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 13.2 Conditional Probability - Questions & Solutions refer to the below link

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

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

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

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. 
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Q.1

Compute P(A|B), if P(B) = 0.5and P(A∩B) =0.32.

Ans.

It is given that P(B) = 0.5 and P(AB) = 0.32 So, P(A|B)=P(AB)P(B)    P(A|B)=0.320.5  =1625

Q.2

If P(A)=0.8,P(B)=0.5and P(B|A)=0.4, find(i)P(AB)(ii)P(A|B)(iii)P(AB)

Ans.

ItisgiventhatP(A)=0.8,P(B)=0.5,andP(B|A)=0.4(i)P(B|A)=0.4P(AB)P(A)=0.4P(AB)0.8=0.4P(AB)=0.32(ii)P(A|B)=P(AB)P(B)  =0.320.5  =0.64(iii)P(AB)=P(A)+P(B)P(AB)  =0.8+0.50.32  =0.98

Q.3

Evaluate P(AB), if 2P( A )=P( B )= 5 13 and P( A|B )= 2 5 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWFfbGaa8NDaiaa=fgacaWFSbGaa8xDaiaa=fgacaWF0bGaa8xzaiaa=bcacaWFqbGaa8hkaiaa=feacqGHQicYcaWFcbGaa8xkaiaa=XcacaWFGaGaa8xAaiaa=zgacaWFGaGaa8Nmaiaa=bfadaqadaqaaiaa=feaaiaawIcacaGLPaaacaWF9aGaa8huamaabmaabaGaa8NqaaGaayjkaiaawMcaaiaa=1dadaWcaaqaaiaa=vdaaeaacaWFXaGaa83maaaacaWFGaGaa8xyaiaa=5gacaWFKbGaa8hiaiaa=bkacaWFqbWaaeWaaeaacaWFbbGaa8hFaiaa=jeaaiaawIcacaGLPaaacaWF9aWaaSaaaeaacaWFYaaabaGaa8xnaaaacaWFUaaaaa@616E@

Ans.

It is given that 2P(A)=P(B)=513P(A)=526,P(B)=513andP(A|B)=25.P(A|B)=P(AB)P(B)      25=P(AB)(513)P(AB)=25×513    =213P(AB)=P(A)+P(B)P(AB)    =526+513213    =1526213    =15426=1126

 

Q.4

If P(A)=611,P (B)=511and (AB =711,find (i)P(AB)(ii) P(A|B) (iii)P(B|A).

Ans.

It is given that P(A)=611, P(B)=511 and P(AB)=711(i)P(AB)=P(A)+P(B)P(AB)      711=611+511P(AB)  =1111P(AB)  P(AB)=1111711=411(ii)Since,  P(A|B)=P(AB)P(B)  =(411)(511)  =45(iii)Since,  P(B|A)=P(AB)P(A)=(411)(611)  =46=23

Q.5 A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(ii) E: at least two heads, F: at most two heads
(iii) E: at most two tails, F: at least one tail
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 seen that the sample space has 8 elements. (i)     E={HHH, HTH, THH, TTH}           F={HHH, HHT} EF={HHH}So, P(E)=48=12,​ P(F)=28=14 and P(EF)=18P(E|F)=P(EF)P(F)=(18)(14)=12(ii))   E={HHH, HHT, HTH, THH} F={HHT, HTH, HTT, THH, THT, TTH, TTT} EF={HHT, HTH, THH}So, ​ P(F)=78 and P(EF)=38 P(E|F)=P(EF)P(F)=(38)(78)=37(iii)      E={HHH, HHT, HTT, HTH, THH, THT, TTH}     F={HHT, HTT, HTH, THH, THT, TTH, TTT}EF={HHT,HTT,HTH,THH,THT,TTH}      P(F)=78 and P(EF)=68So, P(E|F)=P(EF)P(F)    =(68)(78)=67

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

Ans.

If two coins are tossed once, then the sample space S is       S = {HH, HT, TH, TT} (i) E = {HT, TH}       F = {HT, TH} EF={HT, TH}P(F)=24=12 and P(EF)=24=12P(E|F)=P(EF)P(F)=(12)(12)=1(ii)  E={HH},F={TT}and EF=ϕ    P(F)=14,P(EF)=0P(E|F)=P(EF)P(F)      =0(14)=0

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

            E={(1,1,4),(1,2,4),(1,3,4),...  (1,6,4)(2,1,4),(2,2,4),(2,3,4),...(2,6,4)(3,1,4),(3,2,4),(3,3,4),...(3,6,4)(4,1,4),(4,2,4),(4,3,4),...(4,6,4)(5,1,4),(5,2,4),(5,3,4),...(5,6,4)(6,1,4),(6,2,4),(6,3,4),...(6,6,4)}            F={(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}EF={(6,5,4)}So, P(F)=6216 and P(EF)=1216Then,  P(E|F)=P(EF)P(F)        =(1216)(6216)=16

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.

If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be S={MFS, MSF, FMS, FSM, SMF, SFM} E={MFS, FMS, SMF, SFM}F={MFS, SFM} So, EF={MFS, SFM}P(F)=26 and P(EF)=26P(E|F)=P(EF)P(F)=(26)(26)=1

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.

Let the first observation be from the black die and second from the red die.When two dice (one black and another red) are rolled,the sample space S has 6 × 6=36 number of elements. (a) Let A: Obtaining a sum greater than 9 ={(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}B: Black die results in a 5. ={(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}         AB={(5, 5), (5, 6)}  P(B)=636 and P(AB)=236 The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).   P(A|B)=P(AB)P(B) =(236)(636) =26=13(b)) E: Sum of the observations is 8. = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} F: Red die resulted in a number less than 4. ={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)(3,1),(3,2),(3,3),(4,1),(4,2),(4,3)(5,1),(5,2),(5,3),(6,1),(6,2),(6,3)}andEF={(5,3),(6,2)}      P(F)=1836 and P(EF)=236The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, P (E|F)=P(EF)P(F)=2361836=19

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.

When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6} then given events are E={1,3,5}, F={2,3} and G={2,3,4,5}So, EF={3}, then P(EF)=16, P(E)=36, P(F)=26 and P(G)=46(i)P(E|F)=P(EF)P(F)=(16)(26)=12    P(F|E)=P(EF)P(E)=(16)(36)=13(ii)          EG={5}              P(EG)=16              P(E|G)=P(EG)P(G)    =(16)(46)=14            P(G|E)=P(EG)P(E)=(16)(36)=13(iii)          (EF)={1,2,3,5}(EF)G={2,3,5}    P((EF)G)=36=12    P{(EF)|G}=P((EF)G)P(G)    =(12)(46)=34(EF)G={3}  P((EF)G)=16P((EF)|G)=P((EF)G)P(G)=(16)(46)=14

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.

Let B and G represent the boy and the girl child respectively. If a family has two children, the sample space will be S = {(B, B), (B, G), (G, B), (G, G)} Let A= both children are girls={(G, G)}(i) Let B be the event that the youngest child is a girl, then   B={(B, G),(G, G)}AB={(G, G)}P(B)=24=12 and P(AB)=14The conditional probability that both are girls, given that the youngest child is a girl=P (A|B)=P(AB)P(B)=(14)(12)=12Thus, the required probability is 12.(ii) Let E be the event that at least one child is a girl.             E={(B,G),(G,B),(G,G)}Then,  AE={(G,G)}P(AE)=14 and P(E)=34The conditional probability that both are girls, given that at least one child is a girl= P(A|E)        =P(AE)P(E)        =(14)(34)=13Thus, the required probability is 13.

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.

So, Probability of easy and multiple choice questionsP(EM)=5001400        =514          Probability of multiple choice question,P(M)=9001400        =914Probability of a randomly selected an easy question, given that it is a multiple choice question=P(E|M)        =P(EM)P(M)        =(514)(914)=59Thus, required probability is 59.

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.

A={(1,3),(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)}Then,AB={(1,3),(3,1)}P(B)=3036=56 and P(AB)=230=118The probability that the sum of the numbers on the dice is 4,given that the two numbers appearing on throwing the two diceare different=P(A|B)=P(AB)P(B)=(118)(56)=115

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.

The sample space of the experiment is given below:S={(1,H),(1,T),(2,H),(2,T),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,H),(4,T),(5,H),(5,T),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),}Let A = the coin shows a tail ={(1,T),(2,T),(4,T),(5,T)}and B=at least one die shows 3={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)}AB=ϕP(AB)=0P(B)=P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}=136+136+136+136+136+136+136=736Probability of the event that the coin shows a tail, given that at least one die shows 3=P(A|B)=P(AB)P(B)=0(736)=0Thus, the required probability is 0.

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.

Given that P(A)=12, P(B)=0, then P(A|B)=P(AB)P(B)      =P(AB)0Therefore, P (A|B) is not defined. Thus, the correct answer is C.

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.

It is given that    P(A|B)=P(B|A)  P(AB)P(B)=P(AB)P(A)          1P(B)=1P(A)          P(A)=P(B)Thus, option D is correct.

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

Ans.

Since, P(AB)=P(A).P(B)  [As A and B are independent events.]    =35.15    =325Thus, P(AB)=325.

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.

Total number of cards=52Number of black cards=26Let A=A black card in first draw        B=A black card in second drawP(A)=2652=12    P(B)=2551Thus, probability of getting both the cards black     =P(A).P(B)    =12×2551    =25102

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.

Let A, B, and C be the respective events that the first, second, and third drawn orange is good.    Probability that first drawn orange is good, P(A)=1215Probability that second drawn orange is good, P(B)=1114Probability that third drawn orange is good, P(C)=1013The box is approved for sale, if all the three oranges are good.  Thus, probability of getting all the oranges good=P(A).P(B).P(C)  =1215×1114×1013  =4491Therefore, the probability that the box is approved for sale is  4491.

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.

Sample space, when a fair coin and an unbiased die are tossed,                  S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}                  A=Head appears on the coin                  ={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)}                  B=3 on the die                  ={(H,3),(T,3)}          AB={(H,3)}    P(AB)=112P(A)×P(B)=612×212    =112Since,    P(AB)=P(A)×P(B)Therefore, A and B are independent events.

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.

Sample space, when an unbiased die are tossed,                  S={1,2,3,4,5,6}                  A=The number is even                  ={2,  4,  6}                  B=The number is red                  ={4,5,6}          AB={4,6}    P(AB)=26=13P(A)×P(B)=36×36    =14Since,    P(AB)P(A)×P(B)Therefore, A and B are not independent events.

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.

E and F be events with P(E)=35, P(F)=310  and P (EF)=1/5P(E)×P(F)=35×310    =950Since,P (EF)P(E)×P(F)Therefore, A and B are not independent events.

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.

Given that the events A and B are such thatP(A)=12, P(AB)=35  and P(B)=p(i) When the events A and B are mutually exclusive i.e., P(AB)=0    P(AB)=P(A)+P(B)P(AB)35=12+p0p=3512    =6510p=110(ii) When the events A and B are independent i.e., P(AB)=P(A).P(B)    P(AB)=P(A)+P(B)P(AB)      P(AB)=P(A)+P(B)P(A).P(B)35=12+p12×p  3512=p(112)  6510=12pp=110×21=15

 

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.

A and B be independent events with P(A)=0.3 and P(B)=0.4(i)  P(AB)=P(A)×P(B)    =0.3×0.4    =0.12(ii)P(AB)=P(A)+P(B)P(A)×P(B)    =0.3+0.40.12    =0.70.12    =0.58(iii)P(A|B)=P(AB)P(B)    =0.120.4    =0.3(iv)P(B|A)=P(AB)P(A)    =0.120.3    =0.4

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.

A and B are two events such that P(A)=14, P(B)=12  and P(AB)=18P(notA and not B)=P(AB)=P(AB)[By De Morgans Law(AB)=(AB)]=1P(AB)=1{P(A)+P(B)P(A)P(B)}=1{14+1214×12}=1(14+1218)=1(2+418)=158=858=38Thus,P(notA and not B)is  38.

 

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.

Events A and B are such that P(A)=12,P(B)=712  and P(not A or not B) =14P(notA or not B)=P(AB)      14=P(AB)[By De Morgans Law(AB)=(AB)]      14=1P(AB)  P(AB)=114  =34P(A).P(B)=12×712  =724So,  P(AB)P(A).P(B)Thus,A and B are independent events.

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.

Given two independent events A and B such that P(A)=0.3, P(B)=0.6(i) P(A and B)=P(AB)  =P(A).P(B)  =0.3×0.6  =0.18(ii)P(A and not B)  =P(AB)  =P(A)P(AB)  =0.30.18  =0.12(iii) P(A or B)=P(AB)  =P(A)+P(B)P(AB)  =0.3+0.60.18  =0.90.18  =0.72(iv)P(neither A nor B)  =P(AB)  =[P(AB)][By De Morgans Law  AB=(AB)]  =1P(AB)  =10.72  =0.28

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

Ans.

Probability of getting an odd number in a throw of a die=36=12Probability of getting an even number in a throw of a die=36=12    Probability of getting an even number three times=12.12.12        =18Therefore, probability of getting an odd number at least once = 1Probability of getting an odd number in none of the throws = 1Probability of getting an even number thrice      =118      =78

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.

Two balls are drawn at random withreplacement from a box containing 10 blackand 8 red balls.Total balls=10+8=18Black balls=10   Red balls=8(i)        Probability of getting a red ball in the first draw=818=49      The ball is replaced after the first draw. Probability of getting second red ball in the first draw=818=49Probability of getting  both the red balls=49×49        =1681(ii)  Probability of getting a black ball in the first draw=1018      The ball is replaced after the first draw. Probability of getting second red ball in the first draw=818Probability of getting first ball as black and second ball as red        =1018×818        =2081(iii)                Probability of getting first ball as red=818=49The ball is replaced after the first draw.      Probability of getting second ball as black=1018=59Probability of getting first ball as red andsecond ball as black=49×59        =2081Therefore, probability of getting first ball as black and second ball as red=49×59        =2081Therefore, probability that one of them is black and other is red      = 2081+2081        =4081

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.

        Probability of solving specific problem by A, P(A)=12        Probability of solving specific problem by B, P(B)=13Probability of not solving specific problem by A, P(A)=112=12Probability of not solving specific problem by B, P(B)=113=23(i)Probability of Problem solved by both independetly,P(AB)=12×13=16Probability that Problem is solved,P(AB)=P(A)+P(B)P(A)P(B)=12+1312×13=12+1316=3+216=46=23(ii) Probability that exactly one of them solves the problem=P(A)P(B)+P(B)P(A)=12×23+13×12=13+16=2+16=36=12

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.

Total cards in a well shuffled deck=52(i) E:the card drawn is a spade           P(E)=452          =113 F:the card drawn is an ace            P(F)=1352=14    P(EF)=152  P(E).P(F)=113×14  =152P(EF)=P(E).P(F)Thus,the events E and F are independent.(ii) E:the card drawn is black           P(E)=1352          =14 F:the card drawn is aking            P(F)=452  =113  P(E).P(F)=14×113  =152P(EF)=P(E).P(F)Thus,the events E and F are independent.(iii) E:the card drawn is a king or queen           P(E)=852=213 F:the card drawn is aqueen or jack            P(F)=852  =213  P(E).P(F)=213×213  =4169P(EF)P(E).P(F)Thus,the events E and F are not independent.

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.

Let H=the students who read Hindi newspaper and         E=the students who read English newspaper. It is given that,         P(H)=60%=60100=0.6        P(E)=40%=40100=0.4P(HE)=20%=20100=0.2(i) The probability that she reads neither Hindi nor English news papers=P(HE)        =1P(HE)        =1{P(H)+P(E)P(HE)}        =1(0.6+0.40.2)        =10.8        =0.2(ii)If she reads Hindi news paper, the probability that she reads English news paper        =P(E|H)        =P(EH)P(H)        =0.20.6=13(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper        =P(H|E)        =P(HE)P(E)        =0.20.4=12

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.

And,P(A)P(B)=m.n0          P(AB)P(A)P(B)So, A and B are not independent events.Therefore, option A is not correct.(B)Two events A and B are said to be independent, if P(AB)=P(A)×P(B) Consider the result given in alternative B.               P(AB)=[1P(A)][1P(B)]            P(AB)=1P(A)P(B)+P(A)P(B)            P(AB)=1P(A)P(B)+P(A)P(B)[By De Morgans Law]1P(AB)=1P(A)P(B)+P(A)P(B)          P(AB)=P(A)+P(B)P(A)P(B)P(A)+P(B)P(AB)=P(A)+P(B)P(A)P(B)            P(AB)=P(A)P(B)This implies that A and B are independent, ifP(AB)=[1P(A)][1P(B)]Therefore, option B is correct.(C)Let A: Event of getting an odd number on throw of a die={1, 3, 5}B: Event of getting an even number on throw of a die={2, 4, 6}P(A)=36=12 and P(B)=36=12So,AB=ϕP(AB)=0  P(A)P(B)=12.12=140P(AB)P(A)P(B)So, A and B are not independent events.Therefore, option C is not correct.(D)If P(A)+P(B)=1,  then, P(AB)P(A)P(B)So, A and B are not independent events.Therefore, option D is not correct.Thus, option B is correct.

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.

    Number of red balls in urn=5Number of black balls in urn=5Let a red ball be drawn in the first attempt.P(red ball)=510=12If two red balls are added to the urn, then    Number of red balls in urn=7     Total balls in urn=12    P(drawing a red ball)=712Let a black ball be drawn in the first attempt.  P (black ball)=510=12Probability of red ball in second attempt,      P(red ball)=712Therefore, probability of drawing second ball as red=12×712+12×512=12(712+512)=12×1=12

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.

    Number of red balls in first bag=4    Number of black balls in first bag=4    Number of red balls in second bag=2Number of black balls in second bag=6Let E1 and E2 be the events of selecting first bag and second bag respectively.P(E1)=12; P(E2)=12Let A be the event of getting a red ball.       P(A|E1)=P(red ball from first bag)=48=12P(A|E2)=P(red ball from second bag)=28=14The probability of drawing a ball from the first bag, given that it is red  =P(E1|A)  =P(E1).P(A|E1)P(E1).P(A|E1)+P(E2).P(A|E2)[By using Bayestheorem]  =12.1212.12+12.14=1414+18=1438=14×83=23Therefore, the probability of drawing a ball from the first bag, given that it is red is 23.

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.

Let E1=The student is a hostler       E2=The student is a day scholar and   A=The chosen student gets grade A. P(E1)=60%=60100=0.6    P(E2)=40%=40100=0.4P(A|E1)=P(student getting an A grade is a hostler)        =30%=0.3P(A|E2)=P(student getting an A grade is a day scholar)        =20%=0.2The probability that a randomly chosen student is a hostler, given that he has an A grade        =P(E1|A)        =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)[By BayesTheorem]        =0.6×0.30.6×0.3+0.4×0.2        =0.180.18+0.08        =0.180.26        =1826        =913Therefore, the required probability is 913.

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.

Let   E1=The student knows the answerand E2=The student guesses the answerLet   A=The answer is correctP(E1)=34    P(E2)=14The probability that the student answered correctly, given that he knows the answer,P(A|E1)=1P(A|E2)=P(student answered correctly, given that he guessed)        =14The probability that the student knows the answer, given that he answered it correctly        =P(E1|A)        =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)[By BayesTheorem]        =34.134.1+14.14        =3434+116        =3412+116=1213Therefore, the required probability is 1213.

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.

Let E1=A person has disease.E2=A person has no disease.Since E1 and E2 are events complimentary to each other, P (E1) + P (E2)=1P (E2)=1P (E1)=10.1%=10.1100=10.001=0.999Let A be the event that the blood test result is positive. P (E1)=0.1%=0.001P(A|E1)=P(result is positive given the person has disease)=99%=0.99P(A|E2)=P(result is positive given the person has no disease)=0.5%=0.005P(E1|A)=Probability that a person has a disease, given that his test result is positive=P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)×P(A|E2)[By Bayes’ Theorem]=0.000990.005985=9905985=1981197Therefore, the required probability is 1981197.

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.

Let   E1=choosing a two headed coin          E2=choosing a biased coin          E3=choosing an unbiased coinP(E1)=13,  P(E2)=13 and P(E3)=13Let   A=the coin shows headsA twoheaded coin will always show heads. P(A|E1)=1  P(A|E2)=75%  =75100=34The third coin is unbiased, the probability that it shows heads  =12P(A|E3)=12The probability that the coin is twoheaded, given that it shows heads,  =P(E1|A)  =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)×P(A|E2)+P(E3)×P(A|E3)  [By BayesTheorem]  =13.113.1+13.34+13.12  =1313+14+16  =134+3+212  =13912=49Thus, the required probability is 49.

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.

Let   E1=The driver is a scooter driver E2=The driver is a  car driverE3=The driver is a truck driver        A=The person meets with an accident. There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. Total number of drivers = 2000 + 4000 + 6000 = 12000      P(E1)=200012000=16      P(E2)=400012000=13      P(E3)=600012000=12P(A|E1)=P(scooter driver met with an accident)        =0.01=1100P(A|E2)=P(car driver met with an accident)=0.03=3100P(A|E3)=P(truck driver met with an accident)=0.15=15100The probability that the driver is a scooter driver, given that he met with an accident, =P(E1|A)=P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)×P(A|E2)+P(E3)×P(A|E3)[By BayesTheorem]=16×110016×1100+13×3100+12×15100=1616+33+152=161+6+456=152Therefore, the required probability is 152.

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.

Let   E1=Items produced by machines A        E2= Items produced by machines B.         A= The Item was found to be defectiveP(E1)=60%=0.6    P(E2)=40%=0.4Probability that machine A produced defective items,P(A|E1)=2%=0.02Probability that machine B produced defective items,P(A|E2)=1%=0.01The probability that the randomly selected item was from machine B, given that it is defective,P(E2|A)=P(E2)P(A|E2)P(E1)P(A|E1)+P(E2)×P(A|E2)  [By BayesTheorem]        =0.4×0.010.6×0.02+0.4×0.01        =0.0040.012+0.004        =0.0040.016        =416=14Therefore,the probability that the randomly selected item was from machine B, given that it is defective, is 14.

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.

Let    E1=The first group win the competition  E2=The second group win the competition        A=The event of introducing a new product        P(E1)=Probability that the first group wins the competition

=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.

  Let E1=The outcome on the die is 5 or 6and E2=The outcome on the die is 1, 2, 3, or 4. P(E1)=26=13  P(E2)=46=23Let   A=The event of getting exactly one head.P (A|E1)=Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6        =38P (A|E2)=Probability of getting exactly one head in a single throw of coin if she gets 1,2, 3, or 4        =12The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head,        =P (E2|A)        =P(E2)P(A|E2)P(E1)P(A|E1)+P(E2)P(A|E2)  [By BayesTheorem]        =23×1238×13+23×12        =1318+13        =133+824        =131124        =13×2411=811Therefore, the required probability is 811.

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.

  Let E1=the time consumed by machines A for the job   E2=the time consumed by machines B for the joband E3=the time consumed by machines C for the job  P(E1)=50%=0.5  P(E2)=30%=0.3  P(E3)=20%=0.2 A= The event of producing defective itemsP(A|E1)=1%=0.01P(A|E2)=5%=0.05P(A|E3)=7%=0.07The probability that the defective item was produced by A,=P (E1|A)=P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)+P(E3)P(A|E3)[By BayesTheorem]=0.5×0.010.5×0.01+0.3×0.05+0.2×0.07=0.0050.005+0.015+0.014=0.0050.034=534Therefore, the required probability is 534.

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.

Let E1=events of choosing a diamond card E2=events of choosing a card which is not diamondand A = the lost cardOut of 52 cards, 13 cards are diamond and 39 cards are not diamond.P(E1)=1352=14  P(E2)=3952=34When one diamond card is lost, there are 12 diamond cards outof 51 cards. Two cards can be drawn out of 12 diamond cards in12C2 ways.Similarly, 2 diamond cards can be drawn out of 51 cards in51C2 ways. The probability of getting two cards, when one diamond card islost=P(A|E1)  =12C251C2=12!2!10!×2!49!51!  =12×111×151×50  =12×1151×50  =661275When the lost card is not a diamond, there are 13 diamond cardsout of 51 cards. Two cards can be drawn out of 13 diamond cards in 13C2 ways whereas 2 cards can be drawn out of 51 cards in51C2 ways.The probability of getting two cards, when one card is lost which is not diamond  =P(A|E2)   =13C251C2  =13!2!×11!×2!×49!51!  =13×121×151×50  =26425The probability that the lost card is diamond  =P (E1|A)  =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)  [By BayesTheorem]  =14×2242514×22425+34×26425  =1150Therefore, the required probability is 1150.

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.

Let E1 and E2 be the events such that E1: A speaks truth E2: A speaks falseLet A be the event that a head appears. P(E1)=45P(E2)=145[P(E2)=1P(E1)]  =15If a coin is tossed, then it may result in either head (H) or tail (T).The probability of getting a head is 12 whether A speaks truthor not.P(A|E1)=12 and P(A|E2)=12The probability that there is actually a head=P(E1|A)  =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)  [By BayesTheorem]  =45×1245×12+15×12=45Therefore, the required probability is 45.Thus, correct option is A.

Q.48

If A and B are two events such that AB and P(B)0,thenwhich of the following is correct?(A)P(A|B)=P(B)P(A) (B)P(A|B)<P(A)(C)P(A|B)P(A) (D)None of these

Ans.

If AB, then AB=A P (AB)=P(A)...(i)(A)P(A|B)=P(AB)P(B)    =P(A)P(B)[From equation(i)]    P(B)P(A)Therefore, option A is not correct.(B)P(A|B)=P(AB)P(B)    =P(A)P(B)[From equation(i)]P(A|B)<P(A)[P(A)<P(B)]Therefore, option B is not correct.(C)P(A|B)=P(AB)P(B)P(A|B)=P(A)P(B)[From equation(i)]P(A|B)P(A)[P(B)1,so1P(B)1]Therefore, option C is correct.

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.

A coin is tossed six times and X represents the difference between the number of heads and the number of tails. X[6H,0T]=X[6 times H and 0 times T]=|60|=1    X[5H,1T]=X[5 times H and 1 times T]=|51|=4    X[4H,2T]=X[4 times H and 2 times T]=|42|=2    X[3H,3T]=X[3 times H and 3 times T]=|33|=0    X[2H,4T]=X[2 times H and 4 times T]=|24|=2    X[1H,5T]=X[1 times H and 5 times T]=|15|=4    X[0H,6T]=X[0 times H and 6 times T]=|06|=6Therefore, the possible values of X are 0, 1, 2, 4 and 6.

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.

(i) When one coin is tossed twice, the sample space is {HH, HT, TH, TT} Let X represent the number of heads. X(HH)=2,X(HT)=1, X(TH)=1, X(TT)=0Therefore, X can take the value of 0, 1, or 2.P(HH)=14,P(TH)=14,P(HT)=14 and P(TT)=14Then,P(X=0)=P(TT)=14P(X=1)=P(HT)+P(TH)        =14+14=24P(X=2)=P(HH)        =14Thus, the required probability distribution is as follows:

X 0 1 2
P(X) 1 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI0aaaaaaa@3A95@ 1 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIYaaaaaaa@3A93@ 1 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI0aaaaaaa@3A95@

(ii)When three coins are tossed simultaneously, the sample space is    {HHH,HHT,HTH,HTT,THT,THH,TTH,TTT}Let X represent the number of tails.It can be seen that X can take the value of 0, 1, 2, or 3.P(X=0)=P(HHH)=18P(X=1)=P(HHT)+P(HTH)+P(THH)        =18+18+18=38P(X=2)=P(HTT)+P(TTH)+P(THT)        =18+18+18=38P(X=3)=P(TTT)        =18Thus, the probability distribution is as follows:

X 0 1 2 3
P(X) 1 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI4aaaaaaa@3A99@ 3 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaI4aaaaaaa@3A9B@ 3 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaI4aaaaaaa@3A9B@ 1 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI4aaaaaaa@3A99@

(iii) When a coin is tossed four times, the sample space isS={HHHH,HHHT,HHTH,HTHH,THHH,HHTT,HTTH,TTHH,HTHT,  THTH,THHT,  HTTT,  THTT,TTHT,TTTH,  TTTT}Let X be the random variable, which represents the number of heads.It can be seen that X can take the value of 0, 1, 2, 3, or 4.P(X=0)=P(HHHH)=116P(X=1)=P(HHHT)+P(HHTH)+P(HTHH)+P(THHH)        =116+116+116+116        =416=14P(X=2)=P(HHTT)+P(HTTH)+P(TTHH)+P(HTHT)+P(THTH)+P(THHT)        =116+116+116+116+116+116        =616=38P(X=3)=P(HTTT)+P(THTT)+P(TTHT)+P(TTTH)        =116+116+116+116        =416=14P(X=4)=P(TTTT)=116

X 0 1 2 3 4
P(X) 1 16 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIXaGaaGOnaaaaaaa@3B52@ 1 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI0aaaaaaa@3A95@ 3 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaI4aaaaaaa@3A9B@ 1 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI0aaaaaaa@3A95@ 1 16 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIXaGaaGOnaaaaaaa@3B52@

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.

When a die is tossed two times, we obtain (6×6) = 36 numberof observations. Let X be the random variable, which represents the number of successes.(i) Here, success refers to the number greater than 4.P (X=0)=P (number less than or equal to 4 on both the tosses) =46×46=49P (X=1)=P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)=46×26+26×46=49P (X=2)=P (number greater than 4 on both the tosses)=26×26=19Thus, the probability distribution is as follows:

X 0 1 2
P(X) 4 9 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaisdaaeaacaaI5aaaaaaa@3A9D@ 4 9 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaisdaaeaacaaI5aaaaaaa@3A9D@ 1 9 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI5aaaaaaa@3A9A@

(ii) Here, success means six appears on at least one die.P (Y=0)=P(six does not appear on any of the dice)=56×56=2536P(Y=1)=P(six appears on at least one of the dice)=1136Thus, the required probability distribution is as follows:

Y 0 1
P(Y) 25 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdacaaI1aaabaGaaG4maiaaiAdaaaaaaa@3C14@ 11 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdacaaIXaaabaGaaG4maiaaiAdaaaaaaa@3C0F@

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.

Since, it is given that out of 30 bulbs, 6 are defective.Number of nondefective bulbs=306=24P(Nondefective bulb)=2430=65  P(Defective bulb)=630=154 bulbs are drawn from the lot with replacement.Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.P(X=0)=P(4 nondefective and 0 defective)        =C04.45.45.45.45=256625P(X=1)=P(3 nondefective and 1 defective)        =C14.(15).(45)3=256625P(X=2)=P(2 nondefective and 2 defective)        =C24.(15)2.(45)2=96625P(X=3)=P(1 nondefective and 3 defective)        =C34.(15)3.(45)=16625P(X=4)=P(0 nondefective and 4 defective)        =C44.(15)4.(45)0=1625Therefore, the required probability distribution is as follows:

X 0 1 2 3 4
P(X) 256 625 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdacaaI1aGaaGOnaaqaaiaaiAdacaaIYaGaaGynaaaaaaa@3D92@ 256 625 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdacaaI1aGaaGOnaaqaaiaaiAdacaaIYaGaaGynaaaaaaa@3D92@ 96 625 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiMdacaaI2aaabaGaaGOnaiaaikdacaaI1aaaaaaa@3CDA@ 16 625 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdacaaI2aaabaGaaGOnaiaaikdacaaI1aaaaaaa@3CD2@ 1 625 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI2aGaaGOmaiaaiwdaaaaaaa@3C12@

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.

Let the probability of getting a tail in the biased coin be x. P( T )=x P( H )=3x Since, P( T )+P( H )=1 x+3x=1 4x=1 x= 1 4 Therefore,P( T )= 1 4 and P( H )= 3 4 When the coin is tossed twice, the sample space is {HH, TT, HT, TH}. Let X be the random variable representing the number of tails. P( X=0 )=P( no tail ) =P( H )×P( H ) = 3 4 × 3 4 = 9 16 P( X=1 )=P( one tail ) =P( HT )+P( TH ) = 3 4 × 1 4 + 1 4 × 3 4 = 6 16 = 3 8 P( X=2 )=P( two tail ) =P( TT ) = 1 4 × 1 4 = 1 16

Therefore, the required probability distribution is as follows:

X 0 1 2
P(X) 9 16 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiMdaaeaacaaIXaGaaGOnaaaaaaa@3B5A@ 3 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaI4aaaaaaa@3A9B@ 1 16 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIXaGaaGOnaaaaaaa@3B52@

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.

(i) Since sum of probabilities of a probability distribution of random variables is one.    0+k+2k+2k+3k+k2+2k2+7k2+k=1        10k2+9k1=0    (10k1)(k+1)=0          k=1,  110k=1 is not possible as the probability of an event is nevernegative.    k=110(ii)P(x<3)=P(X=0)+P(X=1)+P(X=2)    =0+k+2k    =3k    =3×110=310(iii)P(x<6)=P(X=7)    =7k2+k    =7(110)2+110    =7100+110    =7+10100=17100(iv)P(0<x<3)    =P(X=1)+P(X=2)    =k+2k    =3k    =3(110)=310

Q.57

The random variable X has a probability distribution P(X)of the following form,where k is some number:P(X)={k,ifx=02k,ifx=13k,ifx=20,otherwise(a)Determine the value of k.(b)Find P(X<2),P(X2),P(X2)

Ans.

(a) Since, the sum of probabilities of a probability distribution of random variables is one.i.e., k+2k+3k+0=1    6k=1      k=16(b)P(x<2)=P(X=0)+P(X=1)    =k+2k    =3k    =3×16=12      P(x2)=P(X=0)+P(X=1)+P(X=2)    =k+2k+3k    =6k    =6×16=1      P(x2)=P(X=2)+P(X2)    =3k+0    =3k    =3×16    =12

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

Ans.

Let X denotes the success of getting heads.Therefore, the sample space isS = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}It can be seen that X can take the value of 0, 1, 2, or 3.P(X=0)=P(TTT)  =P(T).P(T).P(T)  =12.12.12=18    P(X=1)=P(HHT)+P(HTH)+P(THH)  =P(H).P(H).P(T)+P(H).P(T).P(H)+P(T).P(H).P(H)  =12.12.12+12.12.12+12.12.12  =18+18+18  =38    P(X=2)=P(HHT)+P(HTH)+P(THH)  =12.12.12+12.12.12+12.12.12  =18+18+18  =38    P(X=3)=P(HHH)  =12.12.12  =18Therefore, the required probability distribution is as follows:

X 0 1 2 3
P(X) null 3 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaI4aaaaaaa@3A9B@ 3 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaI4aaaaaaa@3A9B@ 1 8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI4aaaaaaa@3A99@

Mean of X=E(X)  =i=1nxip(xi)  =0×18+1×38+2×38+3×18  =38+68+38  =128=1.5

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

Ans.

Let X represents the number of sixes obtained when two diceare thrown simultaneously.         P(getting 6)=16P(not getting 6)=56Therefore, X can take the value of 0, 1, or 2.P (X=0)=P(not getting six on any of the dice)    =56×56=2536      P(X=1)=P(six on first die and no six on second die)+      P (no six on first die and sixon second die)            =16×56+56×16=1036      P(X=2)=P(six on both dice)            =136Therefore, the required probability distribution is as follows:

X 0 1 2
P(X) 25 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdacaaI1aaabaGaaG4maiaaiAdaaaaaaa@3C14@ 10 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdacaaIWaaabaGaaG4maiaaiAdaaaaaaa@3C0E@ 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIZaGaaGOnaaaaaaa@3B54@

Then, expectation of X            =E(X)            =i=1nxip(xi)            =0×2536+1×1036+2×136            =1236=13

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.

The two positive integers can be selected from the first six positive integers without replacement in 6×5=30 waysX represents the larger of the two numbers obtained.Therefore, X can take the value of 2, 3, 4, 5, or 6.For X=2, the possible observations are (1, 2) and (2, 1).P(X=2)=230=115For X=3, the possible observations are (1,3), (2,3), (3,1) and (3,2).P(X=3)=430=215For X=4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2) and (4, 1).P(X=4)=630=15For X=5,  the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5,2) and (5, 1).P(X=5)=830=415For X=6,the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6),  (6,5), (6, 4), (6,3), (6, 2) and (6, 1).P(X=6)=1030=13Therefore, the probability distribution of X is

X 2 3 4 5 6
P(X) 1 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIXaGaaGynaaaaaaa@3B51@ 2 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdaaeaacaaIXaGaaGynaaaaaaa@3B52@ 1 5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI1aaaaaaa@3A96@ 4 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaisdaaeaacaaIXaGaaGynaaaaaaa@3B54@ 1 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIZaaaaaaa@3A94@

Then, expectation of X            =E(X)            =i=1nxip(xi)            =2×115+3×215+4×15+5×415+6×13            =215+615+45+2015+63            =2+6+12+20+3015            =7015=143

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.

When two fair dice are rolled, Number of observations=6×6=36P(X=2)=P(1,1)        =136P(X=3)=P(1,2)+P(2,1)        =236P(X=4)=P(2,2)+P(3,1)+P(1,3)        =336P(X=5)=P(2,3)+P(3,2)+P(4,1)+P(1,4)        =436P(X=6)=P(2,4)+P(3,3)+P(4,2)+P(5,1)+P(1,5)        =536P(X=7)=P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(1,6)+P(6,1)        =636P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)        =536P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)        =436P(X=10)=P(4,6)+P(5,5)+P(6,4)        =336P(X=11)=P(5,6)+P(6,5)        =236P(X=12)=P(6,6)        =136Thus, the probability distribution of X is

X 2 3 4 5 6 7 8 9 10 11 12
P(X) 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIZaGaaGOnaaaaaaa@3B54@ 2 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdaaeaacaaIZaGaaGOnaaaaaaa@3B55@ 3 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaIZaGaaGOnaaaaaaa@3B56@ 4 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaisdaaeaacaaIZaGaaGOnaaaaaaa@3B57@ 5 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiwdaaeaacaaIZaGaaGOnaaaaaaa@3B58@ 6 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiAdaaeaacaaIZaGaaGOnaaaaaaa@3B59@ 5 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiwdaaeaacaaIZaGaaGOnaaaaaaa@3B58@ 4 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaisdaaeaacaaIZaGaaGOnaaaaaaa@3B57@ 3 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaIZaGaaGOnaaaaaaa@3B56@ 2 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdaaeaacaaIZaGaaGOnaaaaaaa@3B55@ 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIZaGaaGOnaaaaaaa@3B54@

Then, expectation of X            =E(X)            =i=1nxip(xi)            =2×136+3×236+4×336+5×436+6×536+7×636+8×536+9×436+10×336+11×236+12×136            =7AndE(X2)=  22×136+32×236+42×336+52×436+62×536+72×636+82×536+92×436+(10)2×336+11×236+(12)2×136            =3296=54.833Thus,      Var(X)=E(X2)(E(X))2            =54.833(7)2            =54.83349            =5.833Standard deviation        σ=5.833            =2.415

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.

There are 15 students in the class. Each student has the same chance to be chosen.Therefore, the probability of each student to be selected=115The frequency table of the given data is as follows:

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

P( X=14 )= 2 15 P( X=15 )= 1 15 P( X=16 )= 2 15 P( X=17 )= 3 15 P( X=18 )= 1 15 P( X=19 )= 2 15 P( X=20 )= 3 15 P( X=21 )= 1 15 Thus, the probability distribution of X is MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIXaGaaGinaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGOmaaqaaiaaigdacaaI1aaaaaqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIXaGaaGynaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqaaiaaigdacaaI1aaaaaqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIXaGaaGOnaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGOmaaqaaiaaigdacaaI1aaaaaqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIXaGaaG4naaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maaqaaiaaigdacaaI1aaaaaqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIXaGaaGioaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqaaiaaigdacaaI1aaaaaqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIXaGaaGyoaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGOmaaqaaiaaigdacaaI1aaaaaqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIYaGaaGimaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maaqaaiaaigdacaaI1aaaaaqaaiaadcfadaqadaqaaiaadIfacqGH9aqpcaaIYaGaaGymaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqaaiaaigdacaaI1aaaaaqaaiaadsfacaWGObGaamyDaiaadohacaGGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGaaeOCaiaab+gacaqGIbGaaeyyaiaabkgacaqGPbGaaeiBaiaabMgacaqG0bGaaeyEaiaabccacaqGKbGaaeyAaiaabohacaqG0bGaaeOCaiaabMgacaqGIbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae4BaiaabAgacaqGGaGaaeiwaiaabccacaqGPbGaae4Caaaaaa@A675@

X 14 15 16 17 18 19 20 21
P(X) 2 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdaaeaacaaIXaGaaGynaaaaaaa@3B52@ 1 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIXaGaaGynaaaaaaa@3B51@ 2 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdaaeaacaaIXaGaaGynaaaaaaa@3B52@ 3 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaIXaGaaGynaaaaaaa@3B53@ 1 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIXaGaaGynaaaaaaa@3B51@ 2 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdaaeaacaaIXaGaaGynaaaaaaa@3B52@ 3 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiodaaeaacaaIXaGaaGynaaaaaaa@3B53@ 1 15 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIXaGaaGynaaaaaaa@3B51@

Then, mean of X=E( X ) =14× 2 15 +15× 1 15 +16× 2 15 +17× 3 15 +18× 1 15 +19× 2 15 +20× 3 15 +21× 1 15 = 1 15 ( 28+15+32+51+18+38+60+21 ) = 263 15 =17.53 E( X 2 )= 14 2 × 2 15 + 15 2 × 1 15 + 16 2 × 2 15 + 17 2 × 3 15 + 18 2 × 1 15 + 19 2 × 2 15 + 20 2 × 3 15 + 21 2 × 1 15 = 4683 15 =312.2 Variance( X )=E( X 2 ) [ E( X ) ] 2 =312.2 ( 17.53 ) 2 =312.2307.4177 =4.78234.78 Standard deviation = Variance( X ) = 4.78 =2.19 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D834@

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.

P(X=0)=30%=0.3P(X=1)=70%=0.7So, the probability distribution of X is

X 0 1
P(X) 0.3 0.7

Then,E( X )=0×0.3+1×0.7 =0.7 E( X 2 )= 0 2 ×0.3+ 1 2 ×0.7 =0.7 Var( X )=E( X 2 ) [ E( X ) ] 2 =0.7 ( 0.7 ) 2 =0.70.49 =0.21 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C8FF@

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.

Let X be the random variable representing a number on the die.The total number of observations is six.S={1,  1,  1,  2,  2,  5}P(X=1)=36=12P(X=2)=26=13P(X=5)=16So, the probability distribution of X is

X 1 2 5
P (X) 1 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIYaaaaaaa@3A93@ 1 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaIZaaaaaaa@3A94@ 1 6 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI2aaaaaaa@3A97@

E( X )=1× 1 2 +2× 1 3 +5× 1 6 = 3+4+5 6 = 12 6 =2 Therefore, corect option is B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@726D@

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.

Number of cards in deck=52Number of aces=4Let X denote the number of aces obtained. Therefore, X can take any of the values of 0,1, or 2.In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 nonace cards.P(X=0)=P(0 ace card and 2 nonace cards)  =C04×C248C252  =11281326  P(X=1)=P(1 ace card and 1 nonace cards)  =C14×C148C252  =1921326  P(X=2)=P(2 ace card and 0 nonace cards)  =C24×C048C252  =61326Therefore, the probability distribution of X is

X 0 1 2
P(X) 1128 1326 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdacaaIXaGaaGOmaiaaiIdaaeaacaaIXaGaaG4maiaaikdacaaI2aaaaaaa@3F04@ 192 1326 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdacaaI5aGaaGOmaaqaaiaaigdacaaIZaGaaGOmaiaaiAdaaaaaaa@3E4A@ 6 1326 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiAdaaeaacaaIXaGaaG4maiaaikdacaaI2aaaaaaa@3CD0@

E( X )=0× 1128 1326 +1× 192 1326 +2× 6 1326 = 192 1326 + 12 1326 = 204 1326 = 2 13 Therefore, the correct option is D. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A078@

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.

The repeated tosses of a die are Bernoulli trials. Let X denotesthe number of successes of getting odd numbers in anexperiment of 6 trials.Probability of getting an odd number in a single throw of a die is,                p=36=12              q=1p                  =112=12X has a binomial distribution.Therefore, P (X=x)=Cxnpxqnx, where n=0,1,2,3n                  =Cx6(12)x(12)6x                  =Cx6(12)6(i)P(5 success)=P(X=5)                  =C56(12)6                  =6.(12)6                  =664                  =332(ii)P(at least 5 successes)                  =P(X=5)+P(X=6)                  =C56(12)6+C66(12)6                  =6.164+1.164                  =764(iii)P(at most 5 successes)                  =P(X5)                  =1P(X>5)=1P(X=6)                  =1C66(12)6                  =1164                  =6364

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.

Total outcomes of a pair of dice=6×6=36Number of doublets{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}=6The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in anexperiment of throwing two dice simultaneously four times.Probability of getting doublets in a single throw of the pair ofdice is    p=636=16and  q=1p        =116        =56Here, X has the binomial distribution with n=4, p=16 and q=56        P(X=x)=Cxnpxqnx, where n=0,1,2,3n        =Cx4(16)x(56)4xSo,        P(X=2)=C24(16)2(56)42        =4!2!2!×136×2536        =6×136×2536        =25216

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.

Let X denote the number of defective items in a sample of 10 items drawn successively.Since the drawing is done with replacement, the trials are Bernoulli trials.P(defective item),p=5100    =120q=1p    =1120    =1920X has a binomial distribution with n = 10 and p=120    P(X=x)=Cxnpxqnx, where n=0,1,2,3n    =Cx10(120)x(1920)10xP (not more than 1 defective item)    =P(X1)    =P(X=0)+P(X=1)    =C010(120)0(1920)100+C110(120)1(1920)101    =(1920)10+10.(120)(1920)9    =(1920)9(1920+1020)    =(1920)9(2920)Thus, the required probability is (1920)9(2920).

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?
(iii) none is a spade?

Ans.

Let X=The number of spade cards among the five cards drawn Since the drawing of card is with replacement, the trials are Bernoulli trials.In a well shuffled deck of 52 cards, there are 13 spade cards.        p=1352=14and  q=114=34X has a binomial distribution with n = 5 andp=14P(X=x)=Cxnpxqnx, where n=0,1,2,3n=Cx5(14)x(34)5x(i)P(all 5 cards are spades)=C55(14)5(34)55=1.(14)5(34)0=145(ii) P(only 3 cards are spades)=P(X=3)=C35(14)3(34)53=C35(14)3(34)2=10.164.916=45512(iii) P(none is spade)=P(X=0)=C05(14)0(34)50=1.1.2431024=2431024

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.

Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.Given, p=0.05 and    q=10.05=0.95X has a binomial distribution with n = 5 andp=0.05    P(X=x)=Cxnpxqnx, where n=0,1,2,3n    =Cx5(0.05)x(0.95)5x(i)P(none)=P(X=0)    =C05(0.05)0(0.95)50    =1.1.(0.95)5    =(0.95)5(ii)P(not more than one)    =P(X1)    =P(X=0)+P(X=1)    =C05(0.05)0(0.95)50+C15(0.05)1(0.95)51    =1.1.(0.95)5+5(0.05)(0.95)4    =(0.95)4(0.95+0.25)    =(0.95)4×(1.2)(iii)P(more than one)    =P(X>1)    =1P(X1)    =1(0.95)4×(1.2)(iv)P(at least one)    =P(X1)    =1P(X<1)    =1P(X=0)    =1C05(0.05)0(0.95)50    =1(0.95)5

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.

Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.Since the balls are drawn with replacement, the trials are Bernoulli trials.X has a binomial distribution with n=4 and p=110    q=1110=910X has a binomial distribution with n = 4 andp=110    P(X=x)=Cxnpxqnx, where n=0,1,2,3n    =Cx4(110)x(910)4xP(none marked with 0)    =P(X=0)    =C04(110)0(910)40    =(910)4

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.

Let X represent the number of correctly answered questions outof 20 questions.The repeated tosses of a coin are Bernoulli trails. Since head on a coin represents the true answer and tail represents the false answer, the correctly answered questions areBernoulli trials.So, p=12 and q=112=12X has a binomial distribution with n=20 andp=110    P(X=x)=Cxnpxqnx, where n=0,1,2,3n    =Cx20(12)x(12)20x    =Cx20(12)20P (at least 12 questions answered correctly)    =P(X12)    =P(X=12)+P(X=13)+...+P(X=20)    =C1220(12)20+C1320(12)20+...+Cx20(12)20    =(12)20(C1220+C1320+...+C2020)

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.

X is the random variable whose binomial distribution is (6,  12).Therefore, n=6 and p=12      q=112=12Then,P(X=x)=Cxnpxqnx, where n=0,1,2,3n    =Cx6(12)x(12)6x    =Cx6(12)6P(X=x) will be maximum if Cxn is maximum.Then, C06=C66=1,    C16=C56=6    C26=C46=15,  C36=20The value of C36is maximum. Therefore, for x = 3, P(X = x) is maximum.Thus, X = 3 is the most likely outcome.

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.

The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.Probability of getting a correct answer is, p=13q=113=23X has a binomial distribution with n = 5 andp=13Then,P(X=x)=Cxnpxqnx, where n=0,1,2,3n    =Cx5(13)x(23)5xP(guessing more than 4 correct answers)    =P(X4)    =P(X=4)+P(X=5)    =C45(13)4(23)54+C55(13)5(23)55    =5.(181)(23)+1.(1243).1    =10243+1243    =11243

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.

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.So, X has a binomial distribution with n=50 and p=1100q=11100=99100P(X=x)=Cxnpxqnx, where n=0,1,2,3n  =Cx50(1100)x(99100)50x(a)P(winning at least once)  =P(X1)  =1P(X<1)  =1P(X=0)  =1C050(1100)0(99100)500  =1(99100)50(b)P(winning exactly once)  =P(X=1)  =C150(1100)1(99100)501  =50(1100)(99100)49  =12(99100)49(c)P(at least twice)  =P(X2)  =1P(X<2)  =1P(X=1)P(X=0)  =1C150(1100)1(99100)501C050(1100)0(99100)500  =112(99100)49(99100)50  =1(99100)49(12+99100)  =1(99100)49(149100)

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

Ans.

The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.Probability of getting 5 in a single throw of the die, p=16q=116=56Clearly, X has the binomial distribution with n=7 and p=16   Therefore,P(X=x)=Cxnpxqnx, where n=0,1, 2, 3n       =Cx7(16)x(56)7xP(getting 5 exactly twice)       =P(X=2)       =C27(16)2(56)72       =21.(136)(56)5       =(712)(56)5

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

Ans.

The repeated tossing of the die are Bernoulli trials. Let X represent the number of times of getting sixes in 6 throws of the die.Probability of getting six in a single throw of die, p=16q=116=56Clearly, X has a binomial distribution with n=6 and p=16    Therefore,P(X=x)=Cxnpxqnx, where n=0,1,2,3n        =Cx6(16)x(56)6x  P(at most 2 sixes)=P(X2)        =P(X=0)+P(X=1)+P(X=2)        =C06(16)0(56)60+C16(16)1(56)61+C26(16)2(56)62        =(56)6+6.(16)(56)5+15.(136)(56)4        =(56)4(2536+56+1536)        =(56)4(25+30+1536)        =(56)4(7036)        =3518(56)4

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.

The repeated selections of articles in a random sample space are Bernoulli trails. Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.Clearly, X has a binomial distribution with n=12 and p=10%=110q=1110=910Clearly, X has a binomial distribution with n=12 and p=110    Therefore,P(X=x)=Cxnpxqnx, where n=0,1,2,3n        =Cx12(110)x(910)12xP(9 defective articles)        =P(X=9)=C912(110)9(910)129=220(110)9(910)3=220(931012)=22×931011

Q.79

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is ( A )1 0 1 ( B ) ( 1 2 ) 5 ( C ) ( 9 10 ) 5 ( D ) 9 10 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=LeacaWFUbGaa8hiaiaa=fgacaWFGaGaa8Nyaiaa=9gacaWF4bGaa8hiaiaa=ngacaWFVbGaa8NBaiaa=rhacaWFHbGaa8xAaiaa=5gacaWFPbGaa8NBaiaa=DgacaWFGaGaa8xmaiaa=bdacaWFWaGaa8hiaiaa=jgacaWF1bGaa8hBaiaa=jgacaWFZbGaa8hlaiaa=bcacaWFXaGaa8hmaiaa=bcacaWFHbGaa8NCaiaa=vgacaWFGaGaa8hzaiaa=vgacaWFMbGaa8xzaiaa=ngacaWF0bGaa8xAaiaa=zhacaWFLbGaa8Nlaiaa=bcacaWFubGaa8hAaiaa=vgacaWFGaGaa8hCaiaa=jhacaWFVbGaa8Nyaiaa=fgacaWFIbGaa8xAaiaa=XgacaWFPbGaa8hDaiaa=LhacaWFGaGaa8hDaiaa=HgacaWFHbGaa8hDaiaa=bcacaWFVbGaa8xDaiaa=rhacaWFGaGaa83Baiaa=zgacaWFGaGaa8xyaiaa=bcacaWFZbGaa8xyaiaa=1gacaWFWbGaa8hBaiaa=vgacaWFGaGaa83Baiaa=zgacaWFGaGaa8xnaiaa=bcacaWFIbGaa8xDaiaa=XgacaWFIbGaa83Caiaa=XcacaWFGaGaa8NBaiaa=9gacaWFUbGaa8xzaiaa=bcacaWFPbGaa83Caiaa=bcacaWFKbGaa8xzaiaa=zgacaWFLbGaa83yaiaa=rhacaWFPbGaa8NDaiaa=vgaaeaacaWFGaGaa8xAaiaa=nhaaeaadaqadaqaaiaa=feaaiaawIcacaGLPaaacaWFXaGaa8hmamaaCaaaleqabaGaa8xlaiaa=fdaaaGccaWLjaGaaCzcamaabmaabaGaa8NqaaGaayjkaiaawMcaamaabmaabaWaaSaaaeaacaWFXaaabaGaa8NmaaaaaiaawIcacaGLPaaadaahaaWcbeqaaiaa=vdaaaGccaWLjaGaaCzcamaabmaabaGaa83qaaGaayjkaiaawMcaamaabmaabaWaaSaaaeaacaWF5aaabaGaa8xmaiaa=bdaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacaWF1aaaaOGaaCzcaiaaxMaadaqadaqaaiaa=reaaiaawIcacaGLPaaadaWcaaqaaiaa=LdaaeaacaWFXaGaa8hmaaaaaaaa@B7C5@

Ans.

The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.Probability of getting a defective bulb,p=10100=110q=1110=910Clearly, X has a binomial distribution with n=5 and p=110    Therefore,P(X=x)=Cxnpxqnx, where n=0,1,2,3n        =Cx5(110)x(910)5xP(None of the bulbs is defective)        =P(X=0)        =(910)5Thus, the correct option is C.

Q.80

The probability that a student is not a swimmer is15.Then the probability that out of five students,four areswimmers is(A)C45(45)415 (B)(45)415(C)C4515(45)4 (D)None of these

Ans.

The repeated selection of students who are swimmers are Bernoulli trials. Let X denote the number of students, out of 5 students, who are swimmers.Probability of students who are not swimmers,q=15p=115=45Clearly, X has a binomial distribution with n=5 and p=45    Therefore,P(X=x)=Cxnpxqnx, where n=0,1,2,3n        =Cx5(45)x(15)5xP (four students are swimmers)         =P(X=4)        =C45(45)4(15)54        =C05(110)0(910)50        =C45(15)(45)4Thus, the correct option is A.

Q.81

A and B are two event such that P (A) 0. Find P(B|A), if (i) A is a subset of B (ii) A B = ϕ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=feacaWFGaGaa8xyaiaa=5gacaWFKbGaa8hiaiaa=jeacaWFGaGaa8xyaiaa=jhacaWFLbGaa8hiaiaa=rhacaWF3bGaa83Baiaa=bcacaWFLbGaa8NDaiaa=vgacaWFUbGaa8hDaiaa=nhacaWFGaGaa83Caiaa=vhacaWFJbGaa8hAaiaa=bcacaWF0bGaa8hAaiaa=fgacaWF0bGaa8hiaiaa=bfacaWFGaGaa8hkaiaa=feacaWFPaGaa8hiaiabgcmtext5kaa=bcacaWFWaGaa8Nlaiaa=bcacaWFgbGaa8xAaiaa=5gacaWFKbGaa8hiaiaa=bfacaWFOaGaa8Nqaiaa=XhacaWFbbGaa8xkaiaa=XcacaWFGaGaa8xAaiaa=zgaaeaacaWFOaGaa8xAaiaa=LcacaWFGaGaa8xqaiaa=bcacaWFPbGaa83Caiaa=bcacaWFHbGaa8hiaiaa=nhacaWF1bGaa8Nyaiaa=nhacaWFLbGaa8hDaiaa=bcacaWFVbGaa8Nzaiaa=bcacaWFcbGaa8hiaiaaxMaacaWLjaGaaCzcaiaa=HcacaWFPbGaa8xAaiaa=LcacaWFGaGaa8xqaiabgmtextihlaa=bcacaWFcbGaa8hiaiaa=1dacaWFGaGaeqy1dygaaaa@8911@

Ans.

We are given that P(A)0(i)A is a subset of B.AB=AP(AB)=P(BA)=P(A)    P(B|A)=P(BA)P(A)  =P(A)P(A)=1(ii)    AB=ϕP(AB)=0    P(B|A)=P(BA)P(A)  =0P(A)=0

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.

If a couple has two children, then the sample space isS={(b, b), (b, g), (g, b), (g, g)}, where b=boy and g=girl.(i) Let E=Both children are males            F= At least one of the children is a male.EF={(b, b)}P(EF)=14    P(E)=14 and P(F)=34P(E|F)=P(EF)P(F)      =(14)(34)=13(ii)Let A =Both children are females      and B=The elder child is a female    A={(g,g)}P(A)=14    B={(g,b),(g,g)}P(B)=24=12      AB={(g,g)}P(AB)=14P(A|B)=P(AB)P(B)=(14)(24)=12

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.

Men of grey hair  =5%Women of grey hair  =0.25%Percentage of total people of grey hair  =(5+0.25)%  =5.25%P(Probability that the selected haired person is a male)  =55.25  =500525  =2021

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.

Repeated selection of people who are righthanded or lefthanded is Barnoulli trials. Let X denote the number of right handed people.    p=P(Righthanded people)=90100=910    q=P(Lefthanded people)=1910=110Clearly, X has a binomial distribution with n=10 and p=910Therefore,P(X=x)=Cxnpxqnx, where n=0,1,2,3n        =Cx10(910)x(110)10xTherefore, the probability that at most 6 people are righthanded        = 1P(more than 6 are righthanded)        = 1{P(X7)+P(X8)+P(X9)+P(X10)}        = 1r=710Cr10(910)r(110)10r        = 1r=710Cr10(0.9)r(0.1)10rThus, the required probability is (1r=710Cr10(0.9)r(0.1)10r).

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.

  Total number of balls=25Number of balls bear a markX=10Number of balls bear a markY=15Number of drawn balls=6  p=P(Marked Y)=1525=35  q=P(Marked X)=1025=25Six balls are drawn with replacement. Therefore, the numberof trials are Bernoulli trials.Let W be the random variable that represents the number of balls withYmark on them in the trials.Clearly, W has a binomial distribution with n=6 and p=35.P(W=r)=Cr6q6rpr=Cr6(25)6r(35)r(i) P(all will bearX  mark)=P(W=0)=C06(25)60(35)0=1.(25)6=(25)6(ii)P(not more than 2 bearYmark)=P(W2)=P(Z=0)+P(Z=1)+P(Z=2)=C06(25)60(35)0+C16(25)61(35)1+C26(25)62(35)2=(25)6+6(25)5(35)1+15(25)4(35)2=(25)4{(25)2+6(25)(35)+15(35)2}=(25)4{425+3625+13525}=(25)4(4+36+13525)=(25)4(17525)=7(25)4(iii)P(at least one ball bearsYmark)=P(W1)=1P(W=0)=1(25)6(iv) P (equal number of balls withXmark andYmark)=P (Z=3)=C36(25)3(35)3=6×5×4×3!3!3×2×1×8125×27125=4×825×27125=8643125

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.

The probability of clearing a hurdle is 56.Therefore, the probability of not clearing hurdle is 16.Here, we consider knocking down a hurdle to be a success.Therefore, p=16, q=56 and n=10.Let X be the random variable that represents the number of times the player will knock down the hurdle.Therefore, by binomial distribution, we obtainP(X=x)=Cxnpxqnx=Cx10(16)x(56)10xP (player knocking down less than 2 hurdles)=P (X<2)=P(X=0)+P(X=1)=C010(16)0(56)100+C110(16)1(56)101=(56)10+10(16)(56)9=(56)9[56+106]=(56)9(156)=(56)9(52)=51069×2=51039×210Thus, probability of player knocking down less than 2 hurdles is 51039×210.

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.

The probability of getting a six in a throw of die=16The probability of not getting a six in a throw of die=56Therefore, p=16, q=56 and n=5.Let X be the random variable that represents the number of times the player will knock down the hurdle.Therefore, by binomial distribution, we obtainP(X=x)=Cxnpxqnx=Cx5(16)x(56)5xThe probability that the 2 sixes come in the first five throws of the die=P(X=2)=C25(16)2(56)52=10×5365Probability that third six comes in the sixth throw=10×5365×16=10×5366=62523328Therefore,  Probability that third six comes in the sixth throw is  62523328.

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.

Since, the probability of success is twice the probability of failure.So, let the probability of failure=x probablity of success=2xSince, P(Success)+P(Failure)=12x+x=13x=1x=13So,let the probability of failure=13 probablity of success=23Therefore, p=23,q=13 and n=6.Let X be the random variable that represents the number of successes in six trials.Therefore, by binomial distribution, we obtainP(X=x)=Cnxpxqnx=C6x(23)x(13)6xProbability of at least 4 successes=P(X4)=P(X=4)+P(X=5)+P(X=6)=C64(23)4(13)64+C65(23)5(13)65+C66(23)6(13)66=15.(23)4(13)2+6.(23)5(13)+1.(23)6(13)0=(23)4(15×19+6×23×13+49)=(23)4(159+129+49)=(23)4(319)Therefore, the required probability is 319(23)4.

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.

Here, the probability of getting head and tail is 12 each.Therefore, p=12, q=12.Let the man toss the coin n times. The n tosses are n Bernoulli trials.Let X be the random variable that represents the getting of head.Therefore, by binomial distribution, we obtainP(X=x)=Cxnpxqnx        =Cxn(12)x(12)nx        =Cxn(12)nIt is given that,P(getting at least one head)>90100    P(X1)>90100      1P(X=0)>0.9      1C0n(12)n>0.9      1.(12)n<0.1          12n<110    2n>10The minimum value of n that satisfies the given inequality is 4.Thus, the man should toss the coin 4 or more than 4 timesi.e., n4.

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.

Total outcome in a single throw of a die=6      P(outcome 6)=16  P(outcome other than 6)=56The die is thrown three times. So, there are three cases:(i) When 6 comes in first throw.      P(getting 6 in first throw)=16   The man wins money= Re.1(ii) When 6 comes in second throw.P(getting 6 in second throw)=56×16 The man wins money= Re.(1+1)  =0(iii) When 6 comes in third throw.  P(getting 6 in third throw)=56×56×16 The man wins money= Re.(11+1)  =Re.1The required probability distribution is:

X (Re.) 1 0 -1
P (X) 1 6 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaigdaaeaacaaI2aaaaaaa@3A98@ 5 36 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaiwdaaeaacaaIZaGaaGOnaaaaaaa@3B59@ 25 216 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaaikdacaaI1aaabaGaaGOmaiaaigdacaaI2aaaaaaa@3CCF@

The expected value a man can win  =1×16+0×5361×25216  =1625216  =3625216=Rs.  11216

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.

Let E1=The marble is selected from box A   E2=The marble is selected from box Band E3=The marble is selected from box C R= The event of drawing ball is red    P(R)=1540=38P(E1R)=140P(E2R)=640P(E3R)=840Probability of drawing the red marble from box A is given by P(E1|R).P(R|E1)=P(E1R)P(R)=(140)(38)=115Probability of drawing the red marble from box B is given by P(E2|R).P(R|E2)==P(E2R)P(R)=(640)(38)=25Probability of drawing the red marble from box B is given by P(E3|R).P(R|E3)=P(E3R)P(R)=(840)(38)=815

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.

Let the risk of heart attack to the person having different option to reduce it. E 1 =The selected person followed the course of yoga and meditation =( 10030 )%=70% E 2 =The selected person adopted the drug prescription =( 10025 )%=75% A=A person has a heart attack P( A )=40%=0.4 P( E 1 )=P( E 2 )= 1 2 P( A| E 1 )=0.4×0.7=0.28 P( A| E 2 )=0.4×0.75=0.30 Probability that the patient suffering a heart attack followed a course of meditation and yoga = P (E 1 |A) = P( E 1 ) P(E 1 |A) P( E 1 ) P(E 1 |A)+P( E 2 ) P(E 2 |A) = 1 2 ×0.28 1 2 ×0.28+ 1 2 ×0.30 = 0.28 0.28+0.30 = 0.28 0.58 = 14 29 Thus,the probability that the patient suffering a heart attack followed a course of meditation and yoga is 14 29 . 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Q.94 If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability ½)

Ans.

The total number of determinants of second order with each element being 0 or 1 is (2)4= 16The value of determinant is positive in the following cases:|1001|,|1101|,|1011| P(Positive determinant)=316

Q.95 An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0.2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities
(i) P(A fails |B has failed) (ii) P(A fails alone)

Ans.

Given:    P(A fails)=0.2P(B fails alone)=0.15  P(A​​ and B fail)=0.15Let                    E1=A fails                  E2=B failsSo,               P(E1)=0.2            P(E1E2)=0.15P(B fails alone)=P(E2)P(E1E2)  0.15=P(E2)0.15  P(E2)=0.3        (i)P(E1|E2)=P(E1E2)P(E2)=0.150.3=12(ii)P(A fails alone)=P(E1)P(E1E2)=0.20.15=0.05

Q.96 Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Ans.

Total balls in bag I=3+4=7Total balls in bag II=4+5=9LetE1=red ball is transferred from bag I to II      E2=black ball is transferred from bag I to II        A=red ball drawn from bag II          P(E1)=37 and P(E2)=47When red ball is transferred from bag I to II.    P(E1|A)=510=12When black ball is transferred from bag I to II.    P(E2|A)=410=25P(red ball if transferred ball is black)  =P(E2|A)  =P(E2)P(A|E2)P(E1)P(A|E1)+P(E2)P(A|E2)[By BayesTheorem]  =47×2537×12+47×25  =1615+16  =1631

Q.97 If A and B are two events such that P(A) ≠ 0 and P(B|A) = 1, then which of the following is correct?
(A) A ⊂ B (B) B ⊂ A (C) B = Φ (D) A = Φ

Ans.

Given: P(A)0 and P(B|A)=1Since, P(B|A)=P(AB)P(A)1=P(AB)P(A)  P(A)=P(AB)  A=AB  ABThus, the correct answer is A.

Q.98 If P(A|B) > P(A), then which of the following is correct ?

(A) P(B|A) < P(B) (B) P(A∩B) < P(A).P(B)
(C) P(B|A) > P(B) (D) P(B|A) = P(B)

Ans.

Given:P(A|B)>P(A)Since, P(A|B)=P(AB)P(A)>P(A)P(AB)P(B)>P(A)P(AB)>P(A)P(B)P(AB)P(A)>P(B)P(B|A)>P(B)Thus, the correct answer is C.

Q.99 Choose the correct answer in the following:
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then

(A) P(B|A) = 1 (B) P(A|B) = 1
(C) P(B|A) = 0 (D) P(A|B) = 0

Ans.

Given:P(A)+P(B)P(A and B)=P(A)P(A)+P(B)P(AB)=P(A)P(B)P(AB)=0P(B)=P(AB)1=P(AB)P(B)1=P(A|B)Thus, the correct answer is B.

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FAQs (Frequently Asked Questions)
1. What does Probability explain in detail?

The students are introduced to Probability and its variables with theorems, examples, formulae using real-life examples. The topics are explained in NCERT Solutions Class 12 Mathematics Chapter 13 to give clarity of the concept as a whole.

2. Is Probability theory complex?

 Extramarks NCERT Solutions Class 12 Mathematics Chapter 13 explains the concepts of Probability theory and its key points in detail with Mathematical formulas. Referring to graphical presentations will help you understand the topics better. Extramarks has described every aspect of Probability in more straightforward methods as well.

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