5}{9}\\ \mathrm{Thus},\mathrm{required}\mathrm{probability}\mathrm{is}\frac{5}{9}.\end{array}$

Q.13 Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Ans.

When two dice are thrown, sample space observations = 6 × 6 = 36
Let A = the sum of the numbers on the dice is 4 and
B = the two numbers appearing on throwing the two dice are different.

$\begin{array}{l}\therefore \mathrm{A}=\left\{\left(1,3\right),\left(3,1\right)\right\}\\ \mathrm{B}=\left\{\begin{array}{l}\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right)\\ \left(2,1\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right)\\ \left(3,1\right),\left(3,2\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,5\right),\left(4,6\right)\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,6\right)\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right)\end{array}\right\}\\ \mathrm{Then},\\ \mathrm{A}\cap \mathrm{B}=\left\{\left(1,3\right),\left(3,1\right)\right\}\\ \mathrm{P}\left(\mathrm{B}\right)=\frac{30}{36}=\frac{5}{6}\mathrm{and}\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{2}{30}=\frac{1}{18}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{numbers}\mathrm{on}\mathrm{the}\mathrm{dice}\mathrm{is}4,\\ \mathrm{given}\mathrm{that}\mathrm{the}\mathrm{two}\mathrm{numbers}\mathrm{appearing}\mathrm{on}\mathrm{throwing}\mathrm{the}\mathrm{two}\mathrm{dice}\\ \mathrm{are}\mathrm{different}=\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\\ =\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ =\frac{\left(\frac{1}{18}\right)}{\left(\frac{5}{6}\right)}=\frac{1}{15}\end{array}$

Q.14 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Ans.

$\begin{array}{l}\text{The sample space of the experiment is given below:}\\ \text{S}=\left\{\begin{array}{l}\left(1,H\right),\left(1,T\right),\left(2,H\right),\left(2,T\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\\ \left(4,H\right),\left(4,T\right),\left(5,H\right),\left(5,T\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right),\end{array}\right\}\\ \text{Let A}=\text{the coin shows a tail}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left\{\left(1,T\right),\left(2,T\right),\left(4,T\right),\left(5,T\right)\right\}\text{and}\\ \text{B}=\text{at least one die shows 3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left\{\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(6,3\right)\right\}\\ A\cap B=\varphi ⇒P\left(A\cap B\right)=0\\ P\left(B\right)=P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}+P\left\{\left(3,1\right)\right\}\\ +\text{\hspace{0.17em}}P\left\{\left(3,1\right)\right\}\\ =\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{7}{36}\\ \text{Probability of the event that the coin shows a tail, given that at}\\ \text{least one die shows 3}\\ =\text{P(A|B)}\\ =\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ =\frac{0}{\left(\frac{7}{36}\right)}=0\\ \text{Thus, the required probability is 0}\text{.}\end{array}$

Q.15 If P (A) =1/2, P (B) = 0, then P (A|B) is
(A) 0 (B) 1/2
(C) not defined (D) 1

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2},\mathrm{P}\left(\mathrm{B}\right)=0,\mathrm{}\mathrm{then}\mathrm{}\\ \mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \mathrm{ }=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{0}\\ \mathrm{Therefore},\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\mathrm{is}\mathrm{not}\mathrm{defined}.\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.16 If A and B are events such that P (A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B (B) A = B
(C) A ∩ B = Φ (D) P (A) = P(B)

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that} \mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)\\ ⇒ \mathrm{ }\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\\ ⇒ \mathrm{ }\frac{1}{\mathrm{P}\left(\mathrm{B}\right)}=\frac{1}{\mathrm{P}\left(\mathrm{A}\right)}\\ ⇒ \mathrm{ }\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Thus},\mathrm{option}\mathrm{D}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.17 If P(A) = 3/5 and P (B) = 1/5, find P (A ∩ B) if A and B are independent events.

Ans.

$\begin{array}{l}\mathrm{Since},\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right) \left[\begin{array}{l}\mathrm{As}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent}\\ \mathrm{events}\mathrm{.}\end{array}\right]\\ =\frac{3}{5}.\frac{1}{5}\\ =\frac{3}{25}\\ \mathrm{Thus},\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{3}{25}.\end{array}$

Q.18 Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Ans.

$\begin{array}{l}\mathrm{Total}\mathrm{}\mathrm{number}\mathrm{}\mathrm{of}\mathrm{}\mathrm{cards}=52\\ \mathrm{Number}\mathrm{}\mathrm{of}\mathrm{}\mathrm{black}\mathrm{}\mathrm{cards}=26\\ \mathrm{Let}\mathrm{}\mathrm{A}=\mathrm{A}\mathrm{black}\mathrm{card}\mathrm{in}\mathrm{first}\mathrm{draw}\\ \mathrm{B}=\mathrm{A}\mathrm{black}\mathrm{card}\mathrm{in}\mathrm{second}\mathrm{draw}\\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{26}{52}=\frac{1}{2}\\ \mathrm{ }\mathrm{P}\left(\mathrm{B}\right)=\frac{25}{51}\\ \mathrm{Thus},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{both}\mathrm{the}\mathrm{cards}\mathrm{black}\\ \mathrm{ }=\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{ }=\frac{1}{2}×\frac{25}{51}\\ \mathrm{ }=\frac{25}{102}\end{array}$

Q.19 A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{A},\mathrm{B},\mathrm{and}\mathrm{C}\mathrm{be}\mathrm{the}\mathrm{respective}\mathrm{events}\mathrm{that}\mathrm{the}\mathrm{first},\mathrm{second},\\ \mathrm{and}\mathrm{third}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good}\mathrm{.}\\ \therefore \mathrm{ }\mathrm{Probability}\mathrm{that}\mathrm{first}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good},\mathrm{P}\left(\mathrm{A}\right)=\frac{12}{15}\\ \mathrm{Probability}\mathrm{that}\mathrm{second}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good},\mathrm{P}\left(\mathrm{B}\right)=\frac{11}{14}\\ ⇒\mathrm{Probability}\mathrm{that}\mathrm{third}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good},\mathrm{P}\left(\mathrm{C}\right)=\frac{10}{13}\\ \mathrm{The}\mathrm{box}\mathrm{is}\mathrm{approved}\mathrm{for}\mathrm{sale},\mathrm{if}\mathrm{all}\mathrm{the}\mathrm{three}\mathrm{oranges}\mathrm{are}\mathrm{good}\mathrm{.}\\ \mathrm{ }\mathrm{Thus},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{all}\mathrm{the}\mathrm{oranges}\mathrm{good}=\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right).\mathrm{P}\left(\mathrm{C}\right)\\ \mathrm{ }=\frac{12}{15}×\frac{11}{14}×\frac{10}{13}\\ \mathrm{ }=\frac{44}{91}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{box}\mathrm{is}\mathrm{approved}\mathrm{for}\mathrm{sale}\mathrm{is} \frac{44}{91}.\end{array}$

Q.20 A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Ans.

$\begin{array}{l}\mathrm{Sample}\mathrm{space},\mathrm{when}\mathrm{a}\mathrm{fair}\mathrm{coin}\mathrm{and}\mathrm{an}\mathrm{unbiased}\mathrm{die}\mathrm{are}\mathrm{tossed},\\ \mathrm{ }\mathrm{S}=\left\{\begin{array}{l}\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right)\\ \left(\mathrm{T},1\right),\left(\mathrm{T},2\right),\left(\mathrm{T},3\right),\left(\mathrm{T},4\right),\left(\mathrm{T},5\right),\left(\mathrm{T},6\right)\end{array}\right\}\\ \mathrm{ }\mathrm{A}=\mathrm{Head}\mathrm{appears}\mathrm{on}\mathrm{the}\mathrm{coin}\\ =\left\{\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right)\right\}\\ \mathrm{ }\mathrm{B}=3\mathrm{on}\mathrm{the}\mathrm{die}\\ =\left\{\left(\mathrm{H},3\right),\left(\mathrm{T},3\right)\right\}\\ \mathrm{ }\mathrm{A}\cap \mathrm{B}=\left\{\left(\mathrm{H},3\right)\right\}\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{1}{12}\\ \mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)=\frac{6}{12}×\frac{2}{12}\\ =\frac{1}{12}\\ \mathrm{Since},\mathrm{ }\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Therefore},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.21 A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

Ans.

$\begin{array}{l}\mathrm{Sample}\mathrm{space},\mathrm{when}\mathrm{an}\mathrm{unbiased}\mathrm{die}\mathrm{are}\mathrm{tossed},\\ \mathrm{ }\mathrm{S}=\left\{1,2,3,4,5,6\right\}\\ \mathrm{ }\mathrm{A}=\mathrm{The}\mathrm{number}\mathrm{is}\mathrm{even}\\ =\left\{2, 4, 6\right\}\\ \mathrm{ }\mathrm{B}=\mathrm{The}\mathrm{number}\mathrm{is}\mathrm{red}\\ =\left\{4,5,6\right\}\\ \mathrm{ }\mathrm{A}\cap \mathrm{B}=\left\{4,6\right\}\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{2}{6}=\frac{1}{3}\\ \mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)=\frac{3}{6}×\frac{3}{6}\\ =\frac{1}{4}\\ \mathrm{Since},\mathrm{ }\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Therefore},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.22 Let E and F be events with P (E) = 3/5, P(F) = 3/10 and P (E ∩ F) = 1/5 . Are E and F independent?

Ans.

$\begin{array}{l}\mathrm{E}\mathrm{}\mathrm{and}\mathrm{}\mathrm{F}\mathrm{}\mathrm{be}\mathrm{}\mathrm{events}\mathrm{}\mathrm{with}\mathrm{}\mathrm{P}\left(\mathrm{E}\right)=\frac{3}{5},\mathrm{}\mathrm{P}\left(\mathrm{F}\right)=\frac{3}{10} \mathrm{and}\mathrm{}\mathrm{P}\mathrm{}\left(\mathrm{E}\cap \mathrm{F}\right)=1/5\\ \mathrm{P}\left(\mathrm{E}\right)×\mathrm{P}\left(\mathrm{F}\right)=\frac{3}{5}×\frac{3}{10}\\ =\frac{9}{50}\\ \mathrm{Since},\mathrm{ }\\ \mathrm{P}\mathrm{}\left(\mathrm{E}\cap \mathrm{F}\right)\ne \mathrm{P}\left(\mathrm{E}\right)×\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Therefore},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.23 Given that the events A and B are such that
P(A) = 1/2 , P (A ∪ B) = 3/5 and
P(B) = p. Find p if they are
(i) mutually exclusive (ii) independent.

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{that}\mathrm{the}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{such}\mathrm{that}\\ \mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2},\mathrm{}\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\frac{3}{5} \mathrm{and}\mathrm{P}\left(\mathrm{B}\right)=\mathrm{p}\\ \left(\mathrm{i}\right)\mathrm{When}\mathrm{the}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{mutually}\mathrm{exclusive}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=0\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \frac{3}{5}=\frac{1}{2}+\mathrm{p}-0\\ ⇒\mathrm{p}=\frac{3}{5}-\frac{1}{2}\\ =\frac{6-5}{10}\\ ⇒\mathrm{p}=\frac{1}{10}\\ \left(\mathrm{ii}\right)\mathrm{When}\mathrm{the}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right).\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right).\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)\\ ⇒\frac{3}{5}=\frac{1}{2}+\mathrm{p}-\frac{1}{2}×\mathrm{p}\\ ⇒ \frac{3}{5}-\frac{1}{2}=\mathrm{p}\left(1-\frac{1}{2}\right)\\ ⇒ \frac{6-5}{10}=\frac{1}{2}\mathrm{p}\\ \mathrm{p}=\frac{1}{10}×\frac{2}{1}=\frac{1}{5}\end{array}$

Q.24 Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A ∩ B) (ii) P(A ∪ B)
(iii) P(A|B) (iv) P (B|A)

Ans.

$\begin{array}{l}\mathrm{A}\mathrm{}\mathrm{and}\mathrm{}\mathrm{B}\mathrm{}\mathrm{be}\mathrm{}\mathrm{independent}\mathrm{}\mathrm{events}\mathrm{}\mathrm{with}\mathrm{P}\left(\mathrm{A}\right)=0.3\mathrm{}\mathrm{and}\mathrm{}\mathrm{P}\left(\mathrm{B}\right)=0.4\\ \left(\mathrm{i}\right) \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{ }=0.3×0.4\\ \mathrm{ }=0.12\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{ }=0.3+0.4-0.12\\ \mathrm{ }=0.7-0.12\\ \mathrm{ }=0.58\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \mathrm{ }=\frac{0.12}{0.4}\\ \mathrm{ }=0.3\\ \left(\mathrm{iv}\right)\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\\ \mathrm{ }=\frac{0.12}{0.3}\\ \mathrm{ }=0.4\end{array}$

Q.25 If A and B are two events such that P(A) =1/4, P(B)= 1/2 and P(A∩B)= 1/8, find P(not A and not B).

Ans.

$\begin{array}{l}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{two}\mathrm{events}\mathrm{such}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{1}}{4},\mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{1}}{2} \mathrm{and}\\ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{\mathrm{1}}{8}\\ \mathrm{P}\left(\mathrm{not}\mathrm{ }\mathrm{A}\mathrm{and}\mathrm{not}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}‘\cap \mathrm{B}‘\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)‘\left[\begin{array}{l}\mathrm{By}\mathrm{De}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\\ \left(\mathrm{A}‘\cap \mathrm{B}‘\right)=\left(\mathrm{A}\cup \mathrm{B}\right)‘\end{array}\right]\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ \mathrm{ }=1-\left\{\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\right\}\\ \mathrm{ }=1-\left\{\frac{\mathrm{1}}{4}+\frac{\mathrm{1}}{2}-\frac{\mathrm{1}}{4}×\frac{\mathrm{1}}{2}\right\}\\ \mathrm{ }=1-\left(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right)\\ \mathrm{ }=1-\left(\frac{2+4-1}{8}\right)\\ \mathrm{ }=1-\frac{5}{8}\\ \mathrm{ }=\frac{8-5}{8}\mathrm{ }=\frac{3}{8}\\ \mathrm{Thus},\mathrm{ }\mathrm{P}\left(\mathrm{not}\mathrm{ }\mathrm{A}\mathrm{and}\mathrm{not}\mathrm{B}\right)\mathrm{ }\mathrm{is} \frac{3}{8}.\end{array}$

Q.26 Events A and B are such that P (A) =1/2, P(B)=7/12 and P(not A or not B) = 1/4. State whether A and B are independent?

Ans.

$\begin{array}{l}\mathrm{Events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{such}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{1}}{2}, \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{7}}{12} \mathrm{and}\\ \mathrm{P}\left(\mathrm{not}\mathrm{A}\mathrm{or}\mathrm{not}\mathrm{B}\right)\mathrm{}=\frac{\mathrm{1}}{4}\\ \mathrm{P}\left(\mathrm{not}\mathrm{ }\mathrm{A}\mathrm{or}\mathrm{not}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}‘\cup \mathrm{B}‘\right)\\ \mathrm{ }\frac{\mathrm{1}}{4}\mathrm{ }=\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)‘\left[\begin{array}{l}\mathrm{By}\mathrm{De}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\\ \left(\mathrm{A}‘\cap \mathrm{B}‘\right)=\left(\mathrm{A}\cup \mathrm{B}\right)‘\end{array}\right]\\ \mathrm{ }\frac{\mathrm{1}}{4}=1-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=1-\frac{1}{4}\\ =\frac{3}{4}\\ \mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{1}}{2}×\frac{7}{12}\\ =\frac{7}{24}\\ \mathrm{So}, \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Thus},\mathrm{A}\mathrm{}\mathrm{and}\mathrm{}\mathrm{B}\mathrm{}\mathrm{are}\mathrm{}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.27 Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P(A and B) (ii) P(A and not B)
(iii) P(A or B) (iv) P(neither A nor B)

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{two}\mathrm{independent}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{such}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=0.\mathrm{3},\mathrm{}\\ \mathrm{P}\left(\mathrm{B}\right)=0.\mathrm{6}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)\\ =0.3×0.6\\ =0.18\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{not}\mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}‘\right)\\ =\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ =0.3-0.18\\ =0.12\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{A}\mathrm{or}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ =0.3+0.6-0.18\\ =0.9-0.18\\ =0.72\\ \left(\mathrm{iv}\right)\mathrm{P}\left(\mathrm{neither}\mathrm{}\mathrm{A}\mathrm{}\mathrm{nor}\mathrm{}\mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}‘\cap \mathrm{B}‘\right)\\ =\left[\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)‘\right]\left[\begin{array}{l}\mathrm{By}\mathrm{De}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\\ \mathrm{ }\mathrm{A}‘\cap \mathrm{B}‘=\left(\mathrm{A}\cup \mathrm{B}\right)‘\end{array}\right]\\ =1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ =1-0.72\\ =0.28\end{array}$

Q.28 A die is tossed thrice. Find the probability of getting an odd number at least once.

Ans.

$\begin{array}{l}\mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{in}\mathrm{a}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\frac{3}{6}=\frac{1}{2}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{in}\mathrm{a}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\frac{3}{6}=\frac{1}{2}\\ \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{three}\mathrm{times}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ =\frac{1}{8}\\ \mathrm{Therefore},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{at}\mathrm{least}\mathrm{once}\\ =1-\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{in}\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{throws}\\ =\mathrm{}1-\mathrm{Probability}\mathrm{}\mathrm{of}\mathrm{}\mathrm{getting}\mathrm{}\mathrm{an}\mathrm{}\mathrm{even}\mathrm{}\mathrm{number}\mathrm{}\mathrm{thrice} \mathrm{ }=\mathrm{1}-\frac{1}{8}\\ \mathrm{ }=\frac{7}{8}\end{array}$

Q.29 Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.

Ans.

$\begin{array}{l}\mathrm{Two}\mathrm{}\mathrm{balls}\mathrm{}\mathrm{are}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{at}\mathrm{}\mathrm{random}\mathrm{}\mathrm{with} \mathrm{replacement}\mathrm{}\mathrm{from}\mathrm{}\mathrm{a}\mathrm{}\mathrm{box}\mathrm{}\\ \mathrm{containing}\mathrm{}10\mathrm{}\mathrm{black} \mathrm{and}\mathrm{}8\mathrm{}\mathrm{red}\mathrm{}\mathrm{balls}.\\ \mathrm{ }\mathrm{Total}\mathrm{balls}=10+8=18\\ \mathrm{Black}\mathrm{balls}=10\\ \mathrm{ }\mathrm{Red}\mathrm{balls}=\mathrm{8}\\ \left(\mathrm{i}\right) \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{red}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{8}{18}=\frac{4}{9}\\ \mathrm{ }\mathrm{The}\mathrm{ball}\mathrm{is}\mathrm{replaced}\mathrm{after}\mathrm{the}\mathrm{first}\mathrm{draw}.\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{second}\mathrm{red}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{8}{18}=\frac{4}{9}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting} \mathrm{ }\mathrm{both}\mathrm{the}\mathrm{red}\mathrm{balls}=\frac{4}{9}×\frac{4}{9}\\ \mathrm{ }=\frac{16}{81}\\ \left(\mathrm{ii}\right) \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{black}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{10}{18}\\ \mathrm{ }\mathrm{The}\mathrm{ball}\mathrm{is}\mathrm{replaced}\mathrm{after}\mathrm{the}\mathrm{first}\mathrm{draw}.\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{second}\mathrm{red}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{8}{18}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{black}\mathrm{and}\mathrm{second}\mathrm{ball}\mathrm{as}\mathrm{red}\\ =\frac{10}{18}×\frac{8}{18}\\ =\frac{20}{81}\\ \left(\mathrm{iii}\right) \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{red}=\frac{8}{18}=\frac{4}{9}\\ \mathrm{The}\mathrm{ball}\mathrm{is}\mathrm{replaced}\mathrm{after}\mathrm{the}\mathrm{first}\mathrm{draw}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{second}\mathrm{ball}\mathrm{as}\mathrm{black}=\frac{10}{18}=\frac{5}{9}\\ \therefore \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{red}\mathrm{and}\mathrm{ }\mathrm{second}\\ \mathrm{ball}\mathrm{as}\mathrm{black}=\frac{4}{9}×\frac{5}{9}\\ =\frac{20}{81}\\ \mathrm{Therefore},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{black}\mathrm{and}\mathrm{second}\mathrm{ball}\mathrm{as}\mathrm{red}=\frac{4}{9}×\frac{5}{9}\\ =\frac{20}{81}\\ \mathrm{Therefore},\mathrm{probability}\mathrm{that}\mathrm{one}\mathrm{of}\mathrm{them}\mathrm{is}\mathrm{black}\mathrm{and}\mathrm{other}\mathrm{is}\mathrm{red}\\ =\mathrm{}\frac{20}{81}+\frac{20}{81}\\ =\frac{40}{81}\end{array}$

Q.30 Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.

Ans.

$\begin{array}{l} \mathrm{Probability}\mathrm{of}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{A},\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2}\\ \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{B},\mathrm{P}\left(\mathrm{B}\right)=\frac{1}{3}\\ \mathrm{Probability}\mathrm{of}\mathrm{not}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{A},\mathrm{P}\left(\mathrm{A}‘\right)=1-\frac{1}{2}\\ =\frac{1}{2}\\ \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{not}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{B},\mathrm{P}\left(\mathrm{B}‘\right)=1-\frac{1}{3}\\ =\frac{2}{3}\\ \left(\mathrm{i}\right)\mathrm{Probability}\mathrm{of}\mathrm{Problem}\mathrm{solved}\mathrm{by}\mathrm{both}\mathrm{independetly},\\ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{1}{2}×\frac{1}{3}\\ =\frac{1}{6}\\ \mathrm{Probability}\mathrm{that}\mathrm{Problem}\mathrm{is}\mathrm{solved},\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ =\frac{1}{2}+\frac{1}{3}-\frac{1}{2}×\frac{1}{3}\\ =\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\\ =\frac{3+2-1}{6}\\ =\frac{4}{6}=\frac{2}{3}\\ \left(\mathrm{ii}\right)\mathrm{Probability}\mathrm{that}\mathrm{exactly}\mathrm{one}\mathrm{of}\mathrm{them}\mathrm{solves}\mathrm{the}\mathrm{problem}\\ =\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}‘\right)+\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{A}‘\right)\\ =\frac{1}{2}×\frac{2}{3}+\frac{1}{3}×\frac{1}{2}\\ =\frac{1}{3}+\frac{1}{6}\\ =\frac{2+1}{6}\\ =\frac{3}{6}=\frac{1}{2}\end{array}$

Q.31 One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace’
(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’
(iii) E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’.

Ans.

$\begin{array}{l}\mathrm{Total}\mathrm{cards}\mathrm{in}\mathrm{a}\mathrm{well}\mathrm{shuffled}\mathrm{deck}=52\\ \left(\mathrm{i}\right)\mathrm{}\mathrm{E}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{a}\mathrm{}\mathrm{spade}’\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{4}{52}\\ =\frac{1}{13}\\ \mathrm{}\mathrm{F}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{an}\mathrm{}\mathrm{ace}’\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{13}{52}\mathrm{ }=\frac{1}{4}\\ \mathrm{ }\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{52}\\ \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{1}{13}×\frac{1}{4}\\ \mathrm{ }=\frac{1}{52}\\ \therefore \mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{events}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{independent}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{}\mathrm{E}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{black}’\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{13}{52}\\ =\frac{1}{4}\\ \mathrm{}\mathrm{F}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{a}\mathrm{ }\mathrm{king}’\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{4}{52}\\ \mathrm{ }=\frac{1}{13}\\ \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{1}{4}×\frac{1}{13}\\ \mathrm{ }=\frac{1}{52}\\ \therefore \mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{events}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{independent}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{}\mathrm{E}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{a}\mathrm{king}\mathrm{}\mathrm{or}\mathrm{}\mathrm{queen}’\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{8}{52}=\frac{2}{13}\\ \mathrm{}\mathrm{F}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{a}\mathrm{ }\mathrm{queen}\mathrm{}\mathrm{or}\mathrm{}\mathrm{jack}’\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{8}{52}\\ \mathrm{ }=\frac{2}{13}\\ \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{2}{13}×\frac{2}{13}\\ \mathrm{ }=\frac{4}{169}\\ \therefore \mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)\ne \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{events}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{.}\end{array}$

Q.32 In a hostel, 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi and English news papers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English news papers.
(b) If she reads Hindi news paper, find the probability that she reads English news paper.
(c) If she reads English news paper, find the probability that she reads Hindi news paper.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{H}=\mathrm{the}\mathrm{students}\mathrm{who}\mathrm{read}\mathrm{Hindi}\mathrm{newspaper}\mathrm{and}\\ \mathrm{E}=\mathrm{the}\mathrm{students}\mathrm{who}\mathrm{read}\mathrm{English}\mathrm{newspaper}.\\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that},\\ \mathrm{P}\left(\mathrm{H}\right)=60\mathrm{%}=\frac{60}{100}=0.6\\ \mathrm{P}\left(\mathrm{E}\right)=40\mathrm{%}=\frac{40}{100}=0.4\\ \mathrm{P}\left(\mathrm{H}\cap \mathrm{E}\right)=20\mathrm{%}=\frac{20}{100}=0.2\\ \left(\mathrm{i}\right)\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{she}\mathrm{reads}\mathrm{neither}\mathrm{Hindi}\mathrm{nor}\mathrm{English}\mathrm{news}\mathrm{papers}=\mathrm{P}\left(\mathrm{H}\cup \mathrm{E}\right)‘\\ =1-\mathrm{P}\left(\mathrm{H}\cup \mathrm{E}\right)\\ =1-\left\{\mathrm{P}\left(\mathrm{H}\right)+\mathrm{P}\left(\mathrm{E}\right)-\mathrm{P}\left(\mathrm{H}\cap \mathrm{E}\right)\right\}\\ =1-\left(0.6+0.4-0.2\right)\\ =1-0.8\\ =0.2\\ \left(\mathrm{ii}\right)\mathrm{If}\mathrm{she}\mathrm{reads}\mathrm{Hindi}\mathrm{news}\mathrm{paper},\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{she}\mathrm{reads}\mathrm{English}\mathrm{news}\mathrm{paper}\\ =\mathrm{P}\left(\mathrm{E}|\mathrm{H}\right)\\ =\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{H}\right)}{\mathrm{P}\left(\mathrm{H}\right)}\\ =\frac{0.2}{0.6}=\frac{1}{3}\\ \left(\mathrm{iii}\right)\mathrm{Probability}\mathrm{that}\mathrm{a}\mathrm{randomly}\mathrm{chosen}\mathrm{student}\mathrm{reads}\mathrm{Hindi}\mathrm{newspaper},\mathrm{if}\mathrm{she}\mathrm{reads}\mathrm{English}\mathrm{newspaper}\\ =\mathrm{P}\left(\mathrm{H}|\mathrm{E}\right)\\ =\frac{\mathrm{P}\left(\mathrm{H}\cap \mathrm{E}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ =\frac{0.2}{0.4}=\frac{1}{2}\end{array}$

Q.33 The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0 (B) 1/3 (C) 1/12 (D) 1/36

Ans.

When two dice are rolled, the number of outcomes is 36.The only even prime number is 2. Let E be the event of getting an even prime number on each die.
So, E = {(2, 2)}
P (E) = (1/36)
Thus, the correct option is D.

Q.34 Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P(A′B′) = [1 – P(A)][1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1

Ans.

$\begin{array}{l}\mathrm{And},\mathrm{ }\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)=\mathrm{m}.\mathrm{n}\ne 0\\ \therefore \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{So},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\\ \mathrm{Therefore},\mathrm{option}\mathrm{A}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{B}\right)\mathrm{Two}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{said}\mathrm{to}\mathrm{be}\mathrm{independent},\\ \mathrm{if}\mathrm{P}\left(\mathrm{AB}\right)=\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Consider}\mathrm{the}\mathrm{result}\mathrm{given}\mathrm{in}\mathrm{alternative}\mathrm{B}.\\ \mathrm{P}\left(\mathrm{A}‘\mathrm{B}‘\right)=\left[1-\mathrm{P}\left(\mathrm{A}\right)\right]\left[1-\mathrm{P}\left(\mathrm{B}\right)\right]\\ ⇒ \mathrm{ }\mathrm{P}\left(\mathrm{A}‘\cap \mathrm{B}‘\right)=1-\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ ⇒ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)‘=1-\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \left[\mathrm{By}\mathrm{De}\mathrm{}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\right]\\ ⇒\mathrm{ }1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=1-\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ ⇒ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ ⇒ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{This}\mathrm{implies}\mathrm{that}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent},\mathrm{if}\\ \mathrm{P}\left(\mathrm{A}‘\mathrm{B}‘\right)=\left[1-\mathrm{P}\left(\mathrm{A}\right)\right]\left[1-\mathrm{P}\left(\mathrm{B}\right)\right]\\ \mathrm{Therefore},\mathrm{option}\mathrm{B}\mathrm{is}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{C}\right)\mathrm{Let}\\ \mathrm{A}:\mathrm{Event}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{on}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\left\{1, 3, 5\right\}\\ \mathrm{B}:\mathrm{Event}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{on}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\left\{2, 4, 6\right\}\\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{3}{6}=\frac{1}{2}\mathrm{}\mathrm{and}\mathrm{P}\left(\mathrm{B}\right)=\frac{3}{6}=\frac{1}{2}\mathrm{}\\ \mathrm{So}, \mathrm{A}\cap \mathrm{B}=\mathrm{\varphi }⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=0\\ \mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\ne 0\\ ⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{ }\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{So},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\\ \mathrm{Therefore},\mathrm{option}\mathrm{C}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{D}\right)\mathrm{If}\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)=1,\\ \mathrm{then},\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{So},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\\ \mathrm{Therefore},\mathrm{option}\mathrm{D}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \mathrm{Thus},\mathrm{option}\mathbf{B}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.35 An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Ans.

$\begin{array}{l} \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{balls}\mathrm{in}\mathrm{urn}=5\\ \mathrm{Number}\mathrm{of}\mathrm{black}\mathrm{balls}\mathrm{in}\mathrm{urn}=5\\ \mathrm{Let}\mathrm{a}\mathrm{red}\mathrm{ball}\mathrm{be}\mathrm{drawn}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{attempt}\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{red}\mathrm{}\mathrm{ball}\right)=\frac{5}{10}=\frac{1}{2}\\ \mathrm{If}\mathrm{two}\mathrm{red}\mathrm{balls}\mathrm{are}\mathrm{added}\mathrm{to}\mathrm{the}\mathrm{urn},\mathrm{then}\\ \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{balls}\mathrm{in}\mathrm{urn}=7\mathrm{}\\ \mathrm{ }\mathrm{Total}\mathrm{balls}\mathrm{in}\mathrm{urn}=\mathrm{12}\\ \mathrm{ }\mathrm{P}\left(\mathrm{drawing}\mathrm{a}\mathrm{red}\mathrm{ball}\right)=\frac{7}{12}\\ \mathrm{Let}\mathrm{a}\mathrm{black}\mathrm{ball}\mathrm{be}\mathrm{drawn}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{attempt}\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{black}\mathrm{ball}\right)=\frac{5}{10}=\frac{1}{2}\\ \mathrm{Probability}\mathrm{of}\mathrm{red}\mathrm{ball}\mathrm{in}\mathrm{second}\mathrm{attempt},\\ \mathrm{ }\mathrm{P}\left(\mathrm{red}\mathrm{ball}\right)=\frac{7}{12}\\ \mathrm{Therefore},\mathrm{probability}\mathrm{of}\mathrm{drawing}\mathrm{second}\mathrm{ball}\mathrm{as}\mathrm{red}\\ =\frac{1}{2}×\frac{7}{12}+\frac{1}{2}×\frac{5}{12}\\ =\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)\\ =\frac{1}{2}×1=\frac{1}{2}\end{array}$

Q.36 A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Ans.

$\begin{array}{l}\mathrm{ } \mathrm{ } \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{balls}\mathrm{in}\mathrm{first}\mathrm{bag}=4\\ \mathrm{Number}\mathrm{of}\mathrm{black}\mathrm{balls}\mathrm{in}\mathrm{first}\mathrm{bag}=4\\ \mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{balls}\mathrm{in}\mathrm{second}\mathrm{bag}=2\\ \mathrm{Number}\mathrm{of}\mathrm{black}\mathrm{balls}\mathrm{in}\mathrm{second}\mathrm{bag}=6\\ \mathrm{Let}{\mathrm{E}}_{\mathrm{1}}\mathrm{and}{\mathrm{E}}_{\mathrm{2}}\mathrm{be}\mathrm{the}\mathrm{events}\mathrm{of}\mathrm{selecting}\mathrm{first}\mathrm{bag}\mathrm{and}\mathrm{second}\\ \mathrm{bag}\mathrm{respectively}\mathrm{.}\\ \therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{1}{2};\mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{1}{2}\\ \mathrm{Let}\mathrm{A}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{red}\mathrm{ball}.\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=\mathrm{P}\left(\mathrm{red}\mathrm{ball}\mathrm{from}\mathrm{first}\mathrm{bag}\right)=\frac{4}{8}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=\mathrm{P}\left(\mathrm{red}\mathrm{ball}\mathrm{from}\mathrm{second}\mathrm{bag}\right)=\frac{2}{8}=\frac{1}{4}\\ \mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{ball}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{bag},\mathrm{given}\mathrm{that}\mathrm{it}\\ \mathrm{is}\mathrm{red} =\mathrm{P}\left({\mathrm{E}}_{\mathrm{1}}|\mathrm{A}\right)\\ \mathrm{ }=\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right).\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right).\mathrm{ }\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right).\mathrm{ }\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)}\left[\mathrm{By}\mathrm{using}\mathrm{Bayes}‘\mathrm{theorem}\right]\\ =\frac{\frac{1}{2}.\frac{1}{2}}{\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{4}}\\ \mathrm{ }=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}\\ =\frac{\frac{1}{4}}{\frac{3}{8}}\\ =\frac{1}{4}×\frac{8}{3}=\frac{2}{3}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{ball}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{bag},\mathrm{given}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{red}\mathrm{is}\frac{2}{3}.\end{array}$

Q.37 Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Ans.

$\begin{array}{l}\mathrm{Let}{\mathrm{E}}_{\mathrm{1}}=\mathrm{The}\mathrm{student}\mathrm{is}\mathrm{a}\mathrm{hostler}\\ \mathrm{ }{\mathrm{E}}_{\mathrm{2}}=\mathrm{The}\mathrm{student}\mathrm{is}\mathrm{a}\mathrm{day}\mathrm{scholar}\\ \mathrm{and} \mathrm{A}=\mathrm{The}\mathrm{chosen}\mathrm{student}\mathrm{gets}\mathrm{grade}\mathrm{A}.\\ \therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=60\mathrm{%}=\frac{60}{100}=0.6\\ \mathrm{P}\left({\mathrm{E}}_{2}\right)=40\mathrm{%}=\frac{40}{100}=0.4\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=\mathrm{P}\left(\mathrm{student}\mathrm{getting}\mathrm{an}\mathrm{A}\mathrm{grade}\mathrm{is}\mathrm{a}\mathrm{hostler}\right)\\ =30\mathrm{%}=0.3\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=\mathrm{P}\left(\mathrm{student}\mathrm{getting}\mathrm{an}\mathrm{A}\mathrm{grade}\mathrm{is}\mathrm{a}\mathrm{day}\mathrm{scholar}\right)\\ =20\mathrm{%}=0.2\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{a}\mathrm{randomly}\mathrm{chosen}\mathrm{student}\mathrm{is}\mathrm{a}\mathrm{hostler},\\ \mathrm{given}\mathrm{that}\mathrm{he}\mathrm{has}\mathrm{an}\mathrm{A}\mathrm{grade}\\ =\mathrm{P}\left({\mathrm{E}}_{1}|\mathrm{A}\right)\\ =\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)}\left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ =\frac{0.6×0.3}{0.6×0.3+0.4×0.2}\\ =\frac{0.18}{0.18+0.08}\\ =\frac{0.18}{0.26}\\ =\frac{18}{26}\\ =\frac{9}{13}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{9}{13}.\end{array}$

Q.38 In answering a question on a multiple choice test, a student either knows the answer or guesses. Let (3/4) be the probability that he knows the answer and (1/4) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability (1/4). What is the probability that the student knows the answer given that he answered it correctly?

Ans.

$\begin{array}{l}\mathrm{Let} {\mathrm{E}}_{\mathrm{1}}=\mathrm{The}\mathrm{student}\mathrm{knows}\mathrm{the}\mathrm{answer}\\ \mathrm{and}{\mathrm{E}}_{\mathrm{2}}=\mathrm{The}\mathrm{student}\mathrm{guesses}\mathrm{the}\mathrm{answer}\\ \mathrm{Let} \mathrm{ }\mathrm{A}=\mathrm{The}\mathrm{answer}\mathrm{is}\mathrm{correct}\\ \therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{3}{4}\\ \mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{1}{4}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{student}\mathrm{answered}\mathrm{correctly},\mathrm{given}\mathrm{that}\\ \mathrm{he}\mathrm{knows}\mathrm{the}\mathrm{answer},\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=\mathrm{1}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=\mathrm{P}\left(\mathrm{student}\mathrm{answered}\mathrm{correctly},\mathrm{given}\mathrm{that}\mathrm{he}\mathrm{guessed}\right)\\ =\frac{1}{4}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{student}\mathrm{knows}\mathrm{the}\mathrm{answer},\mathrm{given}\mathrm{that}\mathrm{he}\\ \mathrm{answered}\mathrm{it}\mathrm{correctly}\\ =\mathrm{P}\left({\mathrm{E}}_{1}|\mathrm{A}\right)\\ =\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)}\left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ =\frac{\frac{3}{4}.\mathrm{ }1}{\frac{3}{4}.\mathrm{ }1+\frac{1}{4}.\mathrm{ }\frac{1}{4}}\\ =\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}\\ =\frac{\frac{3}{4}}{\frac{12+1}{16}}=\frac{12}{13}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{12}{13}.\end{array}$

Q.39 A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Ans.

$\begin{array}{l}Let{\text{E}}_{\text{1}}=A\text{person has disease}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{E}}_{\text{2}}=A\text{person has no disease}\text{.}\\ \text{Since E1 and E2 are events complimentary to each other,}\\ \therefore {\text{P (E}}_{\text{1}}{\text{) + P (E}}_{\text{2}}\text{)}=\text{1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{P (E}}_{\text{2}}\text{)}=\text{1}-{\text{P (E}}_{\text{1}}\text{)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1-0.1%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1-\frac{0.1}{100}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1-0.001\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.999\\ \text{Let A be the event that the blood test result is positive}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{P (E}}_{\text{1}}\text{)}=0.1%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.001\\ P\left(A|{E}_{1}\right)=P\left(result\text{is positive given the person has disease}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=99%=0.99\\ P\left(A|{E}_{2}\right)=P\left(result\text{is positive given the person has no disease}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.5%=0.005\\ P\left({E}_{1}|A\right)=\text{Probability that a person has a disease, given that his}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{test result is positive}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{P\left({E}_{1}\right)P\left(A|{E}_{1}\right)}{P\left({E}_{1}\right)P\left(A|{E}_{1}\right)+P\left({E}_{2}\right)×P\left(A|{E}_{2}\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{Bayes’ Theorem}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{0.00099}{0.005985}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{990}{5985}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{198}{1197}\\ \text{Therefore, the required probability is}\frac{198}{1197}.\end{array}$

Q.40 There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Ans.

$\begin{array}{l}\mathrm{Let} \mathrm{ }{\mathrm{E}}_{\mathrm{1}}=\mathrm{choosing}\mathrm{a}\mathrm{two}\mathrm{headed}\mathrm{coin}\\ {\mathrm{E}}_{\mathrm{2}}=\mathrm{choosing}\mathrm{a}\mathrm{biased}\mathrm{coin}\\ {\mathrm{E}}_{\mathrm{3}}=\mathrm{choosing}\mathrm{an}\mathrm{unbiased}\mathrm{coin}\\ \therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{1}{3}, \mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{1}{3}\mathrm{and}\mathrm{P}\left({\mathrm{E}}_{3}\right)=\frac{1}{3}\\ \mathrm{Let} \mathrm{ }\mathrm{A}=\mathrm{the}\mathrm{coin}\mathrm{shows}\mathrm{heads}\\ \mathrm{A}\mathrm{two}–\mathrm{headed}\mathrm{coin}\mathrm{will}\mathrm{always}\mathrm{show}\mathrm{heads}.\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=1\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=75\mathrm{%}\\ \mathrm{ }=\frac{75}{100}\mathrm{ }=\frac{3}{4}\\ \because \mathrm{The}\mathrm{third}\mathrm{coin}\mathrm{is}\mathrm{unbiased},\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{it}\mathrm{shows}\mathrm{heads}\\ \mathrm{ }=\frac{1}{2}\\ \therefore \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{3}\right)=\frac{1}{2}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{coin}\mathrm{is}\mathrm{two}–\mathrm{headed},\mathrm{given}\mathrm{that}\mathrm{it}\mathrm{shows}\\ \mathrm{heads}, \mathrm{ }=\mathrm{P}\left({\mathrm{E}}_{1}|\mathrm{A}\right)\\ \mathrm{ }=\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)×\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)+\mathrm{P}\left({\mathrm{E}}_{3}\right)×\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{3}\right)}\\ \left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ \mathrm{ }=\frac{\frac{1}{3}.\mathrm{ }1}{\frac{1}{3}.\mathrm{ }1+\frac{1}{3}.\mathrm{ }\frac{3}{4}+\frac{1}{3}.\mathrm{ }\frac{1}{2}}\\ \mathrm{ }=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{4}+\frac{1}{6}}\\ \mathrm{ }=\frac{\frac{1}{3}}{\frac{4+3+2}{12}}\\ \mathrm{ }=\frac{\frac{1}{3}}{\frac{9}{12}}=\frac{4}{9}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{4}{9}.\end{array}$

Q.41 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Ans.

$\begin{array}{l}\mathrm{Let} \mathrm{ }{\mathrm{E}}_{\mathrm{1}}=\mathrm{The}\mathrm{driver}\mathrm{is}\mathrm{a}\mathrm{scooter}\mathrm{driver}\\ {\mathrm{E}}_{\mathrm{2}}=\mathrm{The}\mathrm{driver}\mathrm{is}\mathrm{a} \mathrm{car}\mathrm{driver}\\ {\mathrm{E}}_{\mathrm{3}}=\mathrm{The}\mathrm{driver}\mathrm{is}\mathrm{a}\mathrm{truck}\mathrm{driver}\\ \mathrm{A}=\mathrm{The}\mathrm{person}\mathrm{meets}\mathrm{with}\mathrm{an}\mathrm{accident}.\\ \mathrm{There}\mathrm{are}2000\mathrm{scooter}\mathrm{drivers}, 4000\mathrm{car}\mathrm{drivers},\mathrm{and}6000\\ \mathrm{truck}\mathrm{drivers}.\\ \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{drivers}= 2000 + 4000 + 6000 = 12000\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{2000}{12000}=\frac{1}{6}\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{4000}{12000}=\frac{1}{3}\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{3}\right)=\frac{6000}{12000}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=\mathrm{P}\left(\mathrm{scooter}\mathrm{driver}\mathrm{met}\mathrm{with}\mathrm{an}\mathrm{accident}\right)\\ =0.01=\frac{1}{100}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=\mathrm{P}\left(\mathrm{car}\mathrm{driver}\mathrm{met}\mathrm{with}\mathrm{an}\mathrm{accident}\right)\\ =0.03=\frac{3}{100}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{3}\right)=\mathrm{P}\left(\mathrm{truck}\mathrm{driver}\mathrm{met}\mathrm{with}\mathrm{an}\mathrm{accident}\right)\\ =0.15=\frac{15}{100}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{driver}\mathrm{is}\mathrm{a}\mathrm{scooter}\mathrm{driver},\mathrm{given}\mathrm{that}\mathrm{he}\\ \mathrm{met}\mathrm{with}\mathrm{an}\mathrm{accident},\\ =\mathrm{P}\left({\mathrm{E}}_{\mathrm{1}}|\mathrm{A}\right)\\ =\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)×\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)+\mathrm{P}\left({\mathrm{E}}_{3}\right)×\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{3}\right)}\\ \mathrm{ }\left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ =\frac{\frac{1}{6}×\frac{1}{100}}{\frac{1}{6}×\frac{1}{100}+\frac{1}{3}×\frac{3}{100}+\frac{1}{2}×\frac{15}{100}}\\ =\frac{\frac{1}{6}}{\frac{1}{6}+\frac{3}{3}+\frac{15}{2}}\\ =\frac{\frac{1}{6}}{\frac{1+6+45}{6}}\\ =\frac{1}{52}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{1}{52}.\end{array}$

Q.42 A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Ans.

$\begin{array}{l}\mathrm{Let} \mathrm{ }{\mathrm{E}}_{\mathrm{1}}=\mathrm{Items}\mathrm{produced}\mathrm{by}\mathrm{machines}\mathrm{A}\\ {\mathrm{E}}_{\mathrm{2}}=\mathrm{Items}\mathrm{produced}\mathrm{by}\mathrm{machines}\mathrm{B}.\\ \mathrm{ }\mathrm{A}=\mathrm{The}\mathrm{Item}\mathrm{was}\mathrm{found}\mathrm{to}\mathrm{be}\mathrm{defective}\\ \therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=60\mathrm{%}=0.6\\ \mathrm{P}\left({\mathrm{E}}_{2}\right)=40\mathrm{%}=0.4\\ \mathrm{Probability}\mathrm{that}\mathrm{machine}\mathrm{A}\mathrm{produced}\mathrm{defective}\mathrm{items},\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=2\mathrm{%}=0.02\\ \mathrm{Probability}\mathrm{that}\mathrm{machine}\mathrm{B}\mathrm{produced}\mathrm{defective}\mathrm{items},\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=1\mathrm{%}=0.01\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{randomly}\mathrm{selected}\mathrm{item}\mathrm{was}\mathrm{from}\\ \mathrm{machine}\mathrm{B},\mathrm{given}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{defective},\\ \mathrm{P}\left({\mathrm{E}}_{2}|\mathrm{A}\right)=\frac{\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)×\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)} \left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ =\frac{0.4×0.01}{0.6×0.02+0.4×0.01}\\ =\frac{0.004}{0.012+0.004}\\ =\frac{0.004}{0.016}\\ =\frac{4}{16}=\frac{1}{4}\\ \mathrm{Therefore},\mathrm{ }\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{randomly}\mathrm{selected}\mathrm{item}\mathrm{was}\\ \mathrm{from}\mathrm{machine}\mathrm{B},\mathrm{given}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{defective},\mathrm{is}\frac{1}{4}.\end{array}$

Q.43 Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Ans.

$\begin{array}{l}\mathrm{Let} {\mathrm{E}}_{\mathrm{1}}=\mathrm{The}\mathrm{first}\mathrm{group}\mathrm{win}\mathrm{the}\mathrm{competition}\\ \mathrm{ }{\mathrm{E}}_{2}=\mathrm{The}\mathrm{second}\mathrm{group}\mathrm{win}\mathrm{the}\mathrm{competition}\\ \mathrm{ }\mathrm{A}=\mathrm{The}\mathrm{event}\mathrm{of}\mathrm{introducing}\mathrm{a}\mathrm{new}\mathrm{product}\\ \mathrm{P}\left({\mathrm{E}}_{1}\right)=\mathrm{Probability}\mathrm{that}\mathrm{the}\mathrm{first}\mathrm{group}\mathrm{wins}\mathrm{the}\mathrm{competition}\\ \end{array}$

=0.6
P( E 2 )=Probability that the second group wins the competition
=0.4 P (A| E 1 )=Probability of introducing a new product if the first group wins
=
0.7
P (A| E 2 )=Probability of introducing a new product if the second group wins=0.3 The probability that the new product is introduced by the second group is given by
=
P(E 2 |A)
P( E 2 |A )= P( E 2 )P( A| E 2 ) P( E 1 )P( A| E 1 )+P( E 2 )×P( A| E 2 ) [ By Bayes’ Theorem ]
= 0.4×0.3 0.6×0.7+0.4×0.3

=
0.12 0.42+0.12

= 0.12 0.54 = 2 9
Therefore, the required probability is 2 9 .

Q.44 Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Ans.

$\begin{array}{l} \mathrm{Let}{\mathrm{E}}_{\mathrm{1}}=\mathrm{The}\mathrm{outcome}\mathrm{on}\mathrm{the}\mathrm{die}\mathrm{is}5\mathrm{or}6\\ \mathrm{and}{\mathrm{E}}_{\mathrm{2}}=\mathrm{The}\mathrm{outcome}\mathrm{on}\mathrm{the}\mathrm{die}\mathrm{is}1, 2, 3,\mathrm{or}4.\\ \therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{2}{6}=\frac{1}{3}\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{4}{6}=\frac{2}{3}\\ \mathrm{Let} \mathrm{A}=\mathrm{The}\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{exactly}\mathrm{one}\mathrm{head}\mathrm{.}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{\mathrm{1}}\mathrm{\right)}=\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{exactly}\mathrm{one}\mathrm{head}\mathrm{by}\mathrm{tossing}\mathrm{the}\\ \mathrm{coin}\mathrm{three}\mathrm{times}\mathrm{if}\mathrm{she}\mathrm{gets}5\mathrm{or}6\\ \mathrm{ }=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{\mathrm{2}}\mathrm{\right)}=\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{exactly}\mathrm{one}\mathrm{head}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\\ \mathrm{of}\mathrm{coin}\mathrm{if}\mathrm{she}\mathrm{gets}1,2, 3,\mathrm{or}4\\ \mathrm{ }=\frac{1}{2}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{girl}\mathrm{threw}1, 2, 3,\mathrm{or}4\mathrm{with}\mathrm{the}\mathrm{die},\mathrm{if}\mathrm{she}\\ \mathrm{obtained}\mathrm{exactly}\mathrm{one}\mathrm{head},\\ \mathrm{ }=\mathrm{P}\left({\mathrm{E}}_{\mathrm{2}}|\mathrm{A}\right)\\ \mathrm{ }=\frac{\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)} \left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ \mathrm{ }=\frac{\frac{2}{3}×\frac{1}{2}}{\frac{3}{8}×\frac{1}{3}+\frac{2}{3}×\frac{1}{2}}\\ \mathrm{ }=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}\\ \mathrm{ }=\frac{\frac{1}{3}}{\frac{3+8}{24}}\\ \mathrm{ }=\frac{\frac{1}{3}}{\frac{11}{24}}\\ \mathrm{ }=\frac{1}{3}×\frac{24}{11}=\frac{8}{11}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{8}{11}.\end{array}$

Q.45 A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Ans.

$\begin{array}{l} \mathrm{Let}{\mathrm{E}}_{\mathrm{1}}=\mathrm{the}\mathrm{time}\mathrm{consumed}\mathrm{by}\mathrm{machines}\mathrm{A}\mathrm{for}\mathrm{the}\mathrm{job}\\ \mathrm{ }{\mathrm{E}}_{\mathrm{2}}=\mathrm{the}\mathrm{time}\mathrm{consumed}\mathrm{by}\mathrm{machines}\mathrm{B}\mathrm{for}\mathrm{the}\mathrm{job}\\ \mathrm{and}{\mathrm{E}}_{\mathrm{3}}=\mathrm{the}\mathrm{time}\mathrm{consumed}\mathrm{by}\mathrm{machines}\mathrm{C}\mathrm{for}\mathrm{the}\mathrm{job}\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{1}\right)=50\mathrm{%}=0.5\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{2}\right)=30\mathrm{%}=0.3\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{3}\right)=20\mathrm{%}=0.2\\ \mathrm{A}=\mathrm{The}\mathrm{event}\mathrm{of}\mathrm{producing}\mathrm{defective}\mathrm{items}\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=1\mathrm{%}=0.01\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=5\mathrm{%}=0.05\\ \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{3}\right)=7\mathrm{%}=0.07\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{defective}\mathrm{item}\mathrm{was}\mathrm{produced}\mathrm{by}\mathrm{A},\\ =\mathrm{P}\left({\mathrm{E}}_{\mathrm{1}}|\mathrm{A}\right)\\ =\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)+\mathrm{P}\left({\mathrm{E}}_{3}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{3}\right)}\mathrm{ }\\ \mathrm{ }\left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ =\frac{0.5×0.01}{0.5×0.01+0.3×0.05+0.2×0.07}\\ =\frac{0.005}{0.005+0.015+0.014}\\ =\frac{0.005}{0.034}\\ =\frac{5}{34}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{5}{34}.\end{array}$

Q.46 A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Ans.

$\begin{array}{l}\mathrm{Let}{\mathrm{E}}_{\mathrm{1}}=\mathrm{events}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{diamond}\mathrm{card}\\ {\mathrm{E}}_{\mathrm{2}}=\mathrm{events}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{card}\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{diamond}\\ \mathrm{and}\mathrm{A}=\mathrm{the}\mathrm{lost}\mathrm{card}\\ \mathrm{Out}\mathrm{of}52\mathrm{cards}, 13\mathrm{cards}\mathrm{are}\mathrm{diamond}\mathrm{and}39\mathrm{cards}\mathrm{are}\mathrm{not}\\ \mathrm{diamond}\mathrm{.}\\ \therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{13}{52}=\frac{1}{4}\\ \mathrm{ }\mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{39}{52}=\frac{3}{4}\\ \mathrm{When}\mathrm{one}\mathrm{diamond}\mathrm{card}\mathrm{is}\mathrm{lost},\mathrm{there}\mathrm{are}12\mathrm{diamond}\mathrm{cards}\mathrm{out}\\ \mathrm{of}51\mathrm{cards}.\mathrm{Two}\mathrm{cards}\mathrm{can}\mathrm{be}\mathrm{drawn}\mathrm{out}\mathrm{of}12\mathrm{diamond}\mathrm{cards}\mathrm{in}\\ {}^{\mathrm{12}}{\mathrm{C}}_{\mathrm{2}}\mathrm{ways}\mathrm{.}\\ \mathrm{Similarly}, 2\mathrm{diamond}\mathrm{cards}\mathrm{can}\mathrm{be}\mathrm{drawn}\mathrm{out}\mathrm{of}51\mathrm{cards}\mathrm{in}\\ {}^{\mathrm{51}}{\mathrm{C}}_{\mathrm{2}}\mathrm{ways}.\\ \mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{two}\mathrm{cards},\mathrm{when}\mathrm{one}\mathrm{diamond}\mathrm{card}\mathrm{is}\\ \mathrm{ }\mathrm{lost}=\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{\mathrm{1}}\mathrm{\right)}\\ =\frac{{}^{12}{\mathrm{C}}_{2}}{{}^{\mathrm{51}}{\mathrm{C}}_{\mathrm{2}}}=\frac{12!}{2!10!}×\frac{2!49!}{51!}\\ =\frac{12×11}{1}×\frac{1}{51×50}\\ =\frac{12×11}{51×50}\\ =\frac{66}{1275}\\ \mathrm{When}\mathrm{the}\mathrm{lost}\mathrm{card}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{diamond},\mathrm{there}\mathrm{are}13\mathrm{diamond}\mathrm{cards}\\ \mathrm{out}\mathrm{of}51\mathrm{cards}.\\ \mathrm{Two}\mathrm{cards}\mathrm{can}\mathrm{be}\mathrm{drawn}\mathrm{out}\mathrm{of}13\mathrm{diamond}\mathrm{cards}\mathrm{in}{}^{\mathrm{13}}{\mathrm{C}}_{\mathrm{2}}\mathrm{ways}\\ \mathrm{whereas}2\mathrm{cards}\mathrm{can}\mathrm{be}\mathrm{drawn}\mathrm{out}\mathrm{of}51\mathrm{cards}\mathrm{in}{\mathrm{ }}^{51}{\mathrm{C}}_{2}\mathrm{ways}\mathrm{.}\\ \mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{two}\mathrm{cards},\mathrm{when}\mathrm{one}\mathrm{card}\mathrm{is}\mathrm{lost}\mathrm{which}\\ \mathrm{is}\mathrm{not}\mathrm{diamond}\\ =\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{\mathrm{2}}\right)\\ =\frac{{}^{13}{\mathrm{C}}_{2}}{{}^{51}{\mathrm{C}}_{2}}\\ =\frac{13!}{2!\mathrm{ }×11!}×\frac{2!\mathrm{ }×49!}{51!}\\ =\frac{13×12}{1}×\frac{1}{51×50}\\ =\frac{26}{425}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{lost}\mathrm{card}\mathrm{is}\mathrm{diamond}\\ =\mathrm{P}\left({\mathrm{E}}_{\mathrm{1}}|\mathrm{A}\right)\\ =\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)} \left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ =\frac{\frac{1}{4}×\frac{22}{425}}{\frac{1}{4}×\frac{22}{425}+\frac{3}{4}×\frac{26}{425}}\\ =\frac{11}{50}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{11}{50}.\end{array}$

Q.47 Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is
(A) 4/5 (B) 1/2 (C) 1/5 (D) 2/5

Ans.

$\begin{array}{l}\mathrm{Let}{\mathrm{E}}_{\mathrm{1}}\mathrm{and}{\mathrm{E}}_{\mathrm{2}}\mathrm{be}\mathrm{the}\mathrm{events}\mathrm{such}\mathrm{that}\\ {\mathrm{E}}_{\mathrm{1}}:\mathrm{A}\mathrm{speaks}\mathrm{truth}\\ {\mathrm{E}}_{\mathrm{2}}:\mathrm{A}\mathrm{speaks}\mathrm{false}\\ \mathrm{Let}\mathrm{A}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{a}\mathrm{head}\mathrm{appears}.\\ \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{4}{5}\\ \mathrm{P}\left({\mathrm{E}}_{2}\right)=1-\frac{4}{5}\left[\mathrm{P}\left({\mathrm{E}}_{2}\right)=1-\mathrm{P}\left({\mathrm{E}}_{1}\right)\right]\\ =\frac{1}{5}\\ \mathrm{If}\mathrm{a}\mathrm{coin}\mathrm{is}\mathrm{tossed},\mathrm{then}\mathrm{it}\mathrm{may}\mathrm{result}\mathrm{in}\mathrm{either}\mathrm{head}\left(\mathrm{H}\right)\mathrm{or}\mathrm{tail}\left(\mathrm{T}\right)\mathrm{.}\\ \mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{head}\mathrm{is}\frac{1}{2}\mathrm{whether}\mathrm{A}\mathrm{speaks}\mathrm{truth}\\ \mathrm{or}\mathrm{not}\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)=\frac{1}{2}\mathrm{and}\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)=\frac{1}{2}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{there}\mathrm{is}\mathrm{actually}\mathrm{a}\mathrm{head}=\mathrm{P}\left({\mathrm{E}}_{\mathrm{1}}|\mathrm{A}\right)\\ =\frac{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)}{\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{1}\right)+\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left(\mathrm{A}|{\mathrm{E}}_{2}\right)} \left[\mathrm{By}\mathrm{Bayes}‘\mathrm{Theorem}\right]\\ =\frac{\frac{4}{5}×\frac{1}{2}}{\frac{4}{5}×\frac{1}{2}+\frac{1}{5}×\frac{1}{2}}=\frac{4}{5}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{4}{5}.\\ \mathrm{Thus},\mathrm{correct}\mathrm{option}\mathrm{is}\mathbf{A}\mathrm{.}\end{array}$

Q.48

$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{A}\mathrm{}\mathrm{and}\mathrm{}\mathrm{B}\mathrm{}\mathrm{are}\mathrm{}\mathrm{two}\mathrm{}\mathrm{events}\mathrm{}\mathrm{such}\mathrm{}\mathrm{that}\mathrm{}\mathrm{A}\subset \mathrm{B}\mathrm{}\mathrm{and}\mathrm{}\mathrm{P}\left(\mathrm{B}\right)\ne 0,\mathrm{then}\\ \mathrm{which}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{following}\mathrm{}\mathrm{is}\mathrm{}\mathrm{correct}?\\ \left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\mathrm{}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)<\mathrm{P}\left(\mathrm{A}\right)\\ \left(\mathrm{C}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\ge \mathrm{P}\left(\mathrm{A}\right)\mathrm{}\left(\mathrm{D}\right)\mathrm{None}\mathrm{}\mathrm{of}\mathrm{}\mathrm{these}\end{array}$

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{A}\subset \mathrm{B},\mathrm{then}\mathrm{A}\cap \mathrm{B}=\mathrm{A}\\ ⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)...\left(\mathrm{i}\right)\\ \left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \mathrm{ }=\frac{\mathrm{P}\left(\mathrm{A}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\left[\mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{ }\ne \frac{\mathrm{P}\left(\mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\\ \mathrm{Therefore},\mathrm{option}\mathrm{A}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ =\frac{\mathrm{P}\left(\mathrm{A}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\left[\mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\right]\\ ⇒\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\overline{)<}\mathrm{P}\left(\mathrm{A}\right)\left[\because \mathrm{P}\left(\mathrm{A}\right)<\mathrm{P}\left(\mathrm{B}\right)\right]\\ \mathrm{Therefore},\mathrm{option}\mathrm{B}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{C}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ ⇒\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\left[\mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\right]\\ ⇒\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)\ge \mathrm{P}\left(\mathrm{A}\right)\left[\because \mathrm{P}\left(\mathrm{B}\right)\le 1,\mathrm{ }\mathrm{so}\mathrm{ }\frac{1}{\mathrm{P}\left(\mathrm{B}\right)}\ge 1\right]\\ \mathrm{Therefore},\mathrm{option}\mathbf{C}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.49 State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

(i)

 X 0 1 2 P(X) 0.4 0.4 0.2

(ii)

 X 0 1 2 3 4 P(X) 0.1 0.5 0.2 – 0.1 0.3

(iii)

 Y –1 0 1 P(Y) 0.6 0.1 0.2

(iv)

 Z 3 2 1 0 –1 P(Z) 0.3 0.2 0.4 0.1 0.05

Ans.

It is known that the sum of all the probabilities in a probability distribution is one.
(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1
Therefore, the given table is a probability distribution of random variables.
(ii) It can be seen that for X = 3, P (X) = −0.1
It is known that probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.
(iii) Sum of the probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Therefore, the given table is not a probability distribution of random variables.
(iv) Sum of the probabilities
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05
= 1.05 ≠ 1
Therefore, the given table is not a probability distribution of random variables.

Q.50 An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

Ans.

The two balls selected can be represented as BB, BR, RB, RR where B represents a black ball and R represents a red ball. X represents the number of black balls.
So, X (BB) = 2
X(BR) = 1
X(RB) = 1
X(RR) = 0
Therefore, the possible values of X are 0, 1, and 2.
Yes, X is a random variable.

Q.51 Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

Ans.

$\begin{array}{l}\mathrm{A}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{six}\mathrm{times}\mathrm{and}\mathrm{X}\mathrm{represents}\mathrm{the}\mathrm{difference}\\ \mathrm{between}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{heads}\mathrm{and}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{tails}.\\ \therefore \mathrm{X}\left[6\mathrm{H},0\mathrm{T}\right]=\mathrm{X}\left[6\mathrm{times}\mathrm{H}\mathrm{and}0\mathrm{times}\mathrm{T}\right]=|6-0|=1\\ \mathrm{X}\left[5\mathrm{H},1\mathrm{T}\right]=\mathrm{X}\left[5\mathrm{times}\mathrm{H}\mathrm{and}1\mathrm{times}\mathrm{T}\right]=|5-1|=4\\ \mathrm{X}\left[4\mathrm{H},2\mathrm{T}\right]=\mathrm{X}\left[4\mathrm{times}\mathrm{H}\mathrm{and}2\mathrm{times}\mathrm{T}\right]=|4-2|=2\\ \mathrm{X}\left[3\mathrm{H},3\mathrm{T}\right]=\mathrm{X}\left[3\mathrm{times}\mathrm{H}\mathrm{and}3\mathrm{times}\mathrm{T}\right]=|3-3|=0\\ \mathrm{X}\left[2\mathrm{H},4\mathrm{T}\right]=\mathrm{X}\left[2\mathrm{times}\mathrm{H}\mathrm{and}4\mathrm{times}\mathrm{T}\right]=|2-4|=2\\ \mathrm{X}\left[1\mathrm{H},5\mathrm{T}\right]=\mathrm{X}\left[1\mathrm{times}\mathrm{H}\mathrm{and}5\mathrm{times}\mathrm{T}\right]=|1-5|=4\\ \mathrm{X}\left[0\mathrm{H},6\mathrm{T}\right]=\mathrm{X}\left[0\mathrm{times}\mathrm{H}\mathrm{and}6\mathrm{times}\mathrm{T}\right]=|0-6|=6\\ \mathrm{Therefore},\mathrm{the}\mathrm{possible}\mathrm{values}\mathrm{of}\mathrm{X}\mathrm{are}0, 1, 2, 4\mathrm{and}6\mathrm{.}\end{array}$

Q.52 Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{When}\mathrm{one}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{twice},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{is}\\ \left\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\right\}\\ \mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{heads}.\\ \therefore \mathrm{X}\left(\mathrm{HH}\right)=2,\mathrm{ }\mathrm{X}\left(\mathrm{HT}\right)=1,\mathrm{}\mathrm{X}\left(\mathrm{TH}\right)=1,\mathrm{}\mathrm{X}\left(\mathrm{TT}\right)=0\\ \mathrm{Therefore},\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1,\mathrm{or}2\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{HH}\right)=\frac{1}{4},\mathrm{ }\mathrm{P}\left(\mathrm{TH}\right)=\frac{1}{4},\mathrm{ }\mathrm{P}\left(\mathrm{HT}\right)=\frac{1}{4}\mathrm{and}\mathrm{P}\left(\mathrm{TT}\right)=\frac{1}{4}\\ \mathrm{Then},\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{TT}\right)=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HT}\right)+\mathrm{P}\left(\mathrm{TH}\right)\\ =\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HH}\right)\\ =\frac{1}{4}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 P(X) $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{4}$

$\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{When}\mathrm{three}\mathrm{coins}\mathrm{are}\mathrm{tossed}\mathrm{simultaneously},\mathrm{the}\mathrm{sample}\mathrm{space}\\ \mathrm{is} \left\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THT},\mathrm{THH},\mathrm{TTH},\mathrm{TTT}\right\}\\ \mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{tails}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1, 2,\mathrm{or}3\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{HHH}\right)=\frac{1}{8}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)\\ =\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HTT}\right)+\mathrm{P}\left(\mathrm{TTH}\right)+\mathrm{P}\left(\mathrm{THT}\right)\\ =\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(\mathrm{TTT}\right)\\ =\frac{1}{8}\\ \mathrm{Thus},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 3 P(X) $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$

$\begin{array}{l}\left(\mathrm{iii}\right)\mathrm{When}\mathrm{a}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{four}\mathrm{times},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{is}\\ \mathrm{S}=\left\{\begin{array}{l}\mathrm{HHHH},\mathrm{HHHT},\mathrm{HHTH},\mathrm{HTHH},\mathrm{THHH},\mathrm{HHTT},\mathrm{HTTH},\mathrm{TTHH},\\ \mathrm{HTHT}, \mathrm{THTH},\mathrm{THHT}, \mathrm{HTTT}, \mathrm{THTT},\mathrm{TTHT},\mathrm{TTTH}, \mathrm{TTTT}\end{array}\right\}\\ \mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable},\mathrm{which}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{heads}.\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1, 2, 3,\mathrm{or}4\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{HHHH}\right)=\frac{1}{16}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HHHT}\right)+\mathrm{P}\left(\mathrm{HHTH}\right)+\mathrm{P}\left(\mathrm{HTHH}\right)+\mathrm{P}\left(\mathrm{THHH}\right)\\ =\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\\ =\frac{4}{16}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HHTT}\right)+\mathrm{P}\left(\mathrm{HTTH}\right)+\mathrm{P}\left(\mathrm{TTHH}\right)+\mathrm{P}\left(\mathrm{HTHT}\right)+\mathrm{P}\left(\mathrm{THTH}\right)\\ +\mathrm{ }\mathrm{P}\left(\mathrm{THHT}\right)\\ =\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\\ =\frac{6}{16}=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(\mathrm{HTTT}\right)+\mathrm{P}\left(\mathrm{THTT}\right)+\mathrm{P}\left(\mathrm{TTHT}\right)+\mathrm{P}\left(\mathrm{TTTH}\right)\\ =\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\\ =\frac{4}{16}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{X}=4\right)=\mathrm{P}\left(\mathrm{TTTT}\right)=\frac{1}{16}\end{array}$

 X 0 1 2 3 4 P(X) $\frac{1}{16}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{4}$ $\frac{1}{16}$

Q.53 Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die

Ans.

$\begin{array}{l}\mathrm{When}\mathrm{a}\mathrm{die}\mathrm{is}\mathrm{tossed}\mathrm{two}\mathrm{times},\mathrm{we}\mathrm{obtain}\left(6×6\right) = 36\mathrm{number}\\ \mathrm{of}\mathrm{observations}.\mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable},\mathrm{which}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{successes}\mathrm{.}\\ \left(\mathrm{i}\right)\mathrm{Here},\mathrm{success}\mathrm{refers}\mathrm{to}\mathrm{the}\mathrm{number}\mathrm{greater}\mathrm{than}4\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{number}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}4\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{tosses}\right)\\ \mathrm{ }=\frac{4}{6}×\frac{4}{6}=\frac{4}{9}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{number}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}4\mathrm{on}\mathrm{first}\mathrm{toss}\mathrm{and}\\ \mathrm{greater}\mathrm{than}4\mathrm{on}\mathrm{second}\mathrm{toss}\right) +\mathrm{P}\left(\mathrm{number}\mathrm{greater}\\ \mathrm{than}4\mathrm{on}\mathrm{first}\mathrm{toss}\mathrm{and}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}4\mathrm{on}\\ \mathrm{second}\mathrm{toss}\right)\\ \mathrm{ }=\frac{4}{6}×\frac{2}{6}+\frac{2}{6}×\frac{4}{6}=\frac{4}{9}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{number}\mathrm{greater}\mathrm{than}4\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{tosses}\right)\\ \mathrm{ }=\frac{2}{6}×\frac{2}{6}=\frac{1}{9}\\ \mathrm{Thus},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 P(X) $\frac{4}{9}$ $\frac{4}{9}$ $\frac{1}{9}$

$\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{Here},\mathrm{success}\mathrm{means}\mathrm{six}\mathrm{appears}\mathrm{on}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{die}\mathrm{.}\\ \mathrm{P}\mathrm{}\left(\mathrm{Y}=0\right)=\mathrm{P}\left(\mathrm{six}\mathrm{}\mathrm{does}\mathrm{}\mathrm{not}\mathrm{}\mathrm{appear}\mathrm{}\mathrm{on}\mathrm{}\mathrm{any}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{dice}\right)\\ =\frac{5}{6}×\frac{5}{6}=\frac{25}{36}\\ \mathrm{P}\left(\mathrm{Y}=1\right)=\mathrm{P}\mathrm{ }\left(\mathrm{six}\mathrm{}\mathrm{appears}\mathrm{}\mathrm{on}\mathrm{}\mathrm{at}\mathrm{}\mathrm{least}\mathrm{}\mathrm{one}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{dice}\right)=\frac{11}{36}\\ \mathrm{Thus},\mathrm{}\mathrm{the}\mathrm{}\mathrm{required}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{distribution}\mathrm{}\mathrm{is}\mathrm{}\mathrm{as}\mathrm{}\mathrm{follows}:\end{array}$

 Y 0 1 P(Y) $\frac{25}{36}$ $\frac{11}{36}$

Q.54 From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Ans.

$\begin{array}{l}\mathrm{Since},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{out}\mathrm{of}30\mathrm{bulbs}, 6\mathrm{are}\mathrm{defective}\mathrm{.}\\ ⇒\mathrm{Number}\mathrm{of}\mathrm{non}–\mathrm{defective}\mathrm{bulbs}=\mathrm{30}-\mathrm{6}=\mathrm{24}\\ \mathrm{P}\left(\mathrm{Non}-\mathrm{defective}\mathrm{bulb}\right)=\frac{24}{30}=\frac{6}{5}\\ \mathrm{P}\left(\mathrm{Defective}\mathrm{bulb}\right)=\frac{6}{30}=\frac{1}{5}\\ 4\mathrm{bulbs}\mathrm{are}\mathrm{drawn}\mathrm{from}\mathrm{the}\mathrm{lot}\mathrm{with}\mathrm{replacement}\mathrm{.}\\ \mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{that}\mathrm{denotes}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{defective}\mathrm{bulbs}\mathrm{in}\mathrm{the}\mathrm{selected}\mathrm{bulbs}\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(4\mathrm{non}–\mathrm{defective}\mathrm{and}0\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{0}^{4}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(3\mathrm{non}–\mathrm{defective}\mathrm{and}1\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{1}^{4}.\left(\frac{1}{5}\right).{\left(\frac{4}{5}\right)}^{3}=\frac{256}{625}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(2\mathrm{non}–\mathrm{defective}\mathrm{and}2\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{2}^{4}.{\left(\frac{1}{5}\right)}^{2}.{\left(\frac{4}{5}\right)}^{2}=\frac{96}{625}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(1\mathrm{non}–\mathrm{defective}\mathrm{and}3\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{3}^{4}.{\left(\frac{1}{5}\right)}^{3}.\left(\frac{4}{5}\right)=\frac{16}{625}\\ \mathrm{P}\left(\mathrm{X}=4\right)=\mathrm{P}\left(0\mathrm{non}–\mathrm{defective}\mathrm{and}4\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{4}^{4}.{\left(\frac{1}{5}\right)}^{4}.{\left(\frac{4}{5}\right)}^{0}=\frac{1}{625}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 3 4 P(X) $\frac{256}{625}$ $\frac{256}{625}$ $\frac{96}{625}$ $\frac{16}{625}$ $\frac{1}{625}$

Q.55 A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Ans.

$\begin{array}{l}\text{Let the probability of getting a tail in the biased coin be x}\text{.}\\ \therefore \text{}\text{\hspace{0.17em}}\text{}\text{}P\left(T\right)=x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\left(H\right)=3x\\ Since,\text{P}\left(T\right)+P\left(H\right)=1\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+3x=1\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x=1\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{1}{4}\\ Therefore,\text{\hspace{0.17em}}P\left(T\right)=\frac{1}{4}\text{and P}\left(H\right)=\frac{3}{4}\\ \text{When the coin is tossed twice, the sample space is}\\ \text{{HH, TT, HT, TH}}\text{.}\\ \text{Let X be the random variable representing the number of tails}\text{.}\\ \therefore P\left(X=0\right)=P\left(no\text{tail}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=P\left(H\right)×P\left(H\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}×\frac{3}{4}=\frac{9}{16}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\left(X=1\right)=P\left(one\text{tail}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=P\left(HT\right)+P\left(TH\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}×\frac{1}{4}+\frac{1}{4}×\frac{3}{4}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{6}{16}=\frac{3}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\left(X=2\right)=P\left(two\text{tail}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=P\left(TT\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}×\frac{1}{4}=\frac{1}{16}\\ \end{array}$

Therefore, the required probability distribution is as follows:

 X 0 1 2 P(X) $\frac{9}{16}$ $\frac{3}{8}$ $\frac{1}{16}$

Q.56 A random variable X has the following probability distribution:

 X 0 1 2 3 4 5 6 7 P(X) 0 k 2 k 2 k 3 k K2 2k2 7k2+ k

Determine
(i) K (ii) P(X < 3) (iii) P(X > 6) (iv) P(0 < X < 3)

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Since}\mathrm{sum}\mathrm{of}\mathrm{probabilities}\mathrm{of}\mathrm{a}\mathrm{probability}\mathrm{distribution}\mathrm{of}\\ \mathrm{random}\mathrm{variables}\mathrm{is}\mathrm{one}\mathrm{.}\\ \mathrm{ }0+\mathrm{k}+2\mathrm{k}+2\mathrm{k}+3\mathrm{ }\mathrm{k}+{\mathrm{k}}^{2}+2\mathrm{ }{\mathrm{k}}^{2}+7\mathrm{ }{\mathrm{k}}^{2}+\mathrm{k}=1\\ ⇒ \mathrm{ }10\mathrm{ }{\mathrm{k}}^{2}+9\mathrm{ }\mathrm{k}-1=0\\ ⇒ \mathrm{ }\left(10\mathrm{ }\mathrm{k}-1\right)\left(\mathrm{k}+1\right)=0\\ ⇒ \mathrm{ }\mathrm{k}=-1, \frac{1}{10}\\ \mathrm{k}=-1\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{as}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{an}\mathrm{event}\mathrm{is}\mathrm{never}\\ \mathrm{negative}\mathrm{.}\\ \therefore \mathrm{ }\mathrm{k}=\frac{1}{10}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{x}<3\right)=\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =0+\mathrm{k}+2\mathrm{k}\\ =3\mathrm{k}\\ =3×\frac{1}{10}=\frac{3}{10}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{x}<6\right)=\mathrm{P}\left(\mathrm{X}=7\right)\\ =7\mathrm{ }{\mathrm{k}}^{2}+\mathrm{k}\\ =7{\left(\frac{1}{10}\right)}^{2}+\frac{1}{10}\\ =\frac{7}{100}+\frac{1}{10}\\ =\frac{7+10}{100}=\frac{17}{100}\\ \left(\mathrm{iv}\right)\mathrm{P}\left(0<\mathrm{x}<3\right)\\ =\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{k}+2\mathrm{k}\\ =3\mathrm{k}\\ =3\left(\frac{1}{10}\right)=\frac{3}{10}\end{array}$

Q.57

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{random}\mathrm{}\mathrm{variable}\mathrm{}\mathrm{X}\mathrm{}\mathrm{has}\mathrm{}\mathrm{a}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{distribution}\mathrm{}\mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{following}\mathrm{}\mathrm{form},\mathrm{where}\mathrm{}\mathrm{k}\mathrm{}\mathrm{is}\mathrm{}\mathrm{some}\mathrm{}\mathrm{number}:\\ \mathrm{P}\left(\mathbf{X}\right)=\left\{\begin{array}{l}\mathrm{k},\mathrm{ifx}=0\\ 2\mathrm{k},\mathrm{ifx}=1\\ 3\mathrm{k},\mathrm{ifx}=2\\ 0,\mathrm{otherwise}\end{array}\\ \left(\mathbf{a}\right)\mathrm{Determine}\mathrm{}\mathrm{the}\mathrm{}\mathrm{value}\mathrm{}\mathrm{of}\mathrm{}\mathrm{k}.\\ \left(\mathrm{b}\right)\mathrm{Find}\mathrm{}\mathrm{P}\left(\mathrm{X}<2\right),\mathrm{P}\left(\mathrm{X}\le 2\right),\mathrm{P}\left(\mathrm{X}\ge 2\right)\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Since},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{probabilities}\mathrm{of}\mathrm{a}\mathrm{probability}\mathrm{distribution}\\ \mathrm{of}\mathrm{random}\mathrm{variables}\mathrm{is}\mathrm{one}\mathrm{.}\\ \mathrm{i}.\mathrm{e}.,\mathrm{k}+2\mathrm{k}+3\mathrm{k}+0=1\\ ⇒ 6\mathrm{k}=1\\ ⇒ \mathrm{k}=\frac{1}{6}\\ \left(\mathrm{b}\right)\mathrm{P}\left(\mathrm{x}<2\right)=\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)\\ =\mathrm{k}+2\mathrm{k}\\ =3\mathrm{k}\\ =3×\frac{1}{6}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{x}\le 2\right)=\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{k}+2\mathrm{ }\mathrm{k}+3\mathrm{ }\mathrm{k}\\ =6\mathrm{ }\mathrm{k}\\ =6×\frac{1}{6}=1\\ \mathrm{P}\left(\mathrm{x}\ge 2\right)=\mathrm{P}\left(\mathrm{X}=2\right)+\mathrm{P}\left(\mathrm{X}\ge 2\right)\\ =3\mathrm{ }\mathrm{k}+0\\ =3\mathrm{ }\mathrm{k}\\ =3×\frac{1}{6}\\ =\frac{1}{2}\end{array}$

Q.58 Find the mean number of heads in three tosses of a fair coin.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{denotes}\mathrm{the}\mathrm{success}\mathrm{of}\mathrm{getting}\mathrm{heads}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{is}\\ \mathrm{S}= \left\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THH},\mathrm{THT},\mathrm{TTH},\mathrm{TTT}\right\}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1, 2,\mathrm{or}3\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{TTT}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{T}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{8}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{T}\right)+\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{H}\right)+\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{H}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ \mathrm{ }=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\\ \mathrm{ }=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ \mathrm{ }=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\\ \mathrm{ }=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(\mathrm{HHH}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ \mathrm{ }=\frac{1}{8}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 3 P(X) null $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$

$\begin{array}{l}\mathrm{Mean}\mathrm{of}\mathrm{X}=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=0×\frac{1}{8}+1×\frac{3}{8}+2×\frac{3}{8}+3×\frac{1}{8}\\ \mathrm{ }=\frac{3}{8}+\frac{6}{8}+\frac{3}{8}\\ \mathrm{ }=\frac{12}{8}=1.5\end{array}$

Q.59 Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{sixes}\mathrm{obtained}\mathrm{when}\mathrm{two}\mathrm{dice}\\ \mathrm{are}\mathrm{thrown}\mathrm{simultaneously}.\\ \mathrm{P}\left(\mathrm{getting}6\right)=\frac{1}{6}\\ \mathrm{P}\left(\mathrm{not}\mathrm{getting}6\right)=\frac{5}{6}\\ \mathrm{Therefore},\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1,\mathrm{or}2\mathrm{.}\\ \therefore \mathrm{P}\mathrm{}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{not}\mathrm{getting}\mathrm{six}\mathrm{on}\mathrm{any}\mathrm{of}\mathrm{the}\mathrm{dice}\right)\\ =\frac{5}{6}×\frac{5}{6}=\frac{25}{36}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{first}\mathrm{}\mathrm{die}\mathrm{}\mathrm{and}\mathrm{}\mathrm{no}\mathrm{}\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{second}\mathrm{}\mathrm{die}\right)+\\ \mathrm{ }\mathrm{P}\mathrm{}\left(\mathrm{no}\mathrm{}\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{first}\mathrm{}\mathrm{die}\mathrm{}\mathrm{and}\mathrm{}\mathrm{six}\mathrm{ }\mathrm{on}\mathrm{second}\mathrm{die}\right)\\ \mathrm{ }=\frac{1}{6}×\frac{5}{6}+\frac{5}{6}×\frac{1}{6}=\frac{10}{36}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{both}\mathrm{}\mathrm{dice}\right)\\ \mathrm{ }=\frac{1}{36}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 P(X) $\frac{25}{36}$ $\frac{10}{36}$ $\frac{1}{36}$

$\begin{array}{l}\mathrm{Then},\mathrm{expectation}\mathrm{of}\mathrm{X}\\ \mathrm{ }=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=0×\frac{25}{36}+1×\frac{10}{36}+2×\frac{1}{36}\\ \mathrm{ }=\frac{12}{36}=\frac{1}{3}\end{array}$

Q.60 Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{two}\mathrm{positive}\mathrm{integers}\mathrm{can}\mathrm{be}\mathrm{selected}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{six}\\ \mathrm{positive}\mathrm{integers}\mathrm{without}\mathrm{replacement}\mathrm{in}6×\mathrm{5}=30\mathrm{ways}\\ \mathrm{X}\mathrm{represents}\mathrm{the}\mathrm{larger}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{numbers}\mathrm{obtained}\mathrm{.}\\ \mathrm{Therefore},\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}2, 3, 4, 5,\mathrm{or}6\mathrm{.}\\ \mathrm{For}\mathrm{X}=2,\mathrm{the}\mathrm{possible}\mathrm{observations}\mathrm{are}\left(1, 2\right)\mathrm{and}\left(2, 1\right)\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=2\right)=\frac{2}{30}=\frac{1}{15}\\ \mathrm{For}\mathrm{X}=3,\mathrm{the}\mathrm{possible}\mathrm{observations}\mathrm{are}\left(1,3\right), \left(2,3\right), \left(3,1\right)\\ \mathrm{and}\left(3,2\right)\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=3\right)=\frac{4}{30}=\frac{2}{15}\\ \mathrm{For}\mathrm{X}=4,\mathrm{the}\mathrm{possible}\mathrm{observations}\mathrm{are}\left(1, 4\right), \left(2, 4\right), \left(3, 4\right),\\ \left(4, 3\right), \left(4, 2\right)\mathrm{and}\left(4, 1\right)\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=4\right)=\frac{6}{30}=\frac{1}{5}\\ \mathrm{For}\mathrm{}\mathrm{X}=5, \mathrm{the}\mathrm{}\mathrm{possible}\mathrm{}\mathrm{observations}\mathrm{}\mathrm{are}\mathrm{}\left(1,\mathrm{}5\right),\mathrm{}\left(2,\mathrm{}5\right),\mathrm{}\left(3,\mathrm{}5\right),\mathrm{}\\ \left(4,\mathrm{}5\right),\mathrm{}\left(5,\mathrm{}4\right),\mathrm{}\left(5,\mathrm{}3\right),\mathrm{}\left(5,2\right)\mathrm{}\mathrm{and}\mathrm{}\left(5,\mathrm{}1\right).\\ \therefore \mathrm{P}\left(\mathrm{X}=5\right)=\frac{8}{30}=\frac{4}{15}\\ \mathrm{For}\mathrm{}\mathrm{X}=6,\mathrm{the}\mathrm{}\mathrm{possible}\mathrm{}\mathrm{observations}\mathrm{}\mathrm{are}\mathrm{}\left(1,\mathrm{}6\right),\mathrm{}\left(2,\mathrm{}6\right),\mathrm{}\left(3,\mathrm{}6\right),\mathrm{}\\ \left(4,\mathrm{}6\right),\mathrm{}\left(5,\mathrm{}6\right), \left(6,5\right),\left(6,\mathrm{}4\right),\mathrm{}\left(6,3\right),\mathrm{}\left(6,\mathrm{}2\right)\mathrm{}\mathrm{and}\mathrm{}\left(6,\mathrm{}1\right).\\ \therefore \mathrm{P}\left(\mathrm{X}=6\right)=\frac{10}{30}=\frac{1}{3}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 2 3 4 5 6 P(X) $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{4}{15}$ $\frac{1}{3}$

$\begin{array}{l}\mathrm{Then},\mathrm{expectation}\mathrm{of}\mathrm{X}\\ \mathrm{ }=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=2×\frac{1}{15}+3×\frac{2}{15}+4×\frac{1}{5}+5×\frac{4}{15}+6×\frac{1}{3}\\ \mathrm{ }=\frac{2}{15}+\frac{6}{15}+\frac{4}{5}+\frac{20}{15}+\frac{6}{3}\\ \mathrm{ }=\frac{2+6+12+20+30}{15}\\ \mathrm{ }=\frac{70}{15}=\frac{14}{3}\end{array}$

Q.61 Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Ans.

$\begin{array}{l}\mathrm{When}\mathrm{two}\mathrm{fair}\mathrm{dice}\mathrm{are}\mathrm{rolled},\\ \mathrm{Number}\mathrm{of}\mathrm{observations}=\mathrm{6}×\mathrm{6}=\mathrm{36}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(1,1\right)\\ =\frac{1}{36}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(1,2\right)+\mathrm{P}\left(2,1\right)\\ =\frac{2}{36}\\ \mathrm{P}\left(\mathrm{X}=4\right)=\mathrm{P}\left(2,2\right)+\mathrm{P}\left(3,1\right)+\mathrm{P}\left(1,3\right)\\ =\frac{3}{36}\\ \mathrm{P}\left(\mathrm{X}=5\right)=\mathrm{P}\left(2,3\right)+\mathrm{P}\left(3,2\right)+\mathrm{P}\left(4,1\right)+\mathrm{P}\left(1,4\right)\\ =\frac{4}{36}\\ \mathrm{P}\left(\mathrm{X}=6\right)=\mathrm{P}\left(2,4\right)+\mathrm{P}\left(3,3\right)+\mathrm{P}\left(4,2\right)+\mathrm{P}\left(5,1\right)+\mathrm{P}\left(1,5\right)\\ =\frac{5}{36}\\ \mathrm{P}\left(\mathrm{X}=7\right)=\mathrm{P}\left(2,5\right)+\mathrm{P}\left(3,4\right)+\mathrm{P}\left(4,3\right)+\mathrm{P}\left(5,2\right)+\mathrm{P}\left(1,6\right)+\mathrm{P}\left(6,1\right)\\ =\frac{6}{36}\\ \mathrm{P}\left(\mathrm{X}=8\right)=\mathrm{P}\left(2,6\right)+\mathrm{P}\left(3,5\right)+\mathrm{P}\left(4,4\right)+\mathrm{P}\left(5,3\right)+\mathrm{P}\left(6,2\right)\\ =\frac{5}{36}\\ \mathrm{P}\left(\mathrm{X}=9\right)=\mathrm{P}\left(3,6\right)+\mathrm{P}\left(4,5\right)+\mathrm{P}\left(5,4\right)+\mathrm{P}\left(6,3\right)\\ =\frac{4}{36}\\ \mathrm{P}\left(\mathrm{X}=10\right)=\mathrm{P}\left(4,6\right)+\mathrm{P}\left(5,5\right)+\mathrm{P}\left(6,4\right)\\ =\frac{3}{36}\\ \mathrm{P}\left(\mathrm{X}=11\right)=\mathrm{P}\left(5,6\right)+\mathrm{P}\left(6,5\right)\\ =\frac{2}{36}\\ \mathrm{P}\left(\mathrm{X}=12\right)=\mathrm{P}\left(6,6\right)\\ =\frac{1}{36}\\ \mathrm{Thus},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 2 3 4 5 6 7 8 9 10 11 12 P(X) $\frac{1}{36}$ $\frac{2}{36}$ $\frac{3}{36}$ $\frac{4}{36}$ $\frac{5}{36}$ $\frac{6}{36}$ $\frac{5}{36}$ $\frac{4}{36}$ $\frac{3}{36}$ $\frac{2}{36}$ $\frac{1}{36}$

$\begin{array}{l}\mathrm{Then},\mathrm{expectation}\mathrm{of}\mathrm{X}\\ \mathrm{ }=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=2×\frac{1}{36}+3×\frac{2}{36}+4×\frac{3}{36}+5×\frac{4}{36}+6×\frac{5}{36}+7×\frac{6}{36}\\ +8×\frac{5}{36}+9×\frac{4}{36}+10×\frac{3}{36}+11×\frac{2}{36}+12×\frac{1}{36}\\ \mathrm{ }=7\\ \mathrm{And}\mathrm{ }\mathrm{E}\left({\mathrm{X}}^{2}\right)= {2}^{2}×\frac{1}{36}+{3}^{2}×\frac{2}{36}+{4}^{2}×\frac{3}{36}+{5}^{2}×\frac{4}{36}+{6}^{2}×\frac{5}{36}+{7}^{2}×\frac{6}{36}\\ +\mathrm{ }{8}^{2}×\frac{5}{36}+{9}^{2}×\frac{4}{36}+{\left(10\right)}^{2}×\frac{3}{36}+11×\frac{2}{36}+{\left(12\right)}^{2}×\frac{1}{36}\\ \mathrm{ }=\frac{329}{6}=54.833\\ \mathrm{Thus},\\ \mathrm{Var}\left(\mathrm{X}\right)=\mathrm{E}\left({\mathrm{X}}^{2}\right)-{\left(\mathrm{E}\left(\mathrm{X}\right)\right)}^{2}\\ \mathrm{ }=54.833-{\left(7\right)}^{2}\\ \mathrm{ }=54.833-49\\ \mathrm{ }=5.833\\ \mathrm{Standard}\mathrm{deviation}\\ \mathrm{ }\mathrm{\sigma }=\sqrt{5.833}\\ \mathrm{ }=2.415\end{array}$

Q.62 A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Ans.

$\begin{array}{l}\mathrm{There}\mathrm{are}15\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{class}.\mathrm{Each}\mathrm{student}\mathrm{has}\mathrm{the}\mathrm{same}\\ \mathrm{chance}\mathrm{to}\mathrm{be}\mathrm{chosen}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{each}\mathrm{student}\mathrm{to}\mathrm{be}\mathrm{selected}=\frac{1}{15}\\ \mathrm{The}\mathrm{frequency}\mathrm{table}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{data}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 14 15 16 17 18 19 20 21 f 2 1 2 3 1 2 3 1

$\begin{array}{l}P\left(X=14\right)=\frac{2}{15}\\ P\left(X=15\right)=\frac{1}{15}\\ P\left(X=16\right)=\frac{2}{15}\\ P\left(X=17\right)=\frac{3}{15}\\ P\left(X=18\right)=\frac{1}{15}\\ P\left(X=19\right)=\frac{2}{15}\\ P\left(X=20\right)=\frac{3}{15}\\ P\left(X=21\right)=\frac{1}{15}\\ Thus,\text{the probability distribution of X is}\end{array}$

 X 14 15 16 17 18 19 20 21 P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$

$\begin{array}{l}Then,\text{​}\text{mean of X}=E\left(X\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=14×\frac{2}{15}+15×\frac{1}{15}+16×\frac{2}{15}+17×\frac{3}{15}+18×\frac{1}{15}\\ \text{}\text{}\text{}\text{}+19×\frac{2}{15}+20×\frac{3}{15}+21×\frac{1}{15}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{15}\left(28+15+32+51+18+38+60+21\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{263}{15}=17.53\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}E\left({X}^{2}\right)={14}^{2}×\frac{2}{15}+{15}^{2}×\frac{1}{15}+{16}^{2}×\frac{2}{15}+{17}^{2}×\frac{3}{15}+{18}^{2}×\frac{1}{15}\\ \text{}\text{}\text{}\text{}+{19}^{2}×\frac{2}{15}+{20}^{2}×\frac{3}{15}+{21}^{2}×\frac{1}{15}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4683}{15}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=312.2\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Variance\left(X\right)=E\left({X}^{2}\right)-{\left[E\left(X\right)\right]}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=312.2-{\left(17.53\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=312.2-307.4177\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.7823\approx 4.78\\ \text{Standard deviation}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{Variance\left(X\right)}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{4.78}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2.19\end{array}$

Q.63 In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var (X).

Ans.

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=0\right)=30\mathrm{%}=0.3\\ \mathrm{P}\left(\mathrm{X}=1\right)=70\mathrm{%}=0.7\\ \mathrm{So},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 0 1 P(X) 0.3 0.7

$\begin{array}{l}Then,\text{\hspace{0.17em}}E\left(X\right)=0×0.3+1×0.7\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}E\left({X}^{2}\right)={0}^{2}×0.3+{1}^{2}×0.7\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Var\left(X\right)=E\left({X}^{2}\right)-{\left[E\left(X\right)\right]}^{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7-{\left(0.7\right)}^{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7-0.49\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.21\end{array}$

Q.64 The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1 (B) 2 (C) 5 (D) 8/3

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{representing}\mathrm{a}\mathrm{number}\mathrm{on}\mathrm{the}\mathrm{die}\mathrm{.}\\ \mathrm{The}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{observations}\mathrm{is}\mathrm{six}\mathrm{.}\\ \mathrm{S}=\left\{1, 1, 1, 2, 2, 5\right\}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\frac{3}{6}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\frac{2}{6}=\frac{1}{3}\\ \mathrm{P}\left(\mathrm{X}=5\right)=\frac{1}{6}\\ \mathrm{So},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 1 2 5 P (X) $\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{6}$

$\begin{array}{l}E\left(X\right)=1×\frac{1}{2}+2×\frac{1}{3}+5×\frac{1}{6}\\ \text{}=\frac{3+4+5}{6}\\ \text{}=\frac{12}{6}=2\\ Therefore,\text{corect option is B}\text{.}\end{array}$

Q.65 Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
(A) 37/221 (B) 5/13 (C) 1/13 (D) 2/13

Ans.

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{cards}\mathrm{in}\mathrm{deck}=52\\ \mathrm{Number}\mathrm{of}\mathrm{aces}=\mathrm{4}\\ \mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{aces}\mathrm{obtained}.\mathrm{Therefore},\mathrm{X}\mathrm{can}\\ \mathrm{take}\mathrm{any}\mathrm{of}\mathrm{the}\mathrm{values}\mathrm{of}0,1,\mathrm{or}2\mathrm{.}\\ \mathrm{In}\mathrm{a}\mathrm{deck}\mathrm{of}52\mathrm{cards}, 4\mathrm{cards}\mathrm{are}\mathrm{aces}.\mathrm{Therefore},\mathrm{there}\mathrm{are}\\ 48\mathrm{non}–\mathrm{ace}\mathrm{cards}\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(0\mathrm{ace}\mathrm{card}\mathrm{and}2\mathrm{non}–\mathrm{ace}\mathrm{cards}\right)\\ =\frac{\mathrm{C}_{0}^{4}×\mathrm{C}_{2}^{48}}{\mathrm{C}_{2}^{52}}\\ =\frac{1128}{1326}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(1\mathrm{ace}\mathrm{card}\mathrm{and}1\mathrm{non}–\mathrm{ace}\mathrm{cards}\right)\\ =\frac{\mathrm{C}_{1}^{4}×\mathrm{C}_{1}^{48}}{\mathrm{C}_{2}^{52}}\\ =\frac{192}{1326}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(2\mathrm{ace}\mathrm{card}\mathrm{and}0\mathrm{non}–\mathrm{ace}\mathrm{cards}\right)\\ =\frac{\mathrm{C}_{2}^{4}×\mathrm{C}_{0}^{48}}{\mathrm{C}_{2}^{52}}\\ =\frac{6}{1326}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 0 1 2 P(X) $\frac{1128}{1326}$ $\frac{192}{1326}$ $\frac{6}{1326}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\therefore E\left(X\right)=0×\frac{1128}{1326}+1×\frac{192}{1326}+2×\frac{6}{1326}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{192}{1326}+\frac{12}{1326}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{204}{1326}=\frac{2}{13}\\ Therefore,\text{the correct option is D}\text{.}\end{array}$

Q.66 A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{die}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{denotes}\\ \mathrm{the}\mathrm{number}\mathrm{of}\mathrm{successes}\mathrm{of}\mathrm{getting}\mathrm{odd}\mathrm{numbers}\mathrm{in}\mathrm{an}\\ \mathrm{experiment}\mathrm{of}6\mathrm{trials}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}\mathrm{is},\\ \mathrm{p}=\frac{3}{6}=\frac{1}{2}\\ \therefore \mathrm{q}=1-\mathrm{p}\\ \mathrm{ }=1-\frac{1}{2}=\frac{1}{2}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{.}\\ \mathrm{Therefore},\\ \mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{\mathrm{x}}{\left(\frac{1}{2}\right)}^{6-\mathrm{x}}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \left(\mathrm{i}\right)\mathrm{P}\left(5\mathrm{success}\right)=\mathrm{P}\left(\mathrm{X}=5\right)\\ \mathrm{ }=\mathrm{C}_{5}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=6.{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=\frac{6}{64}\\ \mathrm{ }=\frac{3}{32}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{at}\mathrm{least}5\mathrm{successes}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}=5\right)+\mathrm{P}\left(\mathrm{X}=6\right)\\ \mathrm{ }=\mathrm{C}_{5}^{6}{\left(\frac{1}{2}\right)}^{6}+\mathrm{C}_{6}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=6.\frac{1}{64}+1.\frac{1}{64}\\ \mathrm{ }=\frac{7}{64}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{at}\mathrm{most}5\mathrm{successes}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}\le 5\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}>5\right)\\ =1-\mathrm{P}\left(\mathrm{X}=6\right)\\ \mathrm{ }=1-\mathrm{C}_{6}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=1-\frac{1}{64}\\ \mathrm{ }=\frac{63}{64}\end{array}$

Q.67 A pair of dice is thrown 4 times. If getting a doublet is considered a success, Find the probability of two successes.

Ans.

$\begin{array}{l}\mathrm{Total}\mathrm{outcomes}\mathrm{of}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{dice}=6×6\\ =36\\ \mathrm{Number}\mathrm{of}\mathrm{doublets}\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\right\}\\ =6\\ \mathrm{The}\mathrm{repeated}\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{dice}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\\ \mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{doublets}\mathrm{in}\mathrm{an}\\ \mathrm{experiment}\mathrm{of}\mathrm{throwing}\mathrm{two}\mathrm{dice}\mathrm{simultaneously}\mathrm{four}\mathrm{times}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{doublets}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{pair}\mathrm{of}\\ \mathrm{dice}\mathrm{is} \mathrm{p}=\frac{6}{36}=\frac{1}{6}\\ \mathrm{and} \mathrm{ }\mathrm{q}=1-\mathrm{p}\\ =1-\frac{1}{6}\\ =\frac{5}{6}\\ \mathrm{Here},\mathrm{X}\mathrm{has}\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=4,\mathrm{p}=\frac{1}{6}\mathrm{and}\mathrm{q}=\frac{5}{6}\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{4}{\left(\frac{1}{6}\right)}^{\mathrm{x}}{\left(\frac{5}{6}\right)}^{4-\mathrm{x}}\\ \mathrm{So}, \mathrm{ }\mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{C}_{2}^{4}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{4-2}\\ =\frac{4!}{2!2!}×\frac{1}{36}×\frac{25}{36}\\ =6×\frac{1}{36}×\frac{25}{36}\\ =\frac{25}{216}\end{array}$

Q.68 There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{defective}\mathrm{items}\mathrm{in}\mathrm{a}\mathrm{sample}\mathrm{of}10\\ \mathrm{items}\mathrm{drawn}\mathrm{successively}\mathrm{.}\\ \mathrm{Since}\mathrm{the}\mathrm{drawing}\mathrm{is}\mathrm{done}\mathrm{with}\mathrm{replacement},\mathrm{the}\mathrm{trials}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{P}\left(\mathrm{defective}\mathrm{item}\right),\mathrm{p}=\frac{5}{100}\\ =\frac{1}{20}\\ \mathrm{q}=1-\mathrm{p}\\ =1-\frac{1}{20}\\ =\frac{19}{20}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 10\mathrm{and}\mathrm{p}=\frac{1}{20}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{10}{\left(\frac{1}{20}\right)}^{\mathrm{x}}{\left(\frac{19}{20}\right)}^{10-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{not}\mathrm{more}\mathrm{than}1\mathrm{defective}\mathrm{item}\right)\\ =\mathrm{P}\left(\mathrm{X}\le 1\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)\\ =\mathrm{C}_{0}^{10}{\left(\frac{1}{20}\right)}^{0}{\left(\frac{19}{20}\right)}^{10-0}+\mathrm{C}_{1}^{10}{\left(\frac{1}{20}\right)}^{1}{\left(\frac{19}{20}\right)}^{10-1}\\ ={\left(\frac{19}{20}\right)}^{10}+10.\left(\frac{1}{20}\right){\left(\frac{19}{20}\right)}^{9}\\ ={\left(\frac{19}{20}\right)}^{9}\left(\frac{19}{20}+\frac{10}{20}\right)\\ ={\left(\frac{19}{20}\right)}^{9}\left(\frac{29}{20}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}{\left(\frac{19}{20}\right)}^{9}\left(\frac{29}{20}\right).\end{array}$

Q.69 Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{spade}\mathrm{cards}\mathrm{among}\mathrm{the}\mathrm{five}\mathrm{cards}\mathrm{drawn}\\ \mathrm{Since}\mathrm{the}\mathrm{drawing}\mathrm{of}\mathrm{card}\mathrm{is}\mathrm{with}\mathrm{replacement},\mathrm{the}\mathrm{trials}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{In}\mathrm{a}\mathrm{well}\mathrm{shuffled}\mathrm{deck}\mathrm{of}52\mathrm{cards},\mathrm{there}\mathrm{are}13\mathrm{spade}\mathrm{cards}\mathrm{.}\\ \therefore \mathrm{p}=\frac{13}{52}=\frac{1}{4}\\ \mathrm{and} \mathrm{q}=1-\frac{1}{4}=\frac{3}{4}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 5\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{4}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{1}{4}\right)}^{\mathrm{x}}{\left(\frac{3}{4}\right)}^{5-\mathrm{x}}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{all}\mathrm{}5\mathrm{cards}\mathrm{are}\mathrm{spades}\right)\\ =\mathrm{C}_{5}^{5}{\left(\frac{1}{4}\right)}^{5}{\left(\frac{3}{4}\right)}^{5-5}\\ =1.{\left(\frac{1}{4}\right)}^{5}{\left(\frac{3}{4}\right)}^{0}\\ =\frac{1}{{4}^{5}}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{only}3\mathrm{cards}\mathrm{are}\mathrm{spades}\right)=\mathrm{P}\left(\mathrm{X}=3\right)\\ =\mathrm{C}_{3}^{5}{\left(\frac{1}{4}\right)}^{3}{\left(\frac{3}{4}\right)}^{5-3}\\ =\mathrm{C}_{3}^{5}{\left(\frac{1}{4}\right)}^{3}{\left(\frac{3}{4}\right)}^{2}\\ =10.\frac{1}{64}.\frac{9}{16}\\ =\frac{45}{512}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{none}\mathrm{is}\mathrm{spade}\right)=\mathrm{P}\left(\mathrm{X}=0\right)\\ =\mathrm{C}_{0}^{5}{\left(\frac{1}{4}\right)}^{0}{\left(\frac{3}{4}\right)}^{5-0}\\ =1.1.\frac{243}{1024}\\ =\frac{243}{1024}\end{array}$

Q.70 The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{bulbs}\mathrm{that}\mathrm{will}\mathrm{fuse}\mathrm{after}150\\ \mathrm{days}\mathrm{of}\mathrm{use}\mathrm{in}\mathrm{an}\mathrm{experiment}\mathrm{of}5\mathrm{trials}.\mathrm{The}\mathrm{trials}\mathrm{are}\mathrm{Bernoulli}\\ \mathrm{trials}\mathrm{.}\\ \mathrm{Given},\mathrm{p}=\mathrm{0}.05\mathrm{and}\\ \mathrm{ }\mathrm{q}=1-0.05=0.95\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 5\mathrm{and}\mathrm{ }\mathrm{p}=\mathrm{0}.05\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\mathrm{0}.05\right)}^{\mathrm{x}}{\left(0.95\right)}^{5-\mathrm{x}}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{none}\right)=\mathrm{P}\left(\mathrm{X}=0\right)\\ =\mathrm{C}_{0}^{5}{\left(\mathrm{0}.05\right)}^{0}{\left(0.95\right)}^{5-0}\\ =1.1.{\left(0.95\right)}^{5}\\ ={\left(0.95\right)}^{5}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{not}\mathrm{more}\mathrm{than}\mathrm{one}\right)\\ =\mathrm{P}\left(\mathrm{X}\le 1\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)\\ =\mathrm{C}_{0}^{5}{\left(\mathrm{0}.05\right)}^{0}{\left(0.95\right)}^{5-0}+\mathrm{C}_{1}^{5}{\left(\mathrm{0}.05\right)}^{1}{\left(0.95\right)}^{5-1}\\ =1.1.{\left(0.95\right)}^{5}+5\left(0.05\right){\left(0.95\right)}^{4}\\ ={\left(0.95\right)}^{4}\left(0.95+0.25\right)\\ ={\left(0.95\right)}^{4}×\left(1.2\right)\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{more}\mathrm{than}\mathrm{one}\right)\\ =\mathrm{P}\left(\mathrm{X}>1\right)\\ =1-\mathrm{P}\left(\mathrm{X}\le 1\right)\\ =1-{\left(0.95\right)}^{4}×\left(1.2\right)\\ \left(\mathrm{iv}\right)\mathrm{P}\left(\mathrm{at}\mathrm{least}\mathrm{one}\right)\\ =\mathrm{P}\left(\mathrm{X}\ge 1\right)\\ =1-\mathrm{P}\left(\mathrm{X}<1\right)\\ =1-\mathrm{P}\left(\mathrm{X}=0\right)\\ =1-\mathrm{C}_{0}^{5}{\left(\mathrm{0}.05\right)}^{0}{\left(0.95\right)}^{5-0}\\ =1-{\left(0.95\right)}^{5}\end{array}$

Q.71 A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{balls}\mathrm{marked}\mathrm{with}\mathrm{the}\mathrm{digit}0\mathrm{among}\\ \mathrm{the}4\mathrm{balls}\mathrm{drawn}\mathrm{.}\\ \mathrm{Since}\mathrm{the}\mathrm{balls}\mathrm{are}\mathrm{drawn}\mathrm{with}\mathrm{replacement},\mathrm{the}\mathrm{trials}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=4\mathrm{and}\mathrm{p}=\frac{1}{10}\\ \therefore \mathrm{q}=1-\frac{1}{10}=\frac{9}{10}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 4\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{10}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{4}{\left(\frac{1}{10}\right)}^{\mathrm{x}}{\left(\frac{9}{10}\right)}^{4-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{none}\mathrm{marked}\mathrm{with}0\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)\\ =\mathrm{C}_{0}^{4}{\left(\frac{1}{10}\right)}^{0}{\left(\frac{9}{10}\right)}^{4-0}\\ ={\left(\frac{9}{10}\right)}^{4}\end{array}$

Q.72 In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{correctly}\mathrm{answered}\mathrm{questions}\mathrm{out}\\ \mathrm{of}20\mathrm{questions}.\mathrm{The}\mathrm{repeated}\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{coin}\mathrm{are}\mathrm{Bernoulli}\mathrm{trails}.\\ \mathrm{Since}\mathrm{head}\mathrm{on}\mathrm{a}\mathrm{coin}\mathrm{represents}\mathrm{the}\mathrm{true}\mathrm{answer}\mathrm{and}\mathrm{tail}\\ \mathrm{represents}\mathrm{the}\mathrm{false}\mathrm{answer},\mathrm{the}\mathrm{correctly}\mathrm{answered}\mathrm{questions}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{So},\mathrm{p}=\frac{1}{2}\mathrm{and}\mathrm{q}=1-\frac{1}{2}=\frac{1}{2}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=20\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{10}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{20}{\left(\frac{1}{2}\right)}^{\mathrm{x}}{\left(\frac{1}{2}\right)}^{20-\mathrm{x}}\\ =\mathrm{C}_{\mathrm{x}}^{20}{\left(\frac{1}{2}\right)}^{20}\\ \mathrm{P}\left(\mathrm{at}\mathrm{least}12\mathrm{questions}\mathrm{answered}\mathrm{correctly}\right)\\ =\mathrm{P}\left(\mathrm{X}\ge 12\right)\\ =\mathrm{P}\left(\mathrm{X}=12\right)+\mathrm{P}\left(\mathrm{X}=13\right)+...+\mathrm{P}\left(\mathrm{X}=20\right)\\ =\mathrm{C}_{12}^{20}{\left(\frac{1}{2}\right)}^{20}+\mathrm{C}_{13}^{20}{\left(\frac{1}{2}\right)}^{20}+...+\mathrm{C}_{\mathrm{x}}^{20}{\left(\frac{1}{2}\right)}^{20}\\ ={\left(\frac{1}{2}\right)}^{20}\left(\mathrm{C}_{12}^{20}+\mathrm{C}_{13}^{20}+...+\mathrm{C}_{20}^{20}\right)\end{array}$

Q.73 Suppose X has a binomial distribution B (6, 1/2). Show that X = 3 is the most likely outcome.
(Hint: P(X = 3) is the maximum among all P (xi), xi = 0, 1, 2, 3, 4, 5, 6)

Ans.

$\begin{array}{l}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{whose}\mathrm{binomial}\mathrm{distribution}\mathrm{is}\left(6, \frac{1}{2}\right)\mathrm{.}\\ \mathrm{Therefore},\mathrm{n}=6\mathrm{and}\mathrm{p}=\frac{1}{2}\\ \therefore \mathrm{ }\mathrm{q}=1-\frac{1}{2}=\frac{1}{2}\\ \mathrm{Then},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{\mathrm{x}}{\left(\frac{1}{2}\right)}^{6-\mathrm{x}}\\ =\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)\mathrm{will}\mathrm{be}\mathrm{maximum}\mathrm{if}\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{is}\mathrm{maximum}\mathrm{.}\\ \mathrm{Then},\mathrm{}\mathrm{C}_{0}^{6}=\mathrm{C}_{6}^{6}=1, \mathrm{ }\mathrm{C}_{1}^{6}=\mathrm{C}_{5}^{6}=6\\ \mathrm{C}_{2}^{6}=\mathrm{C}_{4}^{6}=15, \mathrm{C}_{3}^{6}=20\\ \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{C}_{3}^{6}\mathrm{ }\mathrm{is}\mathrm{maximum}.\mathrm{Therefore},\mathrm{for}\mathrm{x}= 3,\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)\\ \mathrm{is}\mathrm{maximum}\mathrm{.}\\ \mathrm{Thus},\mathrm{X}= 3\mathrm{is}\mathrm{the}\mathrm{most}\mathrm{likely}\mathrm{outcome}\mathrm{.}\end{array}$

Q.74 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{guessing}\mathrm{of}\mathrm{correct}\mathrm{answers}\mathrm{from}\mathrm{multiple}\mathrm{choice}\\ \mathrm{questions}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{correct}\mathrm{answers}\mathrm{by}\mathrm{guessing}\mathrm{in}\mathrm{the}\mathrm{set}\mathrm{of}5\mathrm{multiple}\mathrm{choice}\\ \mathrm{questions}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{correct}\mathrm{answer}\mathrm{is},\\ \mathrm{p}=\frac{1}{3}\\ \therefore \mathrm{q}=1-\frac{1}{3}=\frac{2}{3}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 5\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{3}\\ \mathrm{Then},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{1}{3}\right)}^{\mathrm{x}}{\left(\frac{2}{3}\right)}^{5-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{guessing}\mathrm{more}\mathrm{than}4\mathrm{correct}\mathrm{answers}\right)\\ =\mathrm{P}\left(\mathrm{X}\ge 4\right)\\ =\mathrm{P}\left(\mathrm{X}=4\right)+\mathrm{P}\left(\mathrm{X}=5\right)\\ =\mathrm{C}_{4}^{5}{\left(\frac{1}{3}\right)}^{4}{\left(\frac{2}{3}\right)}^{5-4}+\mathrm{C}_{5}^{5}{\left(\frac{1}{3}\right)}^{5}{\left(\frac{2}{3}\right)}^{5-5}\\ =5.\left(\frac{1}{81}\right)\left(\frac{2}{3}\right)+1.\left(\frac{1}{243}\right).1\\ =\frac{10}{243}+\frac{1}{243}\\ =\frac{11}{243}\end{array}$

Q.75 A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100. What is the probability that he will win a prize
(a) at least once (b) exactly once (c) at least twice?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{winning}\mathrm{prizes}\mathrm{in}50\mathrm{lotteries}.\\ \mathrm{The}\mathrm{trials}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{So},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=50\mathrm{and}\\ \mathrm{p}=\frac{1}{100}\\ \therefore \mathrm{q}=1-\frac{1}{100}=\frac{99}{100}\\ \therefore \mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{50}{\left(\frac{1}{100}\right)}^{\mathrm{x}}{\left(\frac{99}{100}\right)}^{50-\mathrm{x}}\\ \left(\mathrm{a}\right)\mathrm{P}\left(\mathrm{winning}\mathrm{at}\mathrm{least}\mathrm{once}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}\ge 1\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}<1\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}=0\right)\\ \mathrm{ }=1-\mathrm{C}_{0}^{50}{\left(\frac{1}{100}\right)}^{0}{\left(\frac{99}{100}\right)}^{50-0}\\ \mathrm{ }=1-{\left(\frac{99}{100}\right)}^{50}\\ \left(\mathrm{b}\right)\mathrm{P}\left(\mathrm{winning}\mathrm{exactly}\mathrm{once}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}=1\right)\\ \mathrm{ }=\mathrm{C}_{1}^{50}{\left(\frac{1}{100}\right)}^{1}{\left(\frac{99}{100}\right)}^{50-1}\\ \mathrm{ }=50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}\\ \mathrm{ }=\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}\\ \left(\mathrm{c}\right)\mathrm{P}\left(\mathrm{at}\mathrm{least}\mathrm{twice}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}\ge 2\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}<2\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}=1\right)-\mathrm{P}\left(\mathrm{X}=0\right)\\ \mathrm{ }=1-\mathrm{C}_{1}^{50}{\left(\frac{1}{100}\right)}^{1}{\left(\frac{99}{100}\right)}^{50-1}-\mathrm{C}_{0}^{50}{\left(\frac{1}{100}\right)}^{0}{\left(\frac{99}{100}\right)}^{50-0}\\ \mathrm{ }=1-\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}-{\left(\frac{99}{100}\right)}^{50}\\ \mathrm{ }=1-{\left(\frac{99}{100}\right)}^{49}\left(\frac{1}{2}+\frac{99}{100}\right)\\ \mathrm{ }=1-{\left(\frac{99}{100}\right)}^{49}\left(\frac{149}{100}\right)\end{array}$

Q.76 Find the probability of getting 5 exactly twice in 7 throws of a die.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{tossing}\mathrm{of}\mathrm{a}\mathrm{die}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\\ \mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}5\mathrm{in}7\mathrm{throws}\\ \mathrm{of}\mathrm{the}\mathrm{die}.\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}5\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{die},\\ \mathrm{p}=\frac{1}{6}⇒\mathrm{q}=1-\frac{1}{6}=\frac{5}{6}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=7\mathrm{}\mathrm{and}\mathrm{}\mathrm{p}=\frac{1}{6}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{7}{\left(\frac{1}{6}\right)}^{\mathrm{x}}{\left(\frac{5}{6}\right)}^{7-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{getting}5\mathrm{exactly}\mathrm{twice}\right)\\ =\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{C}_{2}^{7}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{7-2}\\ =21.\left(\frac{1}{36}\right){\left(\frac{5}{6}\right)}^{5}\\ =\left(\frac{7}{12}\right){\left(\frac{5}{6}\right)}^{5}\end{array}$

Q.77 Find the probability of throwing at most 2 sixes in 6 throws of a single die.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{tossing}\mathrm{of}\mathrm{the}\mathrm{die}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\\ \mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{sixes}\mathrm{in}6\mathrm{throws}\\ \mathrm{of}\mathrm{the}\mathrm{die}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{six}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{die},\\ \mathrm{p}=\frac{1}{6}⇒\mathrm{q}=1-\frac{1}{6}=\frac{5}{6}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=6\mathrm{and}\mathrm{p}=\frac{1}{6}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{6}\right)}^{\mathrm{x}}{\left(\frac{5}{6}\right)}^{6-\mathrm{x}}\\ \mathrm{ }\mathrm{P}\left(\mathrm{at}\mathrm{most}2\mathrm{sixes}\right)=\mathrm{P}\left(\mathrm{X}\le 2\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{C}_{0}^{6}{\left(\frac{1}{6}\right)}^{0}{\left(\frac{5}{6}\right)}^{6-0}+\mathrm{C}_{1}^{6}{\left(\frac{1}{6}\right)}^{1}{\left(\frac{5}{6}\right)}^{6-1}\\ +\mathrm{C}_{2}^{6}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{6-2}\\ ={\left(\frac{5}{6}\right)}^{6}+6.\left(\frac{1}{6}\right){\left(\frac{5}{6}\right)}^{5}+15.\left(\frac{1}{36}\right){\left(\frac{5}{6}\right)}^{4}\\ ={\left(\frac{5}{6}\right)}^{4}\left(\frac{25}{36}+\frac{5}{6}+\frac{15}{36}\right)\\ ={\left(\frac{5}{6}\right)}^{4}\left(\frac{25+30+15}{36}\right)\\ ={\left(\frac{5}{6}\right)}^{4}\left(\frac{70}{36}\right)\\ =\frac{35}{18}{\left(\frac{5}{6}\right)}^{4}\end{array}$

Q.78 It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{selections}\mathrm{of}\mathrm{articles}\mathrm{in}\mathrm{a}\mathrm{random}\mathrm{sample}\mathrm{space}\\ \mathrm{are}\mathrm{Bernoulli}\mathrm{trails}.\\ \mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{selecting}\mathrm{defective}\mathrm{articles}\\ \mathrm{in}\mathrm{a}\mathrm{random}\mathrm{sample}\mathrm{space}\mathrm{of}12\mathrm{articles}\mathrm{.}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=12\mathrm{and}\\ \mathrm{p}=10%=\frac{1}{10}\\ \therefore \mathrm{q}=1-\frac{1}{10}=\frac{9}{10}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=12\mathrm{and}\mathrm{p}=\frac{1}{10}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{12}{\left(\frac{1}{10}\right)}^{\mathrm{x}}{\left(\frac{9}{10}\right)}^{12-\mathrm{x}}\\ \mathrm{P}\left(9\mathrm{defective}\mathrm{articles}\right)\\ =\mathrm{P}\left(\mathrm{X}=9\right)\\ =\mathrm{C}_{9}^{12}{\left(\frac{1}{10}\right)}^{9}{\left(\frac{9}{10}\right)}^{12-9}\\ =220{\left(\frac{1}{10}\right)}^{9}{\left(\frac{9}{10}\right)}^{3}\\ =220\left(\frac{{9}^{3}}{{10}^{12}}\right)\\ =\frac{22×{9}^{3}}{{10}^{11}}\end{array}$

Q.79

$\begin{array}{l}\text{I}\text{n}\text{}\text{a}\text{}\text{b}\text{o}\text{x}\text{}\text{c}\text{o}\text{n}\text{t}\text{a}\text{i}\text{n}\text{i}\text{n}\text{g}\text{}100\text{}\text{b}\text{u}\text{l}\text{b}\text{s},\text{}10\text{}\text{a}\text{r}\text{e}\text{}\text{d}\text{e}\text{f}\text{e}\text{c}\text{t}\text{i}\text{v}\text{e}.\text{}\text{T}\text{h}\text{e}\text{}\text{p}\text{r}\text{o}\text{b}\text{a}\text{b}\text{i}\text{l}\text{i}\text{t}\text{y}\text{}\text{t}\text{h}\text{a}\text{t}\text{}\text{o}\text{u}\text{t}\text{}\text{o}\text{f}\text{}\text{a}\text{}\text{s}\text{a}\text{m}\text{p}\text{l}\text{e}\text{}\text{o}\text{f}\text{}5\text{}\text{b}\text{u}\text{l}\text{b}\text{s},\text{}\text{n}\text{o}\text{n}\text{e}\text{}\text{i}\text{s}\text{}\text{d}\text{e}\text{f}\text{e}\text{c}\text{t}\text{i}\text{v}\text{e}\\ \text{}\text{i}\text{s}\\ \left(\text{A}\right)1{0}^{–1}\text{}\text{}\left(\text{B}\right){\left(\frac{1}{2}\right)}^{5}\text{}\text{}\left(\text{C}\right){\left(\frac{9}{10}\right)}^{5}\text{}\text{}\left(\text{D}\right)\frac{9}{10}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{selections}\mathrm{of}\mathrm{defective}\mathrm{bulbs}\mathrm{from}\mathrm{a}\mathrm{box}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{defective}\mathrm{bulbs}\mathrm{out}\\ \mathrm{of}\mathrm{a}\mathrm{sample}\mathrm{of}5\mathrm{bulbs}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{defective}\mathrm{bulb},\\ \mathrm{p}=\frac{10}{100}=\frac{1}{10}⇒\mathrm{q}=1-\frac{1}{10}=\frac{9}{10}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=5\mathrm{and}\mathrm{p}=\frac{1}{10}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{1}{10}\right)}^{\mathrm{x}}{\left(\frac{9}{10}\right)}^{5-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{None}\mathrm{of}\mathrm{the}\mathrm{bulbs}\mathrm{is}\mathrm{defective}\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)\\ ={\left(\frac{9}{10}\right)}^{5}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.80

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{that}\mathrm{}\mathrm{a}\mathrm{}\mathrm{student}\mathrm{}\mathrm{is}\mathrm{}\mathrm{not}\mathrm{}\mathrm{a}\mathrm{}\mathrm{swimmer}\mathrm{}\mathrm{is}\frac{1}{5}.\mathrm{Then}\mathrm{}\mathrm{the}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{that}\mathrm{}\mathrm{out}\mathrm{}\mathrm{of}\mathrm{}\mathrm{five}\mathrm{}\mathrm{students},\mathrm{four}\mathrm{}\mathrm{are}\\ \mathrm{swimmers}\mathrm{}\mathrm{is}\\ \left(\mathrm{A}\right)\mathrm{C}_{4}^{5}{\left(\frac{4}{5}\right)}^{4}\frac{1}{5}\mathrm{}\left(\mathrm{B}\right){\left(\frac{4}{5}\right)}^{4}\frac{1}{5}\\ \left(\mathrm{C}\right)\mathrm{C}_{4}^{5}\frac{1}{5}{\left(\frac{4}{5}\right)}^{4}\mathrm{}\left(\mathrm{D}\right)\mathrm{None}\mathrm{}\mathrm{of}\mathrm{}\mathrm{these}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{selection}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{are}\mathrm{swimmers}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{students},\mathrm{out}\mathrm{of}5\\ \mathrm{students},\mathrm{who}\mathrm{are}\mathrm{swimmers}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{are}\mathrm{not}\mathrm{swimmers},\\ \mathrm{q}=\frac{1}{5}⇒\mathrm{p}=1-\frac{1}{5}=\frac{4}{5}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=5\mathrm{and}\mathrm{p}=\frac{4}{5}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{4}{5}\right)}^{\mathrm{x}}{\left(\frac{1}{5}\right)}^{5-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{four}\mathrm{students}\mathrm{are}\mathrm{swimmers}\right)\\ =\mathrm{P}\left(\mathrm{X}=4\right)\\ =\mathrm{C}_{4}^{5}{\left(\frac{4}{5}\right)}^{4}{\left(\frac{1}{5}\right)}^{5-4}\\ =\mathrm{C}_{0}^{5}{\left(\frac{1}{10}\right)}^{0}{\left(\frac{9}{10}\right)}^{5-0}\\ =\mathrm{C}_{4}^{5}\left(\frac{1}{5}\right){\left(\frac{4}{5}\right)}^{4}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.81

$\begin{array}{l}\text{A}\text{}\text{a}\text{n}\text{d}\text{}\text{B}\text{}\text{a}\text{r}\text{e}\text{}\text{t}\text{w}\text{o}\text{}\text{e}\text{v}\text{e}\text{nt}\text{}\text{}\text{}\text{s}\text{u}\text{c}\text{h}\text{}\text{t}\text{h}\text{a}\text{t}\text{}\text{P}\text{}\left(\text{A}\right)\text{}\ne \text{}0.\text{}\text{F}\text{i}\text{n}\text{d}\text{}\text{P}\left(\text{B}|\text{A}\right),\text{}\text{i}\text{f}\\ \left(\text{i}\right)\text{}\text{A}\text{}\text{i}\text{s}\text{}\text{a}\text{}\text{s}\text{u}\text{b}\text{s}\text{e}\text{t}\text{}\text{o}\text{f}\text{}\text{B}\text{}\text{}\text{}\text{}\left(\text{i}\text{i}\right)\text{}\text{A}\cap \text{}\text{B}\text{}=\text{}\text{ϕ}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{are}\mathrm{given}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)\ne 0\\ \left(\mathrm{i}\right)\mathrm{A}\mathrm{is}\mathrm{a}\mathrm{subset}\mathrm{of}\mathrm{B}\mathrm{.}\\ ⇒\mathrm{A}\cap \mathrm{B}=\mathrm{A}\\ \therefore \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{B}\cap \mathrm{A}\right)=\mathrm{P}\left(\mathrm{A}\right)\\ \mathrm{ }\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{B}\cap \mathrm{A}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\\ \mathrm{ }=\frac{\mathrm{P}\left(\mathrm{A}\right)}{\mathrm{P}\left(\mathrm{A}\right)}=1\\ \left(\mathrm{ii}\right) \because \mathrm{A}\cap \mathrm{B}=\mathrm{\varphi }\\ ⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=0\\ \mathrm{ }\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{B}\cap \mathrm{A}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\\ \mathrm{ }=\frac{0}{\mathrm{P}\left(\mathrm{A}\right)}=0\end{array}$

Q.82 A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{a}\mathrm{couple}\mathrm{has}\mathrm{two}\mathrm{children},\mathrm{then}\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{is}\\ \mathrm{S}=\left\{\left(\mathrm{b},\mathrm{b}\right), \left(\mathrm{b},\mathrm{g}\right), \left(\mathrm{g},\mathrm{b}\right), \left(\mathrm{g},\mathrm{g}\right)\right\},\mathrm{where}\mathrm{b}=\mathrm{boy}\mathrm{and}\mathrm{g}=\mathrm{girl}\mathrm{.}\\ \left(\mathrm{i}\right)\mathrm{Let}\mathrm{E}=\mathrm{Both}\mathrm{children}\mathrm{are}\mathrm{males}\\ \mathrm{F}=\mathrm{At}\mathrm{least}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{children}\mathrm{is}\mathrm{a}\mathrm{male}\mathrm{.}\\ \therefore \mathrm{E}\cap \mathrm{F}=\left\{\left(\mathrm{b},\mathrm{b}\right)\right\}⇒\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{4}\\ \mathrm{ }\mathrm{P}\left(\mathrm{E}\right)=\frac{1}{4}\mathrm{and}\mathrm{P}\left(\mathrm{F}\right)=\frac{3}{4}\\ \mathrm{P}\left(\mathrm{E}|\mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\left(\frac{1}{4}\right)}{\left(\frac{3}{4}\right)}=\frac{1}{3}\\ \left(\mathrm{ii}\right)\mathrm{Let}\mathrm{A}=\mathrm{Both}\mathrm{children}\mathrm{are}\mathrm{females}\\ \mathrm{and}\mathrm{B}=\mathrm{The}\mathrm{elder}\mathrm{child}\mathrm{is}\mathrm{a}\mathrm{female}\\ \mathrm{ }\mathrm{A}=\left\{\left(\mathrm{g},\mathrm{g}\right)\right\}⇒\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{4}\\ \mathrm{ }\mathrm{B}=\left\{\left(\mathrm{g},\mathrm{b}\right),\left(\mathrm{g},\mathrm{g}\right)\right\}⇒\mathrm{P}\left(\mathrm{B}\right)=\frac{2}{4}=\frac{1}{2}\\ \mathrm{A}\cap \mathrm{B}=\left\{\left(\mathrm{g},\mathrm{g}\right)\right\}⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{1}{4}\\ \therefore \mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \mathrm{ }=\frac{\left(\frac{1}{4}\right)}{\left(\frac{2}{4}\right)}=\frac{1}{2}\end{array}$

Q.83 Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Ans.

$\begin{array}{l}\mathrm{Men}\mathrm{of}\mathrm{grey}\mathrm{hair}\\ =5\mathrm{%}\\ \mathrm{Women}\mathrm{of}\mathrm{grey}\mathrm{hair}\\ =0.25\mathrm{%}\\ \mathrm{Percentage}\mathrm{of}\mathrm{total}\mathrm{people}\mathrm{of}\mathrm{grey}\mathrm{hair}\\ =\left(5+0.25\right)\mathrm{%}\\ =5.25\mathrm{%}\\ \therefore \mathrm{P}\left(\mathrm{Probability}\mathrm{that}\mathrm{the}\mathrm{selected}\mathrm{haired}\mathrm{person}\mathrm{is}\mathrm{a}\mathrm{male}\right)\\ =\frac{5}{5.25}\\ =\frac{500}{525}\\ =\frac{20}{21}\end{array}$

Q.84 Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Ans.

$\begin{array}{l}\mathrm{Repeated}\mathrm{selection}\mathrm{of}\mathrm{people}\mathrm{who}\mathrm{are}\mathrm{right}–\mathrm{handed}\mathrm{or}\\ \mathrm{left}–\mathrm{handed}\mathrm{is}\mathrm{Barnoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{right}\\ \mathrm{handed}\mathrm{people}\mathrm{.}\\ \mathrm{ }\mathrm{p}=\mathrm{P}\left(\mathrm{Right}-\mathrm{handed}\mathrm{people}\right)\\ =\frac{90}{100}=\frac{9}{10}\\ \mathrm{ }\mathrm{q}=\mathrm{P}\left(\mathrm{Left}-\mathrm{handed}\mathrm{people}\right)\\ =1-\frac{9}{10}\\ =\frac{1}{10}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=10\mathrm{and}\mathrm{p}=\frac{9}{10}\\ \mathrm{Therefore},\mathrm{ }\\ \mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{10}{\left(\frac{9}{10}\right)}^{\mathrm{x}}{\left(\frac{1}{10}\right)}^{10-\mathrm{x}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{at}\mathrm{most}6\mathrm{people}\mathrm{are}\mathrm{right}–\mathrm{handed}\\ \mathrm{ }=\mathrm{}1-\mathrm{P}\left(\mathrm{more}\mathrm{than}6\mathrm{are}\mathrm{right}–\mathrm{handed}\right)\\ \mathrm{ }=\mathrm{}1-\left\{\mathrm{P}\left(\mathrm{X}-7\right)+\mathrm{P}\left(\mathrm{X}-8\right)+\mathrm{P}\left(\mathrm{X}-9\right)+\mathrm{P}\left(\mathrm{X}-10\right)\right\}\\ \mathrm{ }=\mathrm{}1-\sum _{\mathrm{r}=7}^{10}\mathrm{C}_{\mathrm{r}}^{10}{\left(\frac{9}{10}\right)}^{\mathrm{r}}{\left(\frac{1}{10}\right)}^{10-\mathrm{r}}\\ \mathrm{ }=\mathrm{}1-\sum _{\mathrm{r}=7}^{10}\mathrm{C}_{\mathrm{r}}^{10}{\left(0.9\right)}^{\mathrm{r}}{\left(0.1\right)}^{10-\mathrm{r}}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}\left(1-\sum _{\mathrm{r}=7}^{10}\mathrm{C}_{\mathrm{r}}^{10}{\left(0.9\right)}^{\mathrm{r}}{\left(0.1\right)}^{10-\mathrm{r}}\right).\end{array}$

Q.85 An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear ‘X’ mark.
(ii) not more than 2 will bear ‘Y’ mark.
(iii) at least one ball will bear ‘Y’ mark.
(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Ans.

$\begin{array}{l} \mathrm{ }\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{balls}=\mathrm{25}\\ \mathrm{Number}\mathrm{of}\mathrm{balls}\mathrm{bear}\mathrm{a}\mathrm{mark}‘\mathrm{X}‘=\mathrm{10}\\ \mathrm{Number}\mathrm{of}\mathrm{balls}\mathrm{bear}\mathrm{a}\mathrm{mark}‘\mathrm{Y}‘=\mathrm{15}\\ \mathrm{Number}\mathrm{of}\mathrm{drawn}\mathrm{balls}=\mathrm{6}\\ \mathrm{ }\mathrm{p}=\mathrm{ }\mathrm{P}\left(\mathrm{Marked}‘\mathrm{Y}‘\right)=\frac{15}{25}=\frac{3}{5}\\ \mathrm{ }\mathrm{q}=\mathrm{ }\mathrm{P}\left(\mathrm{Marked}‘\mathrm{X}‘\right)=\frac{10}{25}=\frac{2}{5}\mathrm{ }\\ \mathrm{Six}\mathrm{balls}\mathrm{are}\mathrm{drawn}\mathrm{with}\mathrm{replacement}.\mathrm{Therefore},\mathrm{the}\mathrm{number}\\ \mathrm{of}\mathrm{trials}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{Let}\mathrm{W}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{that}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{balls}\mathrm{with}‘\mathrm{Y}‘\mathrm{mark}\mathrm{on}\mathrm{them}\mathrm{in}\mathrm{the}\mathrm{trials}\mathrm{.}\\ \mathrm{Clearly},\mathrm{W}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=6\mathrm{and}\mathrm{p}=\frac{3}{5}\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{W}=\mathrm{r}\right)=\mathrm{C}_{\mathrm{r}}^{6}{\mathrm{q}}^{6-\mathrm{r}}{\mathrm{p}}^{\mathrm{r}}\\ =\mathrm{C}_{\mathrm{r}}^{6}{\left(\frac{2}{5}\right)}^{6-\mathrm{r}}{\left(\frac{3}{5}\right)}^{\mathrm{r}}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{all}\mathrm{will}\mathrm{bear}‘\mathrm{X}‘ \mathrm{mark}\right)\\ =\mathrm{P}\left(\mathrm{W}=0\right)\\ =\mathrm{C}_{0}^{6}{\left(\frac{2}{5}\right)}^{6-0}{\left(\frac{3}{5}\right)}^{0}\\ =1.{\left(\frac{2}{5}\right)}^{6}\\ ={\left(\frac{2}{5}\right)}^{6}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{not}\mathrm{more}\mathrm{than}2\mathrm{bear}‘\mathrm{Y}‘\mathrm{mark}\right)\\ =\mathrm{P}\left(\mathrm{W}\le 2\right)\\ =\mathrm{P}\left(\mathrm{Z}=0\right)+\mathrm{P}\left(\mathrm{Z}=1\right)+\mathrm{P}\left(\mathrm{Z}=2\right)\\ =\mathrm{C}_{0}^{6}{\left(\frac{2}{5}\right)}^{6-0}{\left(\frac{3}{5}\right)}^{0}+\mathrm{C}_{1}^{6}{\left(\frac{2}{5}\right)}^{6-1}{\left(\frac{3}{5}\right)}^{1}+\mathrm{C}_{2}^{6}{\left(\frac{2}{5}\right)}^{6-2}{\left(\frac{3}{5}\right)}^{2}\\ ={\left(\frac{2}{5}\right)}^{6}+6{\left(\frac{2}{5}\right)}^{5}{\left(\frac{3}{5}\right)}^{1}+15{\left(\frac{2}{5}\right)}^{4}{\left(\frac{3}{5}\right)}^{2}\\ ={\left(\frac{2}{5}\right)}^{4}\left\{{\left(\frac{2}{5}\right)}^{2}+6\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)+15{\left(\frac{3}{5}\right)}^{2}\right\}\\ ={\left(\frac{2}{5}\right)}^{4}\left\{\frac{4}{25}+\frac{36}{25}+\frac{135}{25}\right\}\\ ={\left(\frac{2}{5}\right)}^{4}\left(\frac{4+36+135}{25}\right)\\ ={\left(\frac{2}{5}\right)}^{4}\left(\frac{175}{25}\right)\\ =7{\left(\frac{2}{5}\right)}^{4}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{ball}\mathrm{bears}‘\mathrm{Y}‘\mathrm{mark}\right)\\ =\mathrm{P}\left(\mathrm{W}\ge 1\right)\\ =1-\mathrm{P}\left(\mathrm{W}=0\right)\\ =1-{\left(\frac{2}{5}\right)}^{6}\\ \left(\mathrm{iv}\right)\mathrm{P}\left(\mathrm{equal}\mathrm{number}\mathrm{of}\mathrm{balls}\mathrm{with}‘\mathrm{X}‘\mathrm{mark}\mathrm{and}‘\mathrm{Y}‘\mathrm{mark}\right)\\ =\mathrm{P}\left(\mathrm{Z}=3\right)\\ =\mathrm{C}_{3}^{6}{\left(\frac{2}{5}\right)}^{3}{\left(\frac{3}{5}\right)}^{3}\\ =\frac{6×5×4×3!}{3!3×2×1}×\frac{8}{125}×\frac{27}{125}\\ =4×\frac{8}{25}×\frac{27}{125}\\ =\frac{864}{3125}\end{array}$

Q.86 In a hurdle race, a player has to cross 10 hurdles. The probability that he willclear each hurdle is 5/6 . What is the probability that he will knock down fewer than 2 hurdles?

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{clearing}\mathrm{a}\mathrm{hurdle}\mathrm{is}\frac{5}{6}.\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{not}\mathrm{clearing}\mathrm{hurdle}\mathrm{is}\frac{1}{6}.\\ \mathrm{Here},\mathrm{we}\mathrm{consider}\mathrm{knocking}\mathrm{down}\mathrm{a}\mathrm{hurdle}\mathrm{to}\mathrm{be}\mathrm{a}\mathrm{success}\mathrm{.}\\ \mathrm{Therefore},\mathrm{p}=\frac{1}{6},\mathrm{q}=\frac{5}{6}\mathrm{and}\mathrm{n}=10.\\ \mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{that}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{times}\mathrm{the}\mathrm{player}\mathrm{will}\mathrm{knock}\mathrm{down}\mathrm{the}\mathrm{hurdle}\mathrm{.}\\ \mathrm{Therefore},\mathrm{by}\mathrm{binomial}\mathrm{distribution},\mathrm{we}\mathrm{obtain}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}\mathrm{ }{\mathrm{q}}^{\mathrm{n}-\mathrm{x}}\\ =\mathrm{C}_{\mathrm{x}}^{10}\mathrm{ }{\left(\frac{1}{6}\right)}^{\mathrm{x}}\mathrm{ }{\left(\frac{5}{6}\right)}^{10-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{player}\mathrm{knocking}\mathrm{down}\mathrm{less}\mathrm{than}2\mathrm{hurdles}\right)\\ =\mathrm{P}\left(\mathrm{X}<2\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)\\ =\mathrm{C}_{0}^{10}\mathrm{ }{\left(\frac{1}{6}\right)}^{0}\mathrm{ }{\left(\frac{5}{6}\right)}^{10-0}+\mathrm{C}_{1}^{10}\mathrm{ }{\left(\frac{1}{6}\right)}^{1}\mathrm{ }{\left(\frac{5}{6}\right)}^{10-1}\\ ={\left(\frac{5}{6}\right)}^{10}+10\left(\frac{1}{6}\right){\left(\frac{5}{6}\right)}^{9}\\ ={\left(\frac{5}{6}\right)}^{9}\left[\frac{5}{6}+\frac{10}{6}\right]\\ ={\left(\frac{5}{6}\right)}^{9}\left(\frac{15}{6}\right)\\ ={\left(\frac{5}{6}\right)}^{9}\left(\frac{5}{2}\right)\\ =\frac{{5}^{10}}{{6}^{9}×2}\\ =\frac{{5}^{10}}{{3}^{9}×{2}^{10}}\\ \mathrm{Thus},\mathrm{probability}\mathrm{of}\mathrm{player}\mathrm{knocking}\mathrm{down}\mathrm{less}\mathrm{than}2\mathrm{hurdles}\mathrm{is}\frac{{5}^{10}}{{3}^{9}×{2}^{10}}.\end{array}$

Q.87 A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{six}\mathrm{in}\mathrm{a}\mathrm{throw}\mathrm{of}\mathrm{die}=\frac{1}{6}\\ \mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{not}\mathrm{getting}\mathrm{a}\mathrm{six}\mathrm{in}\mathrm{a}\mathrm{throw}\mathrm{of}\mathrm{die}=\frac{5}{6}\\ \mathrm{Therefore},\mathrm{p}=\frac{1}{6},\mathrm{q}=\frac{5}{6}\mathrm{and}\mathrm{n}=5.\\ \mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{that}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{times}\mathrm{the}\mathrm{player}\mathrm{will}\mathrm{knock}\mathrm{down}\mathrm{the}\mathrm{hurdle}\mathrm{.}\\ \mathrm{Therefore},\mathrm{by}\mathrm{binomial}\mathrm{distribution},\mathrm{we}\mathrm{obtain}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}\mathrm{ }{\mathrm{q}}^{\mathrm{n}-\mathrm{x}}\\ =\mathrm{C}_{\mathrm{x}}^{5}\mathrm{ }{\left(\frac{1}{6}\right)}^{\mathrm{x}}\mathrm{ }{\left(\frac{5}{6}\right)}^{5-\mathrm{x}}\\ \mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}2\mathrm{sixes}\mathrm{come}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{five}\mathrm{throws}\mathrm{of}\\ \mathrm{the}\mathrm{die}=\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{C}_{2}^{5}\mathrm{ }{\left(\frac{1}{6}\right)}^{2}\mathrm{ }{\left(\frac{5}{6}\right)}^{5-2}\\ =10\mathrm{ }×\frac{{5}^{3}}{{6}^{5}}\\ \therefore \mathrm{Probability}\mathrm{that}\mathrm{third}\mathrm{six}\mathrm{comes}\mathrm{in}\mathrm{the}\mathrm{sixth}\mathrm{throw}\\ =10\mathrm{ }×\frac{{5}^{3}}{{6}^{5}}×\frac{1}{6}\\ =10\mathrm{ }×\frac{{5}^{3}}{{6}^{6}}\\ =\frac{625}{23328}\\ \mathrm{Therefore}, \mathrm{Probability}\mathrm{that}\mathrm{third}\mathrm{six}\mathrm{comes}\mathrm{in}\mathrm{the}\mathrm{sixth}\mathrm{throw}\mathrm{is} \frac{625}{23328}.\end{array}$

Q.88 If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

Ans.

In a leap year,
There are 366 days = 52 weeks and 2 days.
In 52 weeks, there are 52 Tuesdays.
Therefore, the probability that the leap year will
contain 53 Tuesdays = the probability that the remaining 2 days will be Tuesdays.
The remaining 2 days can be:
{Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday, Sunday and Monday}
Number of possibilities = 7
Favorable possibilities = 2
So, Probability that a leap year will have 53 Tuesdays = (2/7).

Q.89 An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.

Ans.

$\begin{array}{l}\text{Since, the probability of success is twice the probability of failure}\text{.}\\ \text{So, let the probability of failure}=\text{x}\\ \text{probablity of success}=2x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Since,\text{P}\left(Success\right)+P\left(Failure\right)=1\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+x=1\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x=1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{1}{3}\\ \text{So,let the probability of failure}=\frac{1}{3}\text{}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{probablity of success}=\frac{2}{3}\\ \text{Therefore, p}=\frac{2}{3}\text{,}\text{\hspace{0.17em}}\text{q}=\frac{1}{3}\text{and n}=\text{6}\text{.}\\ \text{Let X be the random variable that represents the number of}\\ \text{successes in six trials}\text{.}\\ \text{Therefore, by binomial distribution, we obtain}\\ \text{\hspace{0.17em}}P\left(X=x\right)=C_{n}{}_{x}\text{\hspace{0.17em}}{p}^{x}\text{\hspace{0.17em}}{q}^{n-x}\\ =C_{6}{}_{x}\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{x}\text{\hspace{0.17em}}{\left(\frac{1}{3}\right)}^{6-x}\\ \text{Probability of at least 4 successes}\\ =\text{P(X}\ge \text{4)}\\ =\text{P(X}=\text{4)}+\text{P(X}=\text{5)}+\text{P(X}=\text{6)}\\ =C_{6}{}_{4}\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{4}\text{\hspace{0.17em}}{\left(\frac{1}{3}\right)}^{6-4}+C_{6}{}_{5}\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{5}\text{\hspace{0.17em}}{\left(\frac{1}{3}\right)}^{6-5}\\ +C_{6}{}_{6}\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{6}\text{\hspace{0.17em}}{\left(\frac{1}{3}\right)}^{6-6}\\ =15.\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{4}\text{\hspace{0.17em}}{\left(\frac{1}{3}\right)}^{2}+6.\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{5}\text{\hspace{0.17em}}\left(\frac{1}{3}\right)+1.\text{\hspace{0.17em}}{\left(\frac{2}{3}\right)}^{6}\text{\hspace{0.17em}}{\left(\frac{1}{3}\right)}^{0}\\ ={\left(\frac{2}{3}\right)}^{4}\left(15×\frac{1}{9}+6×\frac{2}{3}×\frac{1}{3}+\frac{4}{9}\right)\\ ={\left(\frac{2}{3}\right)}^{4}\left(\frac{15}{9}+\frac{12}{9}+\frac{4}{9}\right)\\ ={\left(\frac{2}{3}\right)}^{4}\left(\frac{31}{9}\right)\\ Therefore,\text{the required probability is}\frac{31}{9}{\left(\frac{2}{3}\right)}^{4}.\end{array}$

Q.90 How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Ans.

$\begin{array}{l}\mathrm{Here},\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{head}\mathrm{and}\mathrm{tail}\mathrm{is}\frac{1}{2}\mathrm{each}\mathrm{.}\\ \mathrm{Therefore},\mathrm{p}=\frac{1}{2},\mathrm{q}=\frac{1}{2}\mathrm{.}\\ \mathrm{Let}\mathrm{the}\mathrm{man}\mathrm{toss}\mathrm{the}\mathrm{coin}\mathrm{n}\mathrm{times}.\mathrm{The}\mathrm{n}\mathrm{tosses}\mathrm{are}\mathrm{n}\mathrm{Bernoulli}\\ \mathrm{trials}\mathrm{.}\\ \mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{that}\mathrm{represents}\mathrm{the}\mathrm{getting}\mathrm{of}\mathrm{head}\mathrm{.}\\ \mathrm{Therefore},\mathrm{by}\mathrm{binomial}\mathrm{distribution},\mathrm{we}\mathrm{obtain}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}\mathrm{ }{\mathrm{q}}^{\mathrm{n}-\mathrm{x}}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\left(\frac{1}{2}\right)}^{\mathrm{x}}\mathrm{ }{\left(\frac{1}{2}\right)}^{\mathrm{n}-\mathrm{x}}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\left(\frac{1}{2}\right)}^{\mathrm{n}}\\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that},\\ \mathrm{P}\left(\mathrm{getting}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{head}\right)>\frac{90}{100}\\ \mathrm{P}\left(\mathrm{X}\ge 1\right)>\frac{90}{100}\\ 1-\mathrm{P}\left(\mathrm{X}=0\right)>0.9\\ 1-\mathrm{C}_{0}^{\mathrm{n}}\mathrm{ }{\left(\frac{1}{2}\right)}^{\mathrm{n}}>0.9\\ 1.\mathrm{ }{\left(\frac{1}{2}\right)}^{\mathrm{n}}<0.1\\ \frac{1}{{2}^{\mathrm{n}}}<\frac{1}{10}\\ ⇒ {2}^{\mathrm{n}}>10\\ \mathrm{The}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{n}\mathrm{that}\mathrm{satisfies}\mathrm{the}\mathrm{given}\mathrm{inequality}\\ \mathrm{is}4.\mathrm{Thus},\mathrm{the}\mathrm{man}\mathrm{should}\mathrm{toss}\mathrm{the}\mathrm{coin}4\mathrm{or}\mathrm{more}\mathrm{than}4\mathrm{times}\\ \mathrm{i}.\mathrm{e}.,\mathrm{n}\ge 4.\end{array}$

Q.91 In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.

Ans.

$\begin{array}{l}\mathrm{Total}\mathrm{outcome}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=6\\ \mathrm{ }\mathrm{P}\left(\mathrm{outcome}6\right)=\frac{1}{6}\\ \mathrm{ }\mathrm{P}\left(\mathrm{outcome}\mathrm{other}\mathrm{than}6\right)=\frac{5}{6}\\ \mathrm{The}\mathrm{die}\mathrm{is}\mathrm{thrown}\mathrm{three}\mathrm{times}.\mathrm{So},\mathrm{there}\mathrm{are}\mathrm{three}\mathrm{cases}:\\ \left(\mathrm{i}\right)\mathrm{When}6\mathrm{comes}\mathrm{in}\mathrm{first}\mathrm{throw}\mathrm{.}\\ \mathrm{P}\left(\mathrm{getting}6\mathrm{in}\mathrm{first}\mathrm{throw}\right)=\frac{1}{6}\\ \mathrm{The}\mathrm{man}\mathrm{wins}\mathrm{money}=\mathrm{Re}.1\\ \left(\mathrm{ii}\right)\mathrm{When}6\mathrm{comes}\mathrm{in}\mathrm{second}\mathrm{throw}\mathrm{.}\\ \mathrm{P}\left(\mathrm{getting}6\mathrm{in}\mathrm{second}\mathrm{throw}\right)=\frac{5}{6}×\frac{1}{6}\\ \mathrm{ }\mathrm{The}\mathrm{man}\mathrm{wins}\mathrm{money}=\mathrm{Re}\mathrm{.}\left(-\mathrm{1}+\mathrm{1}\right)\\ =0\\ \left(\mathrm{iii}\right)\mathrm{When}6\mathrm{comes}\mathrm{in}\mathrm{third}\mathrm{throw}\mathrm{.}\\ \mathrm{ }\mathrm{P}\left(\mathrm{getting}6\mathrm{in}\mathrm{third}\mathrm{throw}\right)=\frac{5}{6}×\frac{5}{6}×\frac{1}{6}\\ \mathrm{ }\mathrm{The}\mathrm{man}\mathrm{wins}\mathrm{money}=\mathrm{Re}\mathrm{.}\left(-\mathrm{1}-\mathrm{1}+\mathrm{1}\right) =\mathrm{Re}.\mathrm{ }-\mathrm{1}\\ \mathrm{The}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}:\end{array}$

 X (Re.) 1 0 -1 P (X) $\frac{1}{6}$ $\frac{5}{36}$ $\frac{25}{216}$

$\begin{array}{l}\mathrm{The}\mathrm{expected}\mathrm{value}\mathrm{a}\mathrm{man}\mathrm{can}\mathrm{win}\\ =1×\frac{1}{6}+0×\frac{5}{36}-1×\frac{25}{216}\\ =\frac{1}{6}-\frac{25}{216}\\ =\frac{36-25}{216}\mathrm{ }=\mathrm{Rs}. \frac{11}{216}\end{array}$

Q.92 Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

 BOX Marble Colour Red White Black A 1 6 3 B 6 2 2 C 8 1 1 D 0 6 4

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Ans.

$\begin{array}{l}\mathrm{ }\mathrm{Let}{\mathrm{E}}_{\mathrm{1}}=\mathrm{The}\mathrm{marble}\mathrm{is}\mathrm{selected}\mathrm{from}\mathrm{box}\mathrm{A}\\ \mathrm{ }{\mathrm{E}}_{\mathrm{2}}=\mathrm{The}\mathrm{marble}\mathrm{is}\mathrm{selected}\mathrm{from}\mathrm{box}\mathrm{B}\\ \mathrm{and}{\mathrm{E}}_{\mathrm{3}}=\mathrm{The}\mathrm{marble}\mathrm{is}\mathrm{selected}\mathrm{from}\mathrm{box}\mathrm{C}\\ \mathrm{R}=\mathrm{The}\mathrm{event}\mathrm{of}\mathrm{drawing}\mathrm{ball}\mathrm{is}\mathrm{red}\\ \mathrm{P}\left(\mathrm{R}\right)=\frac{15}{40}=\frac{3}{8}\\ \mathrm{P}\left({\mathrm{E}}_{1}\cap \mathrm{R}\right)=\frac{1}{40}\\ \mathrm{P}\left({\mathrm{E}}_{2}\cap \mathrm{R}\right)=\frac{6}{40}\\ \mathrm{P}\left({\mathrm{E}}_{3}\cap \mathrm{R}\right)=\frac{8}{40}\\ \mathrm{Probability}\mathrm{of}\mathrm{drawing}\mathrm{the}\mathrm{red}\mathrm{marble}\mathrm{from}\mathrm{box}\mathrm{A}\mathrm{is}\mathrm{given}\\ \mathrm{by}\mathrm{P}\left({\mathrm{E}}_{\mathrm{1}}|\mathrm{R}\right)\mathrm{.}\\ \mathrm{P}\left(\mathrm{R}|{\mathrm{E}}_{1}\right)=\frac{\mathrm{P}\left({\mathrm{E}}_{1}\cap \mathrm{R}\right)}{\mathrm{P}\left(\mathrm{R}\right)}=\frac{\left(\frac{1}{40}\right)}{\left(\frac{3}{8}\right)}=\frac{1}{15}\\ \mathrm{Probability}\mathrm{of}\mathrm{drawing}\mathrm{the}\mathrm{red}\mathrm{marble}\mathrm{from}\mathrm{box}\mathrm{B}\mathrm{is}\mathrm{given}\\ \mathrm{by}\mathrm{P}\left({\mathrm{E}}_{\mathrm{2}}|\mathrm{R}\right)\mathrm{.}\\ \mathrm{ }\mathrm{P}\left(\mathrm{R}|{\mathrm{E}}_{2}\right)==\frac{\mathrm{P}\left({\mathrm{E}}_{2}\cap \mathrm{R}\right)}{\mathrm{P}\left(\mathrm{R}\right)}=\frac{\left(\frac{6}{40}\right)}{\left(\frac{3}{8}\right)}=\frac{2}{5}\\ \mathrm{Probability}\mathrm{of}\mathrm{drawing}\mathrm{the}\mathrm{red}\mathrm{marble}\mathrm{from}\mathrm{box}\mathrm{B}\mathrm{is}\mathrm{given}\\ \mathrm{by}\mathrm{P}\left({\mathrm{E}}_{\mathrm{3}}|\mathrm{R}\right)\mathrm{.}\\ \mathrm{P}\left(\mathrm{R}|{\mathrm{E}}_{3}\right)=\frac{\mathrm{P}\left({\mathrm{E}}_{3}\cap \mathrm{R}\right)}{\mathrm{P}\left(\mathrm{R}\right)}\\ =\frac{\left(\frac{8}{40}\right)}{\left(\frac{3}{8}\right)}=\frac{8}{15}\end{array}$

Q.93 Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Ans.

$\begin{array}{l}Let\text{the risk of heart attack to the person having different option}\\ \text{to reduce it}\text{.}\\ {E}_{1}=\text{The selected person followed the course of yoga and}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{meditation}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(100-30\right)%=70%\\ {}_{}\end{array}$