# CBSE Class 12 Maths Revision Notes Chapter 13

**Class 12 Mathematics Chapter 13 Notes**

Mathematics plays an essential role in the success of students appearing for the Class 12 boards and other national level competitive exams. To excel in the subject, students must practice regularly and plan their studies very well. Class 12 Mathematics is further divided into six units. Students can score a maximum of 10 marks from the probability unit. The CBSE syllabus of Mathematics is precise and delivers knowledge in an easy manner.

The Class 12 Mathematics Chapter 13 notes reviews all basic formulas introduced in Class 11 and explain the advanced concepts of probability. These revision notes compile all solutions to the exercise questions from the chapter. Students can access these Class 12 Mathematics Chapter 13 notes to strengthen their fundamental concepts and clear the logic of each problem effectively.

Extramarks, an online learning platform, aims to provide a productive learning experience through academic notes such as the Class 12 Mathematics Chapter 13 notes to all students.

** Class 12 Mathematics Chapter 13 Notes: Key Topics**

The following topics are explained in detail in the class 12 Mathematics chapter 13 notes:

- Introduction
- Basic Terminology
- Conditional Probability
- Properties of conditional probability
- Multiplication Theorem on Probability
- Independent Events
- Bayes’ Theorem
- Random Variables and Their Probability Distribution
- Mean and Variance of a Random Variable
- Bernoulli Trials
- Binomial Distribution

## Class 12 Mathematics chapter 13 notes: An Overview

**Introduction**:

In this chapter, the important concept of the conditional probability of an event with respect to the occurrence of another event is explained in detail, which is helpful in understanding the concepts like Bayes’ theorem, multiplication rule and independent events in a sample space. The concept of a random variable, its probability distribution, mean and variance is very well explained in the Class 12 Mathematics Chapter 13 notes. The last section of the chapter discusses the discrete probability distribution based on Bernoulli trials, called Binomial distribution.

### What is Probability?

Probability measures the chance is there for an expected result to occur. We can say that Probability is equal to the number of favourable outcomes divided by the total no. of outcomes.

Therefore, Probability = number of favourable outcomes / total no. of outcomes.

For example: If a basket has balls of three colours: Red, Green, and Blue, the Probability of picking a red ball will be,

Total number of possibilities = 3 (The ball we pick can be Red, Green or blue)

Total number of favourable possibilities to get a red ball = 1

Therefore, the probability of selecting a red ball is P=1/3

With the help of the Class 12 Mathematics Chapter 13 notes, students may learn the most about probability.

**Basic Terminology**

- Experiment:

An experiment is a function that gives well-defined results.

2. Random Experiment:

A random experiment is carried out where the outcome may not be the same, even in identical conditions.

3. Sample space:

The Sample space, denoted by S, is a set of all the possible outcomes of a random experiment.

- Sample point:

Every element included in the sample space is known as the sample point. For example, When a coin is tossed, the sample space is given as S={HEAD, TAIL} and the elements, namely head or tail, are called the sample points.

- Event:

The set of all favourable outcomes is an event. It is a subspace of the sample set. For, e.g., Suppose you toss a coin and are looking for the head, then the head is the favourable outcome.

- Mutually Exclusive Events:

Any two events are called mutually exclusive if there is no common element present between them.

- Equally Likely Events:

It is also known as the relative property of two events. Consider any two events. They are termed to be equally likely if none of the events will occur in preference to the other. For example, in tossing an unbiased coin, both heads and tails are equally likely to occur.

- Independent Event:

For any two events, when the occurrence or non-occurrence of one event does not depend on the occurrence or non-occurrence of another event, they are said to be independent.

- Exhaustive events:

Exhaustive events are a set of events in a sample space (S) such that at least one of them compulsorily occurs.

- Complement of an Event:

If S is a sample space and B is an event in it, then the complement of B is represented as B’ or B where B’= {x: x ∈ S, x ∉ B}.

Refer to the Class 12 Mathematics Chapter 13 notes to understand each terminology clearly.

Remarks:

- If the probability of a given event is One, it does not indicate that it will surely occur.
- Probability just predicts the occurrence of one event in contrast to other given events.
- These predictions are based on data and the method of analysing it.
- Furthermore, if the probability of a given event is zero, it does not indicate that it will never occur.

**Conditional Probability: **

The conditional Probability is termed as the probability that an event B will take place provided that an event A has previously occurred.

This Probability is denoted by P(B|A) and is read as the probability of B given A. If events A and B are independent, the conditional probability is given as the probability of event B, i.e., P. In this case, the probability of event B, i.e., (P(B)) is independent of event A.

If A and B events are not independent, then the probability that both events will occur is given by

P (A & B) = P(A) P(B|A)

Consider a random experiment having two sample spaces, E and S. These two events belong to the same sample space, then the conditional probability of the event E given that event S has occurred, P(E|S) is defined as P(E|S)=P(E∩S)P(S), provided P(S) ≠ 0.

Conditional probability is also defined as,

P(E|S)= Number of events which are favourable to E∩S Number of events which are favourable to S = n(E∩S)n(S)

The concept of conditional probability is very well explained in the Class 12 Mathematics Chapter 13 notes.

### Properties:

If E and F and events associated with sample space S, then the following properties hold:

#### Property 1:

P(S|F) = 1 and P(F|F) = 1

We know that, P(S|F) = P(S∩F)P(F) = P(F)P(F)= 1

and P(F|F) = P(F∩F)P(F) = P(F)P(F)= 1

Therefore, P(S|F) = 1 and P(F|F) = 1

#### Property 2:

Consider two events, M and N, of a sample space S and let F be an event of the same sample space, such that P(F)≠ 0, then P(M∪N)|F) = P(M|F) + P(N|F) – P(M∩N)|F)

If M and N are disjoint then, since P(M∩N)|F) = 0, as a result, P(M∪N)|F) = P(M|F) + P(N|F).

We know that P(M∪N)|F) = P(M∪N)∩F )P(F)

Using distributive law, we get P(M∪N)|F) = P((M∩F) ∪ (N∩F))P(F) = P(M∩F) + P (N∩F) – P (M∩N∩F) P(F)

This can be written as, P(M∪N)|F) = P(M∩F)P(F)+ P (N∩F)P(F)–P (M∩N∩F)P(F)

Therefore, P(M∪N)|F) = P(M|F) + P(N|F) – P(M∩N)|F)

#### Property 3:

P(E’|F) = 1 – P(E|F)

Using Property 1, P(S|F) = 1, we get

P(E E’ |F) = 1 as S = E E’

Since events E and E’ are disjoint, we can write P (E|F) + P(E’|F) = 1

Therefore, P(E’|F) = 1 – P(E|F)

Students should practice the derivations of each property given in the Class 12 Mathematics Chapter 13 notes.

**Multiplication Theorem**:

Consider two events, E and F, in a sample space of an experiment, the intersection of the two events is given as P(E∩F) = P(E) P(F|E), where P(E) ≠ 0

= P(F) P(E|F), where P(F) ≠ 0

The conditional probability of event given event F has occurred, i.e., P(E|F) is given as

P(E|F) = P(E ∩ F)P(F), provided P(F)0.

As a result, P(E ∩ F)= P(E|F). P(F)….. Equation 1

Now, P(F|E) = P(F ∩ E)P(E), provided P(E)0

Since F ∩ E = E ∩ F, we get

P(F|E) = P(E ∩ F)P(E), provided P(E)0

As a result, P(E ∩ F)= P(F|E). P(E)….. Equation 2

Combining equations 1 and 2, we get

P(E∩F) = P(E) P(F|E), where P(E) ≠ 0

= P(F) P(E|F), where P(F) ≠ 0

Consider three events, E, F and, in a sample space, then the intersection of the three events is given as P(E∩F∩G)= P(E) P(F|E) P(G|E∩F)

= P(E) P(F|E) P(G|EF)

NOTE:

The multiplication rule of probability is extended to more than three events as well.

The Class 12 Mathematics Chapter 13 notes include several practice problems on the multiplication theorem.

**Independent Events:**

Consider two events, M and N, in a sample space (S) of a random experiment. If the probability of occurrence of one of the events is not affected or is independent with respect to the occurrence of the other event, then we can say that the two events, M and N, are independent. Therefore, the events E and F will be independent, if it satisfies these two criteria:

(a) P(N | M) = P(N), provided P (M) ≠ 0

(b) P(M | N) = P(M), provided P (N) ≠ 0

Applying the multiplication theorem on Probability, we can say that P( M ∩ N) = P(M) P(N), then M and N are said to be independent.

The Class 12 Mathematics Chapter 13 notes provide a deep understanding of the concept of independent events for students to score high marks in the exams.

Consider three events M, N and O. They are said to be mutually independent if the following conditions are met:

P(M ∩ N) = P(M) P(N)

P(M ∩ O) = P(M) P(O)

P(N ∩ O) = P(N) P(O)

and P(M ∩ N ∩ O) = P(M) P(N) P(O)

Remarks:

- If the two events M and N depend on each other, i.e, if they are not independent, then P(E∩F)≠ P(E). P(F)
- The term “independent” is defined with respect to event probability, and mutually exclusive is defined with respect to the subset of sample space.
- Mutually exclusive events are defined as a situation when two events do not occur at the same time. Independent event takes place when one event is not affected by the occurrence of the other event.
- Mutually exclusive events will never have shared results, whereas independent events might have shared results.
- Two mutually exclusive events cannot be mutually exclusive, and two mutually exclusive occurrences cannot be mutually exclusive if they have non-zero probabilities of occurrence.
- Two experiments are independent for events E and F, where the first experiment is dependent on the second experiment, then the probability that the events E and F occur simultaneously is given only when both the experiments are performed. It is given as the product of P(E) and P(F), calculated as follows: P(E∩F)=P(E). P(F)

NOTE:

If the two events, E and F, are independent, then we can say that

(a) E′ and F are independent

(b) E and F’ are independent

(c) E′ and F′ are independent

Students must consider referring to the Chapter 13 Mathematics Class 12 notes for several practice questions based on the Independent events.

**Bayes’ Theorem**:

John Bayes is a famous mathematician who used conditional Probability to obtain reverse probability. The ‘Bayes theorem’ was named after him in 1763.

Consider E1, E2,…, En as the partitions associated with sample space. Let E1, E2,…, En be the mutually exclusive and exhaustive events in a sample space. Let A be any event having non-zero probability, then

P(Ei|A) = P(Ei) P(A|Ei)P(Ei) P(A|Ei)

Applying conditional probability and multiplication rule, P(Ei|A)= P(A ∩ Ei)P(A) = P(Ei) P(A|Ei)P(A)

Now using the theorem of total probability, we get P(Ei|A) = P(Ei) P(A|Ei)P(Ei) P(A|Ei)

Students can refer to the exercise answers and solutions to various problems based on the Bayes Theorem included in the Class 12 Mathematics Chapter 13 notes.

Explanation:

- Consider two sacs I and II. Let the Sac I contain 22 white and 33 red balls. The Sac II contains 44 white and 55 red balls. One ball must be drawn at random from the sacs.
- The probability of selecting any sac is given as 12, and the probability of drawing a ball of a particular colour from a particular sac.
- If we know the sac from which the ball is drawn, then the probability that the ball drawn will be of a given colour.
- If the colour of the ball pulled is known, we will find the reverse probability of sac II being selected after the occurrence of an event is known to find the probability that the ball drawn is from a certain sac II.

Remark

When Bayes’ theorem is applied, the nomenclature given below is used:

- Hypotheses are occurrences of events such as E1, E2,…, En
- The probability of the hypotheses Ei is given as P(Ei)
- The posteriori probability of a hypothesis Ei is defined as the conditional Probability P(Ei|A).
- The formula for the occurrence of “causes” is also termed as the “causes formula.”
- As the Ei’s are a subset of the sample space S, only one of the Ei‘s takes place (that is, only one of the events Ei has to occur). As a result, with reference to the occurrence of event A, the foregoing formula gives us the occurrence of a specific Ei.

**Partition of a Sample Space:**

The partition of any sample space S is said to be a set of events E1, E2,…, En.

If Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, …, n

E1∪ E2 ∪…∪ En= S and P (Ei) > 0 for all i = 1, 2, ……,n.

If the events E1, E2,…, En are disjoint, exhaustive, and have non-zero probabilities, then they represent a partition of the given sample space S.

Consider any nonempty event E. Let its complement be E′. The event E and its complement will form a partition associated with the sample space S as they satisfy E ∩ E′ = φ and E ∪ E′ = S.

Also, for any two events E and F associated with the sample space S, then set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′} is also said to be the partition of the sample space S.

The Partition of a sample space is an important concept from the exam point of view. Students can use the Class 12 Mathematics Chapter 13 notes to understand this concept clearly.

**A Theorem of Total Probability:**

Consider {E1, E2,…, En} to be the partition of the sample space S. Suppose that each event E1, E2,…, Enhas a non-zero probability of occurrence. If A is an event linked with sample space S, then the probability of A is given as P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + P(E3) P(A|E3) + ……..+ P(En) P(A|En)

= j=1nP(Ej) P(A|Ef)

We know that S = E1∪ E2 ∪…∪ En and Ei ∩ Ej = φ.

For event A, we have, A = A ∩ S = A ∩ (E1∪ E2 ∪…∪ En )

Therefore, A = (A ∩ E1) ∪ (A ∩ E2) ∪ (A ∩ E3) ∪……. ∪ (A ∩ En)

Thus, P(A) = P[(A ∩ E1) ∪ (A ∩ E2) ∪ (A ∩ E3) ∪……. ∪ (A ∩ En)]

= P(A ∩ E1) + P (A ∩ E2) + P (A ∩ E3) ……. + P(A ∩ En)

Applying multiplication rule, we get P(A ∩ E1)= P(E1) P(A|E1)

Therefore, P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + P(E3) P(A|E3) + ……..+ P(En) P(A|En)

= j=1nP(Ej) P(A|Ef)

Many questions based on the theorem of total probability are included in the Class 12 mathematics Chapter 13 notes.

**Random Variable’s Probability Distribution:**

In the following cases, we were interested in numbers associated with the particular outcome.

- When two dice are tossed, we may be interested in the total sum of the numbers on the dice.
- When an unbiased coin is tossed n times, we may be interested in the number of heads or tails obtained.

In these examples, we have a rule which assigns a real number to each outcome of an experiment. This real number may be different for different outcomes. Therefore, it is called a variable. Also, the value of this variable depends upon the outcome of the random experiment and, hence, it is called a random variable, denoted by X.

Therefore, a random variable X is a real number and a real-valued function associated with the same sample space of the random experiment.

#### Probability distribution of any random variable, say X:

Consider a random variable X. The probability distribution of X is a description which offers the values of random variables and their probabilities.

Therefore, a random variable’s probability distribution can be defined as follows:

Let X be x1, x2,…, xnassociated with probabilities P1, P2,…, Pn, respectively, where pi>0, i=1npi=1 and i=1,2,……,n.

The real numbers x1, x2,…, xn give all the possible values of any random variable X, and pi(i=1,2,…,n) is said to be the probability of a given random variable X by taking the value xi, which means P(X= xi) = pi.

All elements included in the sample space are covered for all values of the random variable X. Therefore, the total probability of X in a probability distribution must be equal to one.

At certain points in the sample, space is X=xi is true.

As a result, the probability that the random variable X takes the value xi is never 0, that is, P(X=xi) ≠ 0

Students can gain practice by solving unlimited questions based on this concept. The mean and variance is an important concept from this chapter which is explained in a stepwise manner in the Class 12 Mathematics Chapter 13 notes.

**Mean of the Random Variable: **

Mean is the measure of a central tendency that is used to roughly locate a middle or an average value of the random variable.

Consider a random variable X, whose possible values are x1, x2,…, xn occur with probabilities P1, P2,…, Pn, respectively. The mean of X, denoted by μ, is defined as the number i=1nxiPi i.e. the mean of X is defined as the weighted average of all the possible values of X as each value is weighted by its probability of occurrence. The mean of a random variable X is also termed as the expectation of X and is denoted by E(X).

Therefore, E(X) = μ = i=1nxiPi = x1P1+ x2P2+ x3P3+……..+ xnPn.

In order words, the sum of all the products of possible values (xi) by their respective probabilities(Pi) is the mean or expectation of any random variable X.

**Variance of a Random Variable:**

- The mean of a random variable X provides no information about the variability of the different values of the random variable.
- If the variance is small, the values of the random variable are near to the mean. Also, the means of random variables X and Y having different probability distributions can be equal.
- To get the difference between X from Y, we require something which measures the extent to which the random variable’s values spread out.
- Variance is a measure of central tendency which calculates the spread and scatter in the data.

Consider a random variable X, whose possible values are x1, x2,…, xn occur with probabilities P1, P2,…, Pn, respectively. Let E(X) = μ = i=1nxiPi = x1P1+ x2P2+ x3P3+……..+ xnPn be the mean of X. Then, the variance of X, denoted by Var (X) or x2, is given as

x2= Var (X) = i=1n(xi-μ)2 p(xi)or x2 = E(X-μ)2

As a result, x= i=1n(xi-μ)2 p(xi)is the standard deviation of the random variable X.

Another Formula for Variance:

Var (X) = i=1n(xi-μ)2 p(xi)= i=1n(xi2+2-2xiμ) p(xi)

Therefore, x2= Var (X) =i=1n(xi2) p(xi)+i=1n2p(xi)–i=1n2xiμ p(xi)

Var (X) = i=1n(xi2) p(xi)+2i=1np(xi)-2μi=1nxi p(xi)

We know that, i=1np(xi)= 1 and i=1nxi p(xi)=

Therefore, Var (X) = i=1n(xi2) p(xi)+2-22

= i=1n(xi2) p(xi)–2

Or x2= Var (X) = i=1n(xi2) p(xi)–(i=1nxiPi)2

Or x2= Var (X) = E(X2) – [E(X)]2

Students will get an idea of all relevant formulas, derivations, important questions and concepts included in the Class 12 Mathematics Chapter 13 notes.

**Bernoulli trials:**

Consider any dichotomous experiments. For example, when the unbiased coin is tossed, it shows a ‘head’ or a ‘tail’, the response to a question can be ‘yes’ or ‘no’, etc. In such cases, the outcomes are either a ‘success’ and ‘failure’ or ‘not success’.

Here, the outcome of any trial does not depend on the other trial. In such trials, the probability of success or failure does not change, or we can say that the probability remains constant. The independent trials, which can have only two outcomes, such as ‘success’ or ‘failure’, are termed to be the Bernoulli trials.

The trials of any random experiment are known as Bernoulli trials if they satisfy the parameters listed below: :

(i) The number of trials should be finite.

(ii) The trials must not depend on each other (independent).

(iii) Each trial should have exactly two outcomes: success or failure.

(iv) The probability of success and failure remains constant in each trial.

The probability of success in any Bernoulli trial is denoted by p, and the probability of failure is denoted by q, such that p + q =1.

The Class 12 Mathematics Chapter 13 notes include various practice questions for students to practice.** **

**Binomial Distribution: **

The binomial distribution is the expansion of the Bernoulli trials. The binomial expansion for the probabilities of n number of successes in n-Bernoulli trials.

In the experiment of n-Bernoulli trials, the probabilities of 0, 1, 2,…, n successes are obtained as 1st, 2nd,…,(n + 1)th terms in the expansion of (q + p)n.

We prove this by finding out the probability of x-successes in an experiment of n-Bernoulli trials.

For x successes, we will have (n-x) failures.

The x successes (S) and (n – x) failures (F) are obtained in n!x! (n-x)! ways. The probability of successes and failures is = P(x successes). P(n–x) failures = P(S). P(S)…… P(S) (x times). P(F). P(F)……P(F) (n-x) times

Therefore, probability = pxqn-x.

Thus, the probability in a binomial probability distribution is given as P(x successes) = n!x! (n-x)! pxqn-x, where x = 0, 1, 2, 3, ….., n and q = 1 – p

Therefore, in the binomial expansion (q + p)n, The probability of success is Cxn pxqn-x. The binomial distribution is denoted by B(n, p).

For any value x= r, the probability distribution is given as

P(X=r)= Crn prqn-r

Visit the links provided below and get the key points and summary in the Class 12 Mathematics Chapter 13 notes associated with the CBSE syllabus.

**Class 12 Mathematics Chapter 13 Notes:** **Exercise & Solutions**

The Class 12 Mathematics Chapter 13 notes provide a strong conceptual understanding of the topics included in the NCERT books. It covers all important questions, formulas, and theorems with stepwise and detailed explanations. They also provide unlimited practice questions. The Class 12 Mathematics Chapter 13 notes will help students in the effective preparation for the Board exams as well as the undergraduate entrance exams such as JEE, BITSAT, etc.

Extramarks, an online learning platform, brings the Class 12 Mathematics Chapter 13 notes, prepared by our subject matter experts by analysing several CBSE past years’ question papers to facilitate precise understanding of all complex concepts. These NCERT Solutions include well explained and step-by-step explanations of all the problems given in the CBSE books.

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**Class 12 Mathematics Chapter 13 Notes: Key Features**

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### Preparation Tips For Class 12 Board Exams:

- Refer to NCERT Books: Students are advised to use the NCERT books for preparing for the board exams. The Class 12 Mathematics Chapter 13 notes can also be studied along with the NCERT textbook.
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**CBSE extra questions**given in the Class 12 Mathematics notes Chapter 13 until you get all the answers correct.

**Q.1 ** **Prove that if E and F are independent events, then so are the events E and F’.**

**Ans**

Since E and F are independent, we have

P(E ∩ F) = P(E) . P(F) …(1)

From the venn diagram

E ∩ F and E ∩ F ’ are mutually exclusive events

and also E =(E ∩ F) ∪ (E ∩ F ’).

Therefore P(E) = P(E ∩ F) + P(E ∩ F’)

or P(E ∩ F’) = P(E) – P(E ∩ F)

= P(E) – P(E).P(F) [by (1)]

= P(E) (1– P(F))

= P(E). P(F’)

Hence, E and F’ are independent events.

**Q.2 ** **A spinner has 4 equal sectors coloured yellow, blue, green and red. What is the probability of landing on red after spinning the spinner?**

**Ans**

The possible outcomes of this experiment are yellow, blue, green, and red.

$\mathrm{p}\text{(red)}=\frac{\text{number of ways to land on red}}{\text{total number of colors}}=\frac{1}{4}.$

**Q.3 ** **A single 6-sided dice is rolled. What is the probability of rolling and even number?**

**Ans**

The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

$\mathrm{p}\text{(even)}=\frac{\text{number of ways to roll an even number}}{\text{total number of outcomes}}=\frac{3}{6}=\frac{1}{2}.$

**Q.4 ** **A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble?**

**Ans**

The possible outcomes of this experiment are red, green, blue and yellow marbles

$\text{p(red marbles) \hspace{0.33em}}=\text{\hspace{0.33em}}\frac{\text{number of ways to choose red marbles}}{\text{total number of marbles}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{6}}{\text{22}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{3}}{\text{11}}\text{.}$

**Q.5 ** **If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find P(A ∩ B).**

**Ans**

P(A ∩ B) = P(B | A). P(A) = 0.4 × 0.8 = 0.32

**Q.6 ** **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 dice is thrown, number of observations in the sample space = 6 × 6 = 36

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.

∴ A = {(1, 3) (2, 2) (3, 1)}

B = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

A ∩ B = {(1, 3), (3, 1)}

∴ P(B) = 30/36 = 5/6

and P(A ∩ B) = 2/36 = 1/18

Let P(A | B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different

∴ P(A | B) = P(A ∩ B)/P(B) = 1/15

**Q.7 ** **If P(A) =**

$\frac{3}{5}$

, P(B) =

$\frac{1}{5}$

, find P(A ∩ B), where A and B are independent events.

**Ans**

P(A ∩ B) = P(A). P(B) =

$\frac{3}{5}.\frac{1}{5}=\frac{3}{25}.$

**Q.8 ** **Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?**

**Ans**

Let K denote the event that the card drawn is king and A be the event that the card drawn is an ace

So,

$\text{P(KKA)}=\text{P(K).P(K|K).P(A|KK)=}\frac{4}{52}.\frac{3}{51}.\frac{4}{50}=\frac{2}{5525}.$

**Q.9 ** **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**

$\begin{array}{|ccccc|}\hline \text{Elder child}& \text{Girl}& \text{Girl}& \text{Boy}& \text{Boy}\\ \text{Yonger child}& \text{Girl}& \text{Boy}& \text{Girl}& \text{Boy}\\ \hline\end{array}$

B = youngest child is a girl

Let A = Both the children are girl

C = At least one girl

$\begin{array}{l}\text{P(A}\cap \text{B)}=\frac{1}{4},\text{\hspace{0.33em}\hspace{0.33em}P(A}\cap \text{C)}=\frac{1}{4},\text{\hspace{0.33em}\hspace{0.33em}P(B)}=\frac{2}{4},\text{\hspace{0.33em}\hspace{0.33em}P(C)}=\frac{3}{4}.\\ \text{(i) P(A | B)}=\frac{\text{P(A}\cap \text{B)}}{\text{P(B)}}=\frac{1}{2}\\ \text{(ii) P(A | C)}=\frac{\text{P(A}\cap \text{C)}}{\text{P(C)}}=\frac{1}{3}.\end{array}$

**Q.10 ** **The probability of a shooter hitting a target is 3/4. How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?**

**Ans**

Let the shooter fire n times. Obviously, n fires are n Bernoulli trials.

In each trial,

p = probability of hitting the target = 3/4

and q = probability of not hitting the target = 1 – p = 1 – 3/4 = 1/4.

$\begin{array}{l}{\text{P (X = r) = n}}_{{\mathrm{C}}_{\mathrm{r}}}{\mathrm{p}}^{\mathrm{r}}{\mathrm{q}}^{\mathrm{n}-\mathrm{r}}={\text{n}}_{{\mathrm{C}}_{\mathrm{r}}}{\left(\frac{3}{4}\right)}^{\mathrm{r}}{\left(\frac{1}{4}\right)}^{\mathrm{n}-\mathrm{r}}={\text{n}}_{{\mathrm{C}}_{\mathrm{r}}}\frac{{3}^{\mathrm{r}}}{{4}^{\mathrm{n}}}\\ \text{Now, given that,}\\ \text{P (hitting the target at least once) > 0.99}\\ \text{i.e. P(r}\ge \text{1) > 0.99}\\ \text{Therefore, 1}-\text{P ( r = 0) > 0.99}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}1}-{\text{n}}_{{\mathrm{C}}_{0}}\frac{1}{{4}^{\mathrm{n}}}>0.99\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n}}_{{\mathrm{C}}_{0}}\frac{1}{{4}^{\mathrm{n}}}<0.01\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{1}{{4}^{\mathrm{n}}}<0.01\\ \text{The minimum value of n to satisfy the inequality is 4.}\\ \text{Thus, the shooter must fire 4 times.}\end{array}$

**Q.11 ** **A dice is thrown. If E is the event ‘the number appearing is a multiple of 3’ and F be the event ‘the number appearing is even’ then find whether E and F are independent?**

**Ans**

We know that the sample space is S = {1, 2, 3, 4, 5, 6}

Now E = {3, 6}, F = {2, 4, 6} and E ∩ F = {6}

Then, P(E) = 2/6 = 1/3, P(F)= 3/6 = 1/2 and P(E ∩ F) = 1/6.

Clearly P(E ∩ F) = P(E). P (F)

Hence E and F are independent events.

**Q.12 ** **Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).**

**Ans**

$\begin{array}{l}\text{P(E|F)}=\frac{\text{P(E}\cap \text{F)}}{\text{P(F)}}=\frac{0.2}{0.3}=\frac{2}{3}\\ \text{P(F|E)}=\frac{\text{P(E}\cap \text{F)}}{\text{P(F)}}=\frac{0.2}{0.6}=\frac{1}{3}.\end{array}$

**Q.13 ** **If 2P(A) = P(B) = 5/13 and P(A|B) = 2/5, evaluate P(A ∪ B). **

**Ans**

P(A ∩ B) = A(A|B).P(B) = 2/5 × 5/13 = 2/13.

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

P(A ∪ B) = 5/26 + 5/13 – 2/13 = 11/26.

**Q.14 ** **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**

3 good oranges may be selected in ^{12}C_{3} ways

3 oranges out of 15 may be selected in ^{15}C_{3} ways

P( box is approved) = P(3 good oranges selected) = ^{12}C_{3}/^{15}C_{3 }= 44/91.

**Q.15 ** **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)

(iv) P(neither A nor B)

**Ans**

(i) P(A and B) = P(A ∩ B) = 0.3 × 0.6 = 0.18

(ii) P(A and not B) = P(A ∩ B) = 0.3 – 0.18 = 0.12

(iii) P(A or B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.6 – 0.18 = 0.72

(iv) P(neither A nor B)

$\text{= P(}\overline{\text{A}}\cap \overline{\text{B}}\text{) = P(}\overline{\text{A}}\text{)}\times \text{P(}\overline{\mathrm{B}}\text{)}$

= [1 – P(A)] × [1 – P(B)]

= (1 – 0.3) × (1 – 0.6)

= 0.7 × 0.4 = 0.28.

**Q.16 ** **A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.**

**Ans**

A be the event that the construction job will be completed on time, P(A)= ?.

B be the event that there will be a strike.

We have P(B) = 0.65, P(no strike) = P(B’) = 1 – P(B) = 1 – 0.65 = 0.35

P(A|B) = 0.32, P(A|B’) = 0.80

P(A) = P(B) P(A|B) + P(B’) P(A|B’)

= 0.65 × 0.32 + 0.35 × 0.8

= 0.208 + 0.28 = 0.488

Thus, the probability that the construction job will be completed in time is 0.488.

**Q.17 ** **Bag I contains 3 red and 4 black balls while another Bag II contains 5 red ****and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.**

**Ans**

Let E_{1} = choosing the bag I, E_{2} = choosing the bag II and A = drawing a red ball.

$\begin{array}{l}{\text{P(E}}_{1}\text{)}=\frac{1}{2}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P(A | E}}_{1}\text{)}=\mathrm{P}\text{(drawing a red ball from bag I)}=\frac{3}{7}\\ {\text{P(E}}_{2}\text{)}=\frac{1}{2}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P(A | E}}_{2}\text{)}=\mathrm{P}\text{(drawing a red ball from bag II)}=\frac{5}{11}\\ \begin{array}{l}\text{Using Bayes\u2019s theorem,}\\ {\text{P(E}}_{2}\text{| A)}=\frac{{\text{P(E}}_{2}{\text{)P(A | E}}_{2}\text{)}}{{\text{P(E}}_{1}{\text{)P(A | E}}_{1}\text{)}+{\text{P(E}}_{2}{\text{)P(A | E}}_{2}\text{)}}\end{array}\\ \text{}=\frac{\frac{1}{2}\times {\displaystyle \frac{5}{11}}}{\frac{2}{2}\times {\displaystyle \frac{3}{7}}+\frac{1}{2}\times \frac{5}{11}}\\ \text{}=\frac{35}{68}.\end{array}$

**Q.18 ** **In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?**

**Ans**

Let events B_{1}, B_{2}, B_{3}

B_{1} = the bolt is manufactured by machine A

B_{2 }= the bolt is manufactured by machine B

B_{3} = the bolt is manufactured by machine C

And E = ‘the bolt is defective’.

P(B_{1}) = 0.25 P(E|B_{1}) = 0.05

P(B_{2}) = 0.35 P(E|B_{2}) = 0.04

P(B_{3}) = 0.40 P(E|B_{3}) = 0.02

By Baye’s theorem,

P(B_{1}|E) = 0.05

$\begin{array}{l}{\text{P(B}}_{2}\text{| E)}=\frac{{\text{P(B}}_{2}{\text{)P(E | B}}_{2}\text{)}}{{\text{P(B}}_{1}{\text{)P(E | B}}_{1}\text{)}+{\text{P(B}}_{2}{\text{)P(E | B}}_{2}\text{)}+{\text{P(B}}_{3}{\text{)P(E | B}}_{3}\text{)}}\\ {\text{P(B}}_{2}\text{| E)}=\frac{\text{0.35 \xd7 0.04}}{0.25\times 0.05+0.35\times 0.04+0.40\times 0.02}\\ \text{}=\frac{0.0140}{0.0345}=\frac{28}{69}.\end{array}$

**Q.19 ** **A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.**

**Ans**

Let S_{1} = six occurs

S_{2} = six does not occur.

and E = the man reports that six occurs in the throwing of the die

$\begin{array}{l}{\text{P(S}}_{1}\text{)}=\frac{1}{6}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P(E | S}}_{1}\text{)}=\frac{3}{4}\\ {\text{P(S}}_{2}\text{)}=\frac{5}{6}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P(E | S}}_{2}\text{)}=\frac{1}{4}\\ \text{Using Bayes\u2019s theorem,}\\ {\text{P(S}}_{1}\text{| E)}=\frac{{\text{P(S}}_{1}{\text{)P(E | S}}_{1}\text{)}}{{\text{P(S}}_{1}{\text{)P(E | S}}_{1}\text{)}+{\text{P(S}}_{2}{\text{)P(E | S}}_{2}\text{)}}\\ \text{}=\frac{\frac{1}{6}\times {\displaystyle \frac{3}{4}}}{\frac{1}{6}\times {\displaystyle \frac{3}{4}}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}.\end{array}$

**Q.20 ** **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**

Let E_{1 }= Student knows the answer

E_{2 }= Student guesses the answer

And A = answer is correct

$\begin{array}{l}{\text{P(E}}_{1}\text{)}=\frac{3}{4}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P(A | E}}_{1}\text{)}=1\\ {\text{P(E}}_{2}\text{)}=\frac{1}{4}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P(A | E}}_{2}\text{)}=\frac{1}{4}\\ \text{Thus, by Bayes\u2019s theorem,}\\ {\text{P(E}}_{1}\text{| A)}=\frac{{\text{P(E}}_{1}{\text{)P(A | E}}_{1}\text{)}}{{\text{P(E}}_{1}{\text{)P(A | E}}_{1}\text{)}+{\text{P(E}}_{2}{\text{)P(A | E}}_{2}\text{)}}\\ \text{}=\frac{\frac{3}{4}\times 1}{\frac{3}{4}\times 1+\frac{1}{4}\times \frac{1}{4}}=\frac{12}{13}.\end{array}$

**Q.21 ** **Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces.**

**Ans**

Let X denote the number of aces in two draws.

Then, X = 0, 1, or 2.

Now, P(X = 0) = P(non-ace and non-ace+

= P(non-ace) × P(non-ace)

= 48/52 × 48/52 = 144/169

P(X = 1) = P(ace and non-ace or non-ace and ace)

= P(ace and non-ace) × P(non-ace and ace)

= 48/52 × 4/52 + 4/52 × 48/52 = 24/169

P(X = 2) = P(ace and ace)

= P(ace) × P(ace)

= 4/52 × 4/52 = 1/169.

Thus, the required probability distribution is

$\begin{array}{|cccc|}\hline \text{X}& \text{0}& \text{1}& \text{2}\\ \text{P(X)}& \frac{\text{144}}{\text{169}}& \frac{\text{24}}{\text{169}}& \frac{\text{1}}{\text{169}}\\ \hline\end{array}$

**Q.22 ** **Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings.**

**Ans**

$\begin{array}{l}\text{Here, two cards are drawn.}\\ \text{Let p = probability of getting king and q = probability of getting non-king}\\ \text{p =}\frac{4}{52}=\frac{1}{13},\text{\hspace{0.33em}}\mathrm{then}\text{\hspace{0.33em}}\mathrm{q}=1-\frac{1}{13}=\frac{2}{13}\\ \text{Now, we know that mean}=\mathrm{np}=2\times \frac{1}{13}=\frac{2}{13}\\ \text{and variance = npq = 2}\times \frac{1}{13}\times \frac{12}{13}=\frac{24}{169}\\ \text{Therefore, standard deviation}=\sqrt{\mathrm{npq}}=\sqrt{2\times \frac{1}{13}\times \frac{12}{13}}=\sqrt{\frac{24}{169}}=\frac{2\sqrt{6}}{13}\end{array}$

**Q.23 ** **Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.**

**Ans**

Let X denote the number of defective eggs in the 10 eggs drawn. Since the drawing is done with replacement, the trials are Bernoulli trials. Here n = 10 and p = 10/100 = 1/10 or q =1 – p = 1 – 1/10 = 9/10.

Now,

P(at least one defective egg) = P(X ≥ 1)

= 1 – P(X = 0)

= 1 – ^{10}C_{0}(9/10)^{10}

= 1 – (9/10)^{10}

**Q.24 ** **A and B throw a dice alternatively till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts first.**

**Ans**

Let S denote the success (getting a ‘6’) and F denote the failure (not getting a ‘6’).

Thus,

$\begin{array}{l}\text{P(S) =}\frac{1}{6}\text{, P(F) =}\frac{5}{6}\\ \text{P (A wins in the first throw) = P(S) =}\frac{1}{6}\\ \text{P (A wins in the 3rd throw) = P(FFS) = P(F)P(F)P(S) =}\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\\ \text{P (A wins in the 5th throw) = P(FFFFS) =}{\left(\frac{5}{6}\right)}^{4}\left(\frac{1}{6}\right)\text{and so on.}\\ \text{P (A wins) =}\frac{1}{6}+{\left(\frac{5}{6}\right)}^{2}\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^{4}\left(\frac{1}{6}\right)+.\dots \\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}\\ \text{P (B wins) =}1-\mathrm{P}\text{(A wins)}=1-\frac{6}{11}=\frac{5}{11}.\end{array}$

**Q.25 ** **In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6 . What is the probability that he will knock down fewer than 2 hurdles?**

**Ans**

There are 10 hurdles in all i.e., n = 10

Probability that the player clear the hurdle = 5/6

Probability that the player knock down the hurdle = 1 – 5/6 = 1/6.

P(player knocking down less than 2 hurdles)

= P(player knocking no hurdle) + P(player knocking 1 hurdle)

$\begin{array}{l}={\left(\frac{5}{6}\right)}^{10}+{10}_{{\mathrm{c}}_{1}}{\left(\frac{5}{6}\right)}^{9}\left(\frac{1}{6}\right)\\ ={\left(\frac{5}{6}\right)}^{9}\left[\frac{5}{6}+10\times \frac{1}{6}\right]={\left(\frac{5}{6}\right)}^{9}\times \frac{15}{6}\\ =\frac{5}{2}{\left(\frac{5}{6}\right)}^{9}\end{array}$

**Q.26 ** **One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be**

**(a) a spade
(b) not a king
(c) a red card
(d) ace of spades **

**(e) a face card**

(f) a queen

(f) a queen

**Ans**

The total number of possible outcomes is 52.

Probability = Number of favourable outcomes/total possible outcomes

(a) No. of spades in a deck of 52 cards = 13

Probability (spade) = 13/52 = 1/4

(b) No. of kings in a deck of 52 cards = 4

Probability (not a king) = (52 – 4)/52 = 48/52 = 12/13.

(c) No. of red cards in a deck of 52 cards = 13 + 13 = 26

Probability (a red card) = 26/52 = 1/2.

(d) No. of ace of spades in a deck of 52 cards = 1

Probability (ace of spades) = 1/52.

(e) No. of face cards in a deck of 52 cards = 12 (4 kings, 4 queens and 4 jacks)

Probability (a face card) = 12/52 = 3/13

(f) No. of queens in a deck of 52 cards = 4

Probability (a queen) = 4/52 =1/13.

**Q.27 ** **Four cards are drawn at random from a pack of 52 playing cards. Find the probability of getting:
(a) all the four cards of the same suit.
(b) two red cards and two black cards.
(c) all cards of the same color.
(d) one card from each suit. **

**Ans**

Four cards can be drawn from a pack of 52 cards in ^{52}C_{4} ways.

(a) There are four suits viz. club, spade, heart and diamond, each of 13 cards. Total number of ways of getting all the four cards of the same suite = ^{13}C_{4} + ^{13}C_{4 }+ ^{13}C_{4} = 4 (^{13}C_{4})

Probability = Number of favourable outcomes/total possible outcomes = 4(^{13}C_{4})/^{52}C_{4 }= 198/20825.

(b) There are 26 red and 26 black cards. Out of 26 red cards, 2 red cards can be drawn in ^{26}C_{2 }ways . Similarly, 2 black cards can be drawn in ^{26}C_{2} ways.

Probability = (^{26}C_{2} × ^{26}C_{2})/^{52}C_{4 }= 4225/10829.

(c) There are 26 red and 26 black cards. Out of 26 red cards, 4 cards can be drawn in ^{26}C_{4 }ways. Similarly, out of 26 black cards, 4 cards can be drawn in ^{26}C_{4 }ways.

Probability = (^{26}C_{4} + ^{26}C_{4})/^{52}C_{4} = 1196/10829.

(d) There are four suits viz. club, spade, heart and club, each of 13 cards. One card from each suit means that there is one diamond, one club, one heart and one spade card.

There are 13 diamond cards out of which 1 can be selected in ^{13}C_{1} ways.

Hence, favourable outcomes = ^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1} = (^{13}C_{1})^{4}

Probability = (^{13}C_{1})^{4}/^{52}C_{4} = 2197/20825.

**Q.28 ** **In a class of 60 students, 32 like Maths, 30 like Biology and 24 like both Maths and Biology. If one of these students is selected at random, find the probability that the selected student **

**(a) likes Maths or Biology**

**(b) likes neither Maths nor Biology**

**(c) likes Maths but not Biology.**

**Ans**

Let the probability that a student likes Maths be denoted by P(M) and Biology be denoted by P(B).It is given that

P(M) = 32/60 = 8/15, P(B) = 30/60 = 1/2, P(M ∩ B) = 24/60 = 2/5

(a) Probability that the selected student likes Maths or Biology

P(M ∪ B) = P(M) + P(B) – P(M ∩ B)

= 8/15 + 1/2 – 2/5 = 19/30.

(b) Probability that the student likes neither Maths nor Biology = 1 – P(M ∪ B)

= 1 – 19/30 = 11/30

(c) Probability that a student likes Maths but not Biology = P(M) – P(M ∩ B)

= 8/15 – 2/5 = 2/15.

**Q.29 ** **P, Q, R shot to hit a target. If P hits it 3 times in 4 trials, Q hits it 2 times in ****3 trials and R hits it 4 times in 5 trials, what is the probability that the **** target is hit by at least two persons?**

**Ans**

Let E_{1}, E_{2} and E_{3 }be independent events that P, Q and R hit the target respectively. Then

P(E_{1}) = 3/4 , P(E_{2}) = 2/3 and P(E_{3}) = 4/5

P(E_{1}^{c}) = 1 – 3/4 = 1/4 , P(E_{2}^{c}) = 1 – 2/3 = 1/3 and P(E_{3}^{c}) = 1 – 4/5 = 1/5 .

The target is hit by at least two persons in the following mutually exclusive ways:

(a) P hits, Q hits and R does not hit i.e., E_{1} ∩ E_{2 }∩ E_{3}^{c}

(b) P hits, Q does not hit and R hits i.e., E_{1} ∩ E_{2}^{c }∩ E_{3}

(c) P does not hit, Q hits and R hits i.e., E_{1}^{c }∩ E_{2 }∩ E_{3}

(d) P hits, Q hits and R hits i.e., E_{1} ∩ E_{2 }∩ E_{3}

Required Probability = P[(a) ∪ (b) ∪ (c) ∪ (d)] = P(a) + P(b) + P(c) + P(d)

= (3/4 × 2/3 × 1/5) + (1/4 × 2/3 × 4/5) + (3/4 × 2/3 × 4/5)

= 25/30 = 5/6.

**Q.30 ** **A group of two persons is to be selected from two boys and two girls. What is ****the probability that the group will have**

**(a) two boys
(b) one boy
(c) only girls **

**Ans**

Total number of persons = 2 + 2 = 4. Two persons can be selected in ^{4}C_{2} ways.

(a) Two boys can be selected in ^{2}C_{2} ways.

Hence, probability = ^{2}C_{2} /^{4}C_{2 }= 1/6.

(b) It means that there is one boy and one girl in the group. One boy out of 2 boys can be selected in ^{2}C_{1} ways. One girl out of 2 girls can be selected in ^{2}C_{1} ways.

Hence, probability = ^{2}C_{1 }× ^{2}C_{1}/^{4}C_{2} = 2 × 2/2 × 3 = 2/3.

(c) Two girls can be selected from 2 girls in ^{2}C_{2} ways.

Hence, probability = ^{2}C_{2}/^{4}C_{2} = 1/6.

**Q.31 ** **Three dice are thrown simultaneously. Find the probability that :**

**(a) all of them show the same face.**

**(b) all show different faces.**

**(c) two of them show the same face.**

**Ans**

The sample space of this experiment contains 6 × 6 × 6 = 216 sample points.

(a) All the dice will show the same face in 6 mutually exclusive ways :

{1, 1, 1},{2, 2, 2}…{6, 6, 6}

Probability = 6/216 = 1/36..

(b) The total number of ways in which all the dice shows different faces is equal to the number of ways of arranging 6 objects by taking 3 at a time

Probability = (^{6}C_{3} × 3!)/216 = 5/9.

(c) Suppose any number from 1 to 6 comes on two dice. This can be done in ^{6}C_{1}ways. Now, the remaining dice shows a different number. This can be done in ^{5}C_{1} ways. Now, the digits on the three dices occur as 2, 2, 4 ; 3, 4, 3; etc. Thus, three digits can be arranged in 3!/2! ways.

Probability = ^{6}C_{3} × ^{5}C_{1 }× (3!/2!)/216 = 5/12.

**Q.32 ** **Given that E and F are events such that P(E) = 0.5, P(F) = 0.4 and P(E ∩ F) = 0.1, find P (E | F) and f (F | E).**

**Ans**

$\begin{array}{l}\text{Since, P(E | F)}=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ \therefore \text{P(E | F)}=\frac{0.1}{0.4}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{1}{4}\\ \text{And P(F | E)}=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{E}\right)}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{0.1}{0.5}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{1}{5}\end{array}$

**Q.33 ** **Compute P (A | B), if P(B) = 0.4 and P(A ∩ B) = 0.22.**

**Ans**

$\begin{array}{l}\text{It is given that P(B) = 0.4 and P(A}\cap \text{B) = 0.22}\\ \text{So, P(A | B)}=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}\\ \therefore \text{P(A | B)}=\frac{0.22}{0.4}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{11}{20}\end{array}$

**Q.34 ** **A coin is tossed three times. Two events E and F are defined as follows:
E: head on third toss
F: heads on first two tosses.
Determine P (E|F)**

**Ans**

If a coin is tossed three times, then the sample space S is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be see that the sample space has 8 elements.

(i)

E = {HHH, HTH, THH, TTH}

F = {HHH, HHT}

∴ E ∩ F = {HHH}

$\begin{array}{l}\text{So, P(E)}=\frac{4}{8}=\frac{1}{2},\text{\hspace{0.33em}}\mathrm{P}\text{(F)}=\frac{2}{8}=\frac{1}{4}\text{and P(E}\cap \text{F)}=\frac{1}{8}\\ \therefore \text{P(E|F)}=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{\left(\frac{1}{8}\right)}{\left(\frac{1}{4}\right)}=\frac{1}{2}\end{array}$

**Q.35 ** **Let a pair of dice numbered 1, 2, 3, 4, 4, 5 be thrown and the random variable X be the sum of the numbers that appear on the two dice. Find the mean or expectation of X.**

**Ans**

The sample space of experiment consists of 36 elementary events in the form of ordered

pairs (x_{i}, y_{i}), where x_{i} = 1, 2, 3, 4, 4, 5 and y_{i} = 1, 2, 3, 4, 4, 5.

The random variable, X = the sum on both dice

= 2, 3, 4, 5, 6, 7, 8, 9 or 10

$\begin{array}{l}\text{Now,}\\ \mathrm{P}\text{(}\mathrm{X}=2\text{)}=\mathrm{P}\left(\left\{\right(1,1\left)\right\}\right)=\frac{1}{36}\\ \mathrm{P}\text{(}\mathrm{X}=3\text{)}=\mathrm{P}\left(\left\{\right(1,1),\text{\hspace{0.33em}}(2,1\left)\right\}\right)=\frac{2}{36}=\frac{1}{18}\\ \mathrm{P}\text{(}\mathrm{X}=4\text{)}=\mathrm{P}\left(\left\{\right(1,3),\text{\hspace{0.33em}}(2,2),\text{\hspace{0.33em}}(3,1\left)\right\}\right)=\frac{3}{36}=\frac{1}{12}\\ \mathrm{P}\text{(}\mathrm{X}=5\text{)}=\mathrm{P}\left(\left\{\right(1,4),\text{\hspace{0.33em}}(1,4),\text{\hspace{0.33em}}(2,3),\text{\hspace{0.33em}}(3,2),\text{\hspace{0.33em}}(4,1),\text{\hspace{0.33em}}(4,1\left)\right\}\right)=\frac{6}{36}=\frac{1}{6}\\ \mathrm{P}\text{(}\mathrm{X}=6\text{)}=\mathrm{P}\left(\left\{\right(1,5),\text{\hspace{0.33em}}(2,4),\text{\hspace{0.33em}}(2,4),\text{\hspace{0.33em}}(3,3),\text{\hspace{0.33em}}(4,2),\text{\hspace{0.33em}}(4,2),\text{\hspace{0.33em}}(5,1\left)\right\}\right)=\frac{7}{36}\\ \mathrm{P}\text{(}\mathrm{X}=7\text{)}=\mathrm{P}\left(\left\{\right(2,5),\text{\hspace{0.33em}}(3,4),\text{\hspace{0.33em}}(3,4),\text{\hspace{0.33em}}(4,3),\text{\hspace{0.33em}}(4,3),\text{\hspace{0.33em}}(5,2\left)\right\}\right)=\frac{6}{36}=\frac{1}{6}\\ \mathrm{P}\text{(}\mathrm{X}=8\text{)}=\mathrm{P}\left(\left\{\right(3,5),\text{\hspace{0.33em}}(4,4),\text{\hspace{0.33em}}(4,4),\text{\hspace{0.33em}}(4,4),\text{\hspace{0.33em}}(4,4),\text{\hspace{0.33em}}(5,3\left)\right\}\right)=\frac{6}{36}=\frac{1}{6}\\ \mathrm{P}\text{(}\mathrm{X}=9\text{)}=\mathrm{P}\left(\left\{\right(4,5),\text{\hspace{0.33em}}(4,5),\text{\hspace{0.33em}}(5,4),\text{\hspace{0.33em}}(5,4\left)\right\}\right)=\frac{4}{36}=\frac{1}{9}\\ \mathrm{P}\text{(}\mathrm{X}=10\text{)}=\mathrm{P}\left(\left\{\right(5,5\left)\right\}\right)=\frac{1}{36}\\ \text{The probability distribution of X is}\end{array}$ $\begin{array}{|cccccccccc|}\hline {\text{X or x}}_{\text{i}}& \text{2}& \text{3}& \text{4}& \text{5}& \text{6}& \text{7}& \text{8}& \text{9}& \text{10}\\ {\text{P(X) or P}}_{\text{i}}& \frac{\text{1}}{\text{36}}& \frac{\text{1}}{\text{18}}& \frac{\text{1}}{\text{12}}& \frac{\text{1}}{\text{6}}& \frac{\text{7}}{\text{36}}& \frac{\text{1}}{\text{6}}& \frac{\text{1}}{\text{6}}& \frac{\text{1}}{\text{9}}& \frac{\text{1}}{\text{36}}\\ \hline\end{array}$ $\begin{array}{l}\text{Mean}=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{6}+6\times \frac{7}{36}+7\times \frac{1}{6}+8\times \frac{1}{6}+9\times \frac{1}{9}+10\times \frac{1}{36}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{2+3\times 2+4\times 3+5\times 6+6\times 7+7\times 6+8\times 6+9\times 4+10\times 1}{36}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{2+6+12+30+42+42+48+36+10}{36}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\frac{228}{36}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\text{6.33}\\ \text{Thus, the mean of the sum of the numbers that appear on throwing two dice is 6.33.}\end{array}$

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