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NCERT Solution for Class 12 Mathematics Chapter 13, “Probability,” is a comprehensive study material designed to help students understand probability theory’s fundamental concepts and principles. The chapter covers conditional probability, Bayes’ theorem, random variables, probability distributions, and mathematical expectations. The NCERT solutions provide step-by-step explanations and solved examples that will help students develop a deeper understanding of the subject. With the help of NCERT Solution for Class 12 Mathematics Chapter 13, students can improve their problem-solving skills and gain the confidence to tackle complex probability problems. These solutions are also helpful for students who are preparing for competitive exams like JEE, NEET, and other entrance exams. Overall, NCERT Solution for Class 12 Mathematics Chapter 13 is an essential resource for students who want to excel in mathematics and build a strong foundation in probability theory.

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Answers of Mathematics NCERT solutions for class 12 Chapter 13 Probability

Chapter 13

Probability

Exercise 13.1

Question 1:

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

Answer

It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2
zbqSsCqm9xiWcxNBErf8Nh4WRx362tdHhuxpQ

Question 2:

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

Answer

It is given that P(B) = 0.5 and P(A ∩ B) = 0.32

Question 3:

If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find

(i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

Answer

It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4 (i) P (B|A) = 0.4

(i)

DOEWVuv0x9zuYQccdVDJkzJ0fZBZTNtlrOf9hiSErDYRVOt9dLxEU8aBh4l6qmn3jYd415peyxiZPH vlBWLQtGFyvrhR6UVs8scYkJebiqHRCD86eggi1 tmPSdO0aPa 9la6w

oa1mPiWsfiaw8Si tBnhPpqpnCW54BGHJYDy0gcuwgLr Dl4ECXKB8wkAQCLP3zGhKrMRqW8zOrLiMuIj2qK4EXV6eHnEUZdkWTv

(ii)

ueSj vLCJsn5pwuupb4xEm0jFEObN17yoCnSaPlurcQdl5NcTDfCGu8Suj8oN0 QpoP4 hW4rcc4 eWyvuR8LXUFenkFpJk1wCqZm0x42iZ80HvvV nLfd7xEAndXNqJp7 Xkw

(iii)NvkCl6ip0tCbJTBqcU2owm2VQzIJwRqrPRZCBkpQ9XqsNldbLlcRueRJVjpMAK Y18MGIZV

Question 4:

Evaluate P (A ∪ B), if 2P (A) = P (B) = and P(A|B) =

AnswerakYpij9Kzvq5k7hR0TrwA2gEkN6aHhetzBw AZyxJACwaipgCmJL4zVM45 jEA cdvBYlcMt6QT G lkafU3wW3Hr g3W48ZCC0RJK1FN4N U6cByvQBtBYXgCQeaNlsZSW24dgmv SEaJAYoWj u35kb2EPU0vzspN0wp3HodbO1S2RiL0 s7wOv5rAmOOy9qypE a4vhPLYafDhPTJ1JAVKdn90sWlqBbNLhRttck5KiFmgVXholy9FYQPwZXgNNYrRfUIhisPYwiL83D8Qk9hr1ZT8Zpkl5Zr4Nbc0hIfW4M8zjUNZ7X C5Oee1 XrajkvnnADhMJvw0d8

It is known that,

IEIXXQhbSIGCmgPLbnsK9ii XQhjTc4mu3e8gZdZTLZdQbB4NoeDJm6lpIka8BIAy2dcPLqWrbPJch6kp1JKan14LsOghwP33wi8rDGopk8gq0wyonbkMHdcaiKo6POkyJeUIM4
xgeY8aGvc6EXzg3PhibrpOwpIMpkX WKOZK5gz Oq2ggE44eGrqrr3b3BQcomT272 T8Fyh59e6B7U6JWLZT8h7abO7EHTL5q2vqV6K06oiOIOVjnXCLe2 tg0ufTdGAlDIO9ac

Question 5:

If and find

(i) P(A ∩ B) (ii) P(A|B) (iii) P(B|A) 

Answer

It is given that

(i) It is known that,

KiGxEBRuoE7rrxt vjOM6S3OmYpX25aYrlFQBLwrh0iRYf4F4cH8glQgXZmM YSImm6cLFhbcasQGg4jeEXXuTR8AO3Yzy98QCXrt8rnc2hHNsH ZUsotCFXg33eG6p0IcgbkCM

lCyizZJi40Vgol ybUB KER0OTiYGbyt0kRSL61nLTQ8mW EWjg8 G4A6Lduvfb2vv FrFeidCW4 tVFyYztlApVUCdrOuocKGeabN96ISbfKMIjOYv0dfaDdhF kjNFQ87

(iii) It is known that,

hz2W5ra683iW0eI7Vg rMA 6G0eYNDYqvgxq hUCMPKhRmkfEhEBwaFCyWKNwyEstAq7BhPu w3qDyWhzTxQnsj7iOviSq3S1EA4wWklUjBGzrM8jOGXu aY4nul7Gumvn6SMHo

Question 6:

A coin is tossed three times, where

E: head on third toss, F: heads on first two tosses

E: at least two heads, F: at most two heads

E: at most two tails, F: at least one tail 

Answer

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.

E = {HHH, HTH, THH,TTH} F = {HHH,HHT}

E ∩ F = {HHH}wrRhgaT muFbsdq65l11G RcwJLh

E = {HHH, HHT, HTH,THH}

F = {HHT, HTH, HTT, THH, THT, TTH, TTT} E ∩ F = {HHT, HTH, THH}

Clearly,
p zBfoeYyoF87F7Ws8f9m6VtKxVl2BCPjB9 2Aw6GgXa6ChNLIYMqesN4m0O5PFQWRCfKnlNiSnkiw2S15O8Z7SmriMZlSfVgPXLEeDSDI0ieMHVsaAAu chkORG8K9NAgPt3uk

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

AynxMgAKm2UKpJv3cvbtfPmVhEcilaH5hIidJFFmmQnUb4M7VRdehpIxRJnruDYvv9 Wf215VE QwhdAVKOPJev74zR9twhE5QXs3sCn6BiTyhMlJoprvEEYLBcQuqJVOBWyz6o

Question 7:

Two coins are tossed once, where

E: tail appears on one coin, F: one coin shows head

E: not tail appears, F: no head appears Answer

Answer

If two coins are tossed once, then the sample space S is S = {HH, HT, TH, TT}

E = {HT, TH} F = {HT,TH}e7W0F0 9dBrXn0 MzdoO92V7WyTzZYjj1C8rpxDT9eUu0dp0oEaM4fQXqEzSxlmMhwsj5RNA z32FgvvzLvmeLdq9 G948tbDHGXrd bMXVnWlv5tAP6V7wS jxEDFoOdMn R14

E = {HH} F ={TT}

∴E ∩ F = Φ

P (F) = 1 and P (E ∩ F) = 0

∴P(E|F) =qFyyn9PpfvKKHtDgAjYYJLb5MjaFrohsUpns6lh9N4xNL0tC1ACN 9dvfMMPG45xAICZ6IdqDxq7VKw uKLgk 8EqbAbQ0beeMq232apFgDCutjXr0pKmktTgY4pmOSd34 AMRc

Question 8:

A die is thrown three times,

E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses 

Answer

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

2v2qtUqCV5sHc889zMzKTKwpZqKmxHnu2 b6ibHPcYAe2zt3l vpfWrfhadoyEnQcSy4lXJ7fIa410tZI1y2dG6ggMp8BJNYUb rMTttQkpNl08i6PD yb nThLdbjteoMnGkjQBgFaCdWJVGb3O1JUnTJ54b8pmGQRgo uK8rsZo3Gd4K3kK aJTALIE9l2LX4l7lAxYLIGJxSzP2H7aKW0TNmJUWfd4 HTd8F0Wu BXqX5 FqQ8IVngnmpt6zNQ9DQYoH5k bH0w

Question 9:

Mother, father and son line up at random for a family picture E: son on one end, F: father in middle

Answer

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}

∴ E ∩ F = {MFS, SFM}

5gnwysNz0ljBQwgVWR2sEhhN9mDuV0yfyRcxnQB6Nw1C X9XRNkxdVTWhMPLoGTNF4vUNgajwuXbtuv1ccN66XVm8j6ddGx8NdZ

Question 10:

A black and a red dice are rolled.

(i) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(ii) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Answer

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.

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)}

∴ A ∩ B = {(5, 5), (5, 6)}

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).88OijQ bggdqFkPO9rzwV826iIKm BptrYda8fuqdqqxLvq1teOxfvdzd7zLcQFlG vPh6S0XGE7hLgkYVdTePIJpEyjwN pqFv QCbHJMAcfeKo8sNkW3Y0lGe7oKME3uG3lnw

(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.

UcM98ILmmdHm mdm2FERzE7WinF1 LlVR4X2F7tZzFQ sgTvNnGmcHBjTPP48Jt9ugLCY ScpukXqMfz9bcyaxOZMV9xcdtbE3KoHQ7yvraHJAMvKoqTGaPg RvUwp6LX8jSSDE

The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F).

7geBZMXJFb8iY mi5202WUvMq2hI75TayqxkX09 7vMdIDa0zJ kFmtx9O g0mar eoG0K5pMEgCVloONYl sokoKIDw0K3 5kiIu1HgwDPG6n0QCu8fRrBEm2uN8AyZPRQ63E

Question 11:

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 ∩G)|G)

Answer

When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6}

It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}

mnRH7 xU8DP4XWr7f4BDoeOj3oSF5tQndmS gYkQGedyToJmX0ZIcU Ozs3FUKcfANC VAlvq3bA2nfFn1ha70Xu2QklKv4qnREn9qWmGcodwX6tXEXmd8G9B2TQp8AQ0DqQC 0

(i) E ∩ F = {3}

1TYwiNvUXWpzirAQ85ycVN1HR1cvm6Aec

(ii) E ∩ G = {3, 5}

vCbgs8sWrf5VjL4QQ7YAl0EDxxv4K7fiEmU1ObXs7W0E0cFPCYHkKhcTurm0kxV6d0m22U95nhCl3LDWhS m3

(iii) E ∪ F = {1, 2, 3, 5}

(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}

E ∩ F = {3}

(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}

6ohKjwA MKkt 2nCtuL0EAn

Question 12:

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?

Answer

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 be the event that both children are girls.
sRT0tnoGb KPf9HOLvgq 5OvZOaYx 9WGOAVpIXGR oBUyfCC 7T0GLAsOltQEmqSVB99akP6j61 dUfmWWGZqQ6 lXkI8bHkd3zUjvnbDNgiVLHMFNQClisbLq3onRO5SjEfXI

Let B be the event that the youngest child is a girl.

W2X5AXlVWDKKzyJa p1sjnIiRxUnWQmsV3irQn8zg0S3uCWn9r5BlVhsDCfyceSuBoIp

The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).JD3H

Therefore, the required probability is

.

Let C be the event that at least one child is a girl.

The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).

d3KfBbOTp9lIjWgK J06sRxrc1hvaNCDJ n50w ODFCz4ixqjcLL6Taogu2Qpzn83pOE29wt3JW7q gfBIz5te6edU1Gwmtt7CcUQawfFnIr0MMnEzKXgEJ6F0cI24 4WT2fOrQ

Question 13:

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?

Answer

The given data can be tabulated as

True/FalseMultiple choiceTotal
Easy300500800
Difficult200400600
Total5009001400

Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions

Total number of questions = 1400

Total number of multiple choice questions = 900

Therefore, probability of selecting an easy multiple choice question is

P (E ∩ M) =

Probability of selecting a multiple choice question, P(M),is

P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

vmp1S4iISSiAluwqeaaEhip3kFfyQcne8oCXH OZ2IlICf7VVpOaMqK5qvDs4d12y OVETxYoS5pP

Therefore, the required probability is

Question 14:

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

Answer

When dice is thrown, number of observations in the sample space = 6 × 6 =36

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.vg9LHz9OOGQx6YPxlFj HF95zWnf3fUJ8K5cuLZn2B6PsrdC 4ArqDl3fhGvdRkr3V22qVZWbvczuogn 0Y

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

Therefore, the required probability is

Question 15:

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’.

Answer

The outcomes of the given experiment can be represented by the following tree diagram.

The sample space of the experiment is,
YllKUJzTrb9DIbs19XKr4ZOAxZ2tL6taQ0awu5I3R UrPKQSKmA8ZR7t3Pz2exl0Cm zTQCmOrVGb047SgiajUhwMHyC3N48dWanSaC X1G ytvUKuJKmimyb0yRqskpnFCrwZ0

Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.

rAOo0eVh5p05sJYzIMICkRU3YEWqCVotFndqU43FoorVlPd36Gr78y0RZiYPcWhGv8iuhnKzaCaGhWbcjGRAuKTy255zN7 SiTxg5Ky4L V4krGR tp fO5QwHW9L6Vir 1BRAQ

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).

Therefore,
pglNlDoGSXXtunGQ7JJhZv UM 4LL7 Rpxk2rMiQvPF6QPcocFestFcPaKGQ5LOItLVxSphtRkP2wHV5J J1qLRcNR2vGBBj7OxIy2yS QU8yV3pRAcU2wugf0oLqkZQQodCV2k

Question 16:

If

0(B)

(C) not defined (D) 1 

Answer

It is given that and

wZ7VKLBbu7rmVjAChDRPAw ZcBVDND36cQmn7a8zXBPViZ6uvI7VeTpe0LLwuTVSd72XeSzG1wnoN 4K3O5erHP4jBkeoBMwCEwUs5PlSKMVngkbJJox4KM8 DAB1tvw7PPV XY

Therefore, P (A|B) is not defined. Thus, the correct answer is C.

Question 17:

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) 

Answer

It is given that, P(A|B) = P(B|A)
3N5e5LPqOvCNA9FdB9N5OKDriZIukHGNK4pExQayLM2lXnRKXc1irVQbC2WIJiD0s9U jbqctn00H778Tgwt5e6QpDSztvMQII yao JADC0 WLhpAcyWwmu89U2IOJ5lnDz h8

⇒ P (A) = P (B)

Thus, the correct answer is D.

Exercise 13.2

Question 1:

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

Answer

It is given that

A and B are independent events. Therefore,
UNiiWBcfic6SWyOlbRBJ nP f e6jFfCwQ76ailrNv8f5fcpsqko8Bpn7PTjOSryjKGkzngqje2U zhwDYrgFdpjXMIuLa0eJ6OnoHi3BeV9qDRRqM5wO7pIRV80GZYE3wjp68k

*Question 2:

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.

Answer

There are 26 black cards in a deck of 52 cards.

Let P (A) be the probability of getting a black card in the first draw.
nZ7d6g4h45PpZ3qlOc7d05rXyU6sT7stZ3K9Hu 52Laa5dAY zR6u2irR3Jh1vLALrl8dMycPW

Let P (B) be the probability of getting a black card on the second draw. Since the card is not replaced,

7rg0c0m5RWHaQn8dWGZ76Pr1FMCSeHuGn4Q0f f69BsKfJ KBZrCvaJKSpdR3O5ayOWmAYuNUzS29mnYV9SHUsVPpfaT7 SWw37g1sj AtWoVcDkcDiBytuvlgIu9Gc9UQoVKWM

Thus, probability of getting both the cards black=

*Question 3:

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.

Answer

Let A, B, and C be the respective events that the first, second, and third drawn orange is good.

Therefore, probability that first drawn orange is good, P(A)

The oranges are not replaced.

Therefore, probability of getting second orange good, P(B)=

Similarly, probability of getting third orange good, P(C)

The box is approved for sale, if all the three oranges are good. 

Thus, probability of getting all the oranges good

Therefore, the probability that the box is approved for sale is

Question 4:

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

Answer

If a fair coin and an unbiased die are tossed, then the sample space S is given by,

Let A: Head appears on the coin
FsdswTPa3eTFe2aptTTGuu5qbtKTyLwd1KkjKDLpeW gtvQRUM5HMB47ue KkkRj1rnzSRBJsw66SSPReVK27Ltsi1m5lqQrzElgCKJyEG1 5lVffY iQjq7Hy0UyAY078fdBHIzi76lNDQ4dhHgGw46bACcS2yRd4RhpEsgSW3TMRKcgkXEo8cDzCZre7FrJa4nhMhI7jnTPPEi5iKppVoOWHM ZM6DMpfZ7fKASK6YPEBUNY7qA3n5eynGKFlQgGF2 petIzF7c

B: 3 on dieRpDUV9BL3ktGFrSy5WZEv5
d lHf0RMv1WPS2HjKjtKwGTa6QHK7SvjmAM1AWYleklXzE4Nh48kZpREIgpGYSytIQyIn5JzITs1RNfr8A4TilomZlyZmfMmXWfvfxyEPN4ntSXF5KA yEYfUPFqA7Yt3rmrzEs

Go2Y1q5ezkFmvS53taHLEs6OsB2aLNWe5ofepQYcB9TcUwUfjPFMQpFeENOP61OWGH6CDMipY 5Ms1BxYCC2H5fu

Question 5:

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? 

Answer

When a die is thrown, the sample space (S) is S = {1, 2, 3, 4, 5, 6}

Let A: the number is even = {2, 4, 6}

B: the number is red = {1, 2, 3}

∴ A ∩ B = {2}

7I bqQhZJ72BPc8aCQ77ZBx

9wHy siNoiATA7iNJeCMZjXgl IeF4PvoR1fsdXpyDS6ZHbKFqugr Xu3W 4RjwLgh6lyYEiUE 3OBTzP Ie8LRx5gDEh2r6 SIsv0Okd5MG8pzZPNrh85NI27oxS M8cBAINR85mEK Fr NMIc 8OTZwuA wLes6uijt8fp8lEHZlPuSZXHBnj9uV4YyEwvTn7hMLmNjXYWrSCvBESOiiVyeArnNtxcu4JANJNXp07dcazw1Y

Therefore, A and B are not independent.

Question 6:

Let E and F be event swith . Are E  and F independent?

Answer

It is given that , and
iJOw8ZST3VnUf1XFdk2Hmz pM3fuEA2xViIo9Cua1L bvzmJWloQMNFtMIjkHDd8pXW5QSR7b7v5aA WXJP62vEuQRxTeFq3 9a8ewWI 1jif0We2j5yLgC9TXE69O7D07m0MJk

Therefore, E and F are not independent.

Question 7:

Given that the events A and B are such that and P(B)=p. Find p if they are (i) mutually exclusive (ii) independent. 

Answer

It is given that

When A and B are mutually exclusive, A ∩ B =Φ

∴P (A ∩ B) = 0

It is known that,IfvMZaGBe79WjcPOTpulbMJuj Sojl0sgmG77mTUOeXMNoya88I7nWUcwz2 rba0 Y2lhDBj0AyU5l8XRhmD3EFKV6zOVAnHUTXYVS2rWPTJ qTZUg3jgRxyY9 t SgXsBoyHM
FT175fIjW0WvfrBt0BXMgGww33V1l7geeIIlQLkc2 ribTwyCaqtuitIkDfXN09AXV0A7B

When A and B are independent,XyrCSWaUnhVMBEwdF9tXUKWeSpOHG83zvm8fN27i2sGFEG88vsNuBJZviGndaO7bSWsvHwZZZHdVkDjEw83VcpcDz92JsYWYBHlyoUW6i0mGSOg53lak1k4lCj1MvyBtgu40gAg

It is known that,IfvMZaGBe79WjcPOTpulbMJuj Sojl0sgmG77mTUOeXMNoya88I7nWUcwz2 rba0 Y2lhDBj0AyU5l8XRhmD3EFKV6zOVAnHUTXYVS2rWPTJ qTZUg3jgRxyY9 t SgXsBoyHM
bKyr6EN3C TArvEVv2EcJTuhjV9YevPEnTQ5pyzWvEafrZm4Ublp5c581VtOl6tGU0WMa77bKWtvg9mwoJ7JsvqHXHpYept1cbMDSN83ZsmqJZzvFT2GESdJdjys5truS3FXYSg

Question 8:

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) 

Answer

It is given that P (A) = 0.3 and P (B) = 0.4

If A and B are independent events, then
5fEeGmEUfpjQzDUsZNlKlGgungd4s2hVSf1GefQRH4JH1wsaOajTyRO6G9JZbkBSNrnHY8DJEMiIb5DAYiX3dZvZp2juE

It is known that,etVpW30y2 BDCxPa7mq0oBdsO1LYAKz uhdv8ZiKoOihZlMtrF2xlm7OzPfd5 uK9HdJvKNWt0B7xVXiM83b63GeFM77W8BiKi5OzJlWSmk

lCyizZJi40Vgol ybUB KER0OTiYGbyt0kRSL61nLTQ8mW EWjg8 G4A6Lduvfb2vv FrFeidCW4 tVFyYztlApVUCdrOuocKGeabN96ISbfKMIjOYv0dfaDd
JDehEq6hf3tFqagqEXhiQcv O2FZe9fG6MV Jb6UuUBfoyxGunUJBWq2yHb3PPhT0V32rmajH4AVwZibaDbFweVEyCCmFwovznAkj W1RFOcL7yq6IsCy6RNVkJ dgbbZEm3P6g

r9OXkYXG676OZPpkZq5sDD tebBlpSbmh3Nk9u YFfH63WH2Djfriq1zRgBcf8iPK5fNSqj8XYEDec3Cu2durajxduwkvfeZtkU2qO PC4ff17H1v u1G dMbLqvo2hJzsSiKcU

Question 9:

If A and B are two event s such that , find P (not A and not B).

Answer

It is given that,

P(not on A and not on B)= P(not on A and not on B)

Question 10:

Events A and B are such that .State

whether A and B are independent?

Answer

It is given that 7N2 uOPavuoLyiFZUVImWqqX7UVBU7aKOMkaShqmuKJdU0Ikrr8oMnf4 FCw0MD9CE v83t1h5TqM BOBX

Therefore, A and B are independent events.

Question 11:

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) 

Answer

It is given that P (A) = 0.3 and P (B) = 0.6 Also, A and B are independent events.

(i) jcjR3mCOI4EyszXbCFXEjAqe7O9dtum y5LSDez7KCARIesCOs r5vha7ioi7WnKpnadK9U82JDreMPmpdqYT5w28HgYcSwKr1eRU5f83ScmcccmZ2epd4r

P (A and not B)=

P (A or B)=

P (neither A nor B)=

DVfi3RPVATzVCaJGzCw6ABDytBK6HWLpzgnXrau9BV4fbRx9h92ZF1T5xftsMDz7NjqJfy8gUiT1pdqynWLYSKBfVZT hsJWe6yELyw3RMXbp32Za8 1t S AN9Nic6HFtr9p w

Question 12:

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

Answer

Probability of getting an odd number in a single throw of a die = Similarly, probability of getting an even number =

Probability of getting an even number three times = Therefore, probability of getting an odd number at least once

= 1 − Probability of getting an odd number in none of the throws

= 1 − Probability of getting an even number thrice
8X0SA9KUWg2JMMb9 FTDhRqhrAvL43b etvGwqdexlHJwxLuXcX0XQOYMe0NUrlWiMUq5JP23i2TySlcFfbnx

Question 13:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that both balls are red, first ball is black and second is red, one of them is black and other is red. 

Answer

Total number of balls = 18 Number of red balls = 8 Number of black balls = 10

Probability of getting a red ball in the first draw=

The ball is replaced after the first draw.

∴ Probability of getting a red ball in the second draw =

Therefore, probability of getting both the balls red=

Probability of getting first ball black =

The ball is replaced after the first draw.

Probability of getting second ball as red =

Therefore, probability of getting first ball as black and second ball as red 

=

Probability of getting first ball as red =

The ball is replaced after the first draw.
T4uy25rVgntQQC6RvZ6VMsffhT DOYL0txyPlvjDM0 unvOwu0O H17dSiJsRpW3pak94Cc5ULWuosEp9z8UZiLIPhCT8zDFwUaA7WdgHtfwbE X7u gSRsuYhkEZmmE4RkK7w

Probability of getting second ball as black =

Therefore, probability of getting first ball as black and second ball as red = Therefore, probability that one of them is black and other is red

= Probability of getting first ball black and second as red + Probability of getting first ball red and second ball blackTw6w7KhAVAYcJtL9P BASWCpcs1sGx0NZpzSjBCCI6HBdldGmkIkv1qhdAO3VrsAaQmFirzX1kX68NrIz93BhkW4RG01TXqZG0RiLaO o6172qD5IK8kJs4ArJygDZKQaMz8rBw

Question 14:

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

Answer

Probability of solving the problem by A, P (A)=

Probability of solving the problem by B, P(B)=

Since the problem is solved independently by A and B,

Probability that the problem is solved = P (A ∪B)

= P (A) + P (B) − P (AB)

RkcPsalUSqFRZ4am7VfwQvENA63i0IbxXMaGHW17QBVilMIM80ZkJFd0spNkuH7ehu28ugos44tEIvvOmCeP4XOS149xsHAzoDr5zbTAQ4cQku5yKjz XOFWHlti5D9C4wgKH M

Probability that exactly one of them solves the problem is given by,
xFsAtsqgC5JoML7gXov2KoiiA0a75bnqt dXT2c I6fwxpTB x9XhnOCtr8ZiAuWq4dQarbL IJGRhmpf HmnuW9bQCVH1lNunTmGUaWQX u7v6k 54NY5WVdLBbvoIkbM iKkQ

Question 15:

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?

E: ‘the card drawn is aspade’ F: ‘the card drawn is anace’

E: ‘the card drawn is black’ F: ‘the card drawn is a king’

E: ‘the card drawn is a king or queen’ F: ‘the card drawn is a queen or jack’

Answer

In a deck of 52 cards, 13 cards are spades and 4 cards areaces.
w2Tf9aE Kr3y6OxccnoC10VM17kdK34auw5VhI16zpGNDDSb7fnGugeZicMWBh3bw5W5nyx5cgx HoRx7 C1RsUrZgCr lcnQczlBePDtX2Q viB0mNKFEBtXRNHBjw7M385wFo

∴P(E) = P(the card drawn is a spade) =

∴P(F) = P(the card drawn is an ace) =kJaxqKV2vwv 7q6 WjG1w49ZAG T3NfGt4SJWsIH4067eAPHONL8Z5JNa0T6NODx2E9myDtNesUm5oUxE7LHAG99Ttd9rQx6e OBqgTVUeRZhv3qE5Ch8 9C qUrHDvJCdDY3Dc

In the deck of cards, only 1 card is an ace of spades.

P(EF) = P(the card drawn is spade and anace)=P(E) × P(F)=

⇒P(E) × P(F) = P(EF)

Therefore, the events E and F are independent.

In a deck of 52 cards, 26 cards are black and 4 cards are kings.

∴P(E) = P(the card drawn is black) =

∴P(F) = P(the card drawn is a king) =

In the pack of 52 cards, 2 cards are black as well as kings.

∴ P (EF) = P(the card drawn is a black king)

P(E) × P(F)

Therefore, the given events E and F are independent.

In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.

∴  P(E) = P(the card drawn is a king or a queen) = 

∴  P(F) = P(the card drawn is a queen or a jack) = 

There are 4 cards which are king or queen and queen or jack.

∴P(EF) = P(the card drawn is a king or a queen, or queen or a jack)

=

P(E) × P(F)=

Therefore, the given events E and F are not independent.

*Question 16:

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English news papers. A student is selected at random.

Find the probability that she reads neither Hindi nor English newspapers.

If she reads Hindi newspaper, find the probability that she reads English news paper.

If she reads English newspaper, find the probability that she reads Hindi newspaper.

Answer

Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.

It is given that,
Cj3tZYCotjcAbAmCNDrojfVYpWGCSiG3sHlgkuXyr9uuy Up5IlyoMVb0I DrwMiTJrC2oVc3pZbVct8DGIDW3tWHfzrhC1YOnu yZ1oHqmbBDD9tItzpLJKV98EM7Dm32tyBfk

Probability that a student reads Hindi or English newspaper is,
VvYS01BOQw1ycLsV g08owoQovR9 T7ZfeNo2yXacQGINwC3xhiw1uceedRyXlNkLGtJ8

Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by P(E|H).

Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P(H|E).

DDoxwEdNe1 RV3sOABnPcFOjvJBuF8WO5ZpIqygzsStfoqRxKxBJvWanNnymfui yOojAfNEmLbRNLNN M 4

Question 17:

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(A) 0 (B) (C) (D)

Answer

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.

∴ E = {(2, 2)}
gNKBAJQkJckZYIYvL6RjBr3gKCoLQk U

Therefore, the correct answer is D.

Question 18:

Two events A and B will be independent, if

(A) A and B are mutually exclusive 

(B)

(C) P(A) = P(B)

(D) P(A) + P(B) = 1

Answer

Two events A and B are said to be independent, if P(AB) = P(A) × P(B) a3l9cs1NiiHni1ivcHDPhk1 aj5d28tqdIkkrpO4emXjPwtZQ RhZivv5w7WyswnOeQhybeewfQovccBHPCKdntypP5LudqsSZlN4oDnO u0iNozEdHpf1IWpsLsGskuKceaRSQ

Consider the result given in alternative B.

This implies that A and B are independent, if

Distracter Rationale

A. Let P (A) = m, P (B) = n, 0 <m, n < 1 A and B are mutually exclusive.

vRIYl5 LNgo3LNaSmuZ7ZfOJ8RXNnb73m eNr6QsoXYWGLdN43kJq7ptTySj4KjMQdAnqUBZoiiPtaetOCX7qqoZtA3s4izncg9kg34uER72EvXR3L3Sr04wZ4nVlApWVuv Gw0

C. Let A: Event of getting an odd number on throw of a die = {1, 3, 5}
bUs8YueNyY6kvPRnp55biwz YcEXk ulP1 6aWfvWGGDIE9Q 68cDqB1WibNx

B: Event of getting an even number on throw of a die = {2, 4, 6}
q2NBNa3Btl9Tbj2ZBcCFFzV8HMh1iAQre5x6kS7d4YR9KbT1QC xW6OnygKm1WBRHyPwLwhZz3OjGL7VC kQbc yrQbapWxuSA4Baqv4ITn3NkO4oH LML

Here,

D. From the above example, it can be seen that,

AlP4LvteD82PmjRkUXdPPIFlGagJsGaE6uS9kEwR1Ov1QIJM3J1ZCzLIS1 8c

However, it cannot be inferred that A and B are independent. Thus, the correct answer is B.

Exercise 13.3

*Question 1:

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 ball so f 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?

Answer

The urn contains 5 red and 5 black balls. Let a red ball be drawn in the first attempt.

P(drawing a red ball)

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.

P (drawing a red ball)

Let a black ball be drawn in the first attempt.

P (drawing a black ball in the first attempt)

If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.

P (drawing a red ball)

Therefore, probability of drawing second ball as red is
tTuApVqXeA5H6lBqd1MUo4aQk GruU28wAQ

Question 2:

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.

Answer

Let E1 and E2 be the events of selecting first bag and second bag respectively.
cEUcv1NOApPGxSQO7OchKGBNK21v08ClTglX3PAtislVMRA9RAe4zPZnYkUvrg et24eXKK335iX3sNX0ONusIK0ojyCkWVd AOGnZn2wjIyk1ThhVq Kybi7rcCutxc0o0 cqEI9CAAmdB8S9H7IyGhsy7kH8hRBX4KG9Ni3sHjm8RXzd8PMG3HNTODYcbww E8gJ93

Let A be the event of getting a red ball.

The probability of drawing a ball from the first bag, given that it is red, is given by P (E2|A).

By using Bayes’ theorem, we obtain
rykoViDXEpBgVjm9PixpHYT3L t4d2Zb9 3hpp5l 4OTl106XEFTO7lkMugWxOatR9UUwPnnpsqUJ8AuXydix DJjbCTkfvwNww LTZwJSQF4LYFV3UbBoUT2yTy9Fihc03T4ks

Question 3:

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 hostler?

Answer

Let E1 and E2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.tCjcbRstm37y3IzOIDZDZ vhzOR35P0HcqzA4PUgI7Uef1SB5xMEumKeKHlRUb1eNcp Vst2BvYwigNpspgAcektnFvEB ePBJpMV20Xu0YP3tcjKY5ozjQ49VB5YMFogflMUd4R2zl S5fprGq NUpeneWYJbpbNs8lZjvTcXXWdNdFpXPpFvOb0zQze4ssA3VfS8zHTAyklcmrOTEGpT8yZM7pyjBbajuZBLE9yGNmdZaduHB4LLiwj T4sk37LCs910y5Pej9FQ
a3iIEGKzwYK1lLLkbYtE3uLMV3MY8JxEMc2Jmjl9Cx4P 9BbRXw6pYEsvW6VSH7uWUTIxZ38hEHdf5TfimQN643VNvX 0CssppjrkswfjV8XZFI2PBkHSfu9b58cAp3VCnwY2ws

The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by

By using Bayes’ theorem, we obtain

PvwcWPdRgEbr0NC5m5et5Q GsiPcF3BVbKIBTUeUYi2nr55TXOkZND02VBeOJRebT BNXV5HA3a1GyWktQJrD4DZysTvMpqxnDqFpVtVS2jVIGKmSwKn7wGps

Question 4:

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

Answer

Let E1 and E2 be the respective events that the student knows the answer and he guesses the answer.

Let A be the event that the answer is correct.
9 Xgdq0OSjpnUpaIR HCspRyX8FhlC9Xiv729AAdbnfSmq5bvCzCJw

The probability that the student answered correctly, given that he knows the answer, is 1.

∴ P (A|E1) = 1

Probability that the student answered correctly, given that he guessed, is .

The probability that the student knows the answer, given that he answered it correct l, is given by

v9jiU

By using Bayes’ theorem, we obtain

Question 5:

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 per son tested (that is, if a healthy per son 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 per son has the disease given that his test result is positive? 

Answer:

Let E1 and E2 be the respective events that a per son has a disease and a per son has no disease.

Since E1 and E2 are events complimentary to each other,

∴ P (E1) + P (E2) = 1

⇒ P (E2) = 1 − P (E1) = 1 − 0.001 = 0.999

Let A be the event that the blood test result is positive.
hH8nH6EOV apA5OeiY73U94Cb0aYplOrnzgHJP0 vebGL6QAl1j42QCI TUpri66IIeeGRS0ZQCIF7O1VTPUu0GziyS12JxfZULLijAzY1BXx1u4cu zU3KKLCoJRkIi1uGo1eC 1DcIVkT1uTvDqbZEh1Ich5gdLeOKBAi0ijbflVdKS4SjGe1JA g9xfU7u9DzRMGlcQdj 81N uK H2ZFgA

cv3QvSfDBUX7m7mXxz4elC48ZGhDvtznFv0t3meiM7Bb8VQEmqX4Aj0a6W1Crt3DjCgeG9fPxbUR0obhD2Aoa9cdgTI0HFqtNwCG71FkVfjCw deHp 3KRDjXrrHKZTPp

Probability that a person has a disease, given that his test result is positive, is given by P (E1|A).

By using Bayes’ theorem, we obtain
yoEIMrPWQeiy2uesBOAM9DtcafyOpE 6BKx9cpBy uggL62AfAQTu13VzIbv6ix6VVvzperioNDWx07AxGrhLTDbuHuvBgQ35fyVdndHMjnP7xyPLBDyK k9JP12jwddOJqemY

Question 6:

There are three coins. One is 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?

Answer

Let E1, E2, and E3 be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.Xb dNXKCAi5lvEANCK5gVx6jLMJgjugU9PJVHHVOehgZ6zEIK0idxCt0vOP ZTGSTuzgmpG951aPIncRLiqb dNlHOTZZfPsc2sY18GQGx2B 0Qn3iHEDOhO wtEfTUQoqGJS8E

Let A be the event that the coin shows heads. A two-headed coin will always show heads.jiUMD0QWMTL8A9775mwCe40h0PMiL16xDcRTPZXQ aqB1NXG doR8 f86WY3r3azT8gxso2nennVAMJF22qnvPsyhf5QVfAi3rVN0GWz3soasxD jNGwDwYTuIzmPzkU wmz8pA

Probability of heads coming up, given that it is a biased coin= 75%Pd0TjmgU5DjYZG1gMjH4HJHV Y8Dz1hk3PRVGpMgLp4u7j Km5JpG5IZmwYySWBZs00N K bcZ9jATQiAU

Since the third coin is unbiased, the probability that it shows heads is alwayshDtcHGtDCoU10UZwkz55qjpWpmBPJKA5ZwEmPlBPx1sSrEXmiEb7gkk0 pf9TZmv3LUfDuuXJjF05REovGBWmPqGxj KDT5tKxS jQEUtEyZXAvOze96AFgt fOj2H6uCjS0yo
vPBuhQQRYmPwWlxlPllM7LDjWfjrHHAgHGaPXDb 0L3aYA4wRoB h0xC NS dq21KrgjRQP9gI4G8yuRDXv

The probability that the coin is two-headed, given that it shows heads, is given by P (E1|A).

By using Bayes’ theorem, we obtain

Ii7Dc3mzdalOT7urTQA oEPaEAWBWF7EFnu M8bsux25JVi1IYDHbO7LO5kP7tqEhKQrZrgs3lOhL76UK2qN765AUCTLgNX5nnT G2PvLktMTTX0l9O 1st9hKazLXIOo89yC8

*Question 7:

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of 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?

Answer

Let E1, E2, and E3 be the respective events that the driver is a scooter driver, a car driver, and a truck driver.

Let A be the event that 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) = P (driver is a scooter driver) P (E2) = P (driver is a car driver) 

3KEO54jOZjG7xkhiGImHDjgDexqWrUQ dB1Cv9DLWKLrwHa1WwAlmdr7kVczChlsh 9EwvvuqqCQH4O20Uir6nRfgGU2wBnht WpSS1sFdzWmL205eTQoPfxA

P (E3) = P (driver is a truck driver)

59ZrpQ4lz odb3ey76X54HvwmGXC6smXgt43Ihjc1wl0rVN5jHIIXNkDf0Ej1YlpsJu13pfhVuoLhspDgw8Zq0QgePqz7Oe4nkvooIRYfyrZ18Rx8Rq3xv9gWR Cagtnj357aII

The probability that the driver is a scooter driver, given that he met with an accident, is given by P (E1|A).

By using Bayes’ theorem, we obtain
PJhxIfqeAkangvZU5Dyz98eERRMzhG lzU2yZ 7ekLNMKtrm3rdZysO91hwxJjIrPK4JAn5Jc8AeAemQ21DbEgfVC dsHTOHrSlICLr8oqTVe fJAastsBiwCItFWBdo6imkGK4

8

Question 8:

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 was produced by machine B?

Answer

Let E1 and E2 be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.

∴ Probability of items produced by machine A, P (E1)

Probability of items produced by machine B, P(E2)

Probability that machine A produced defective items, P(X|E1)Probability that machine B produced defective items, P(X|E2)

The probability that the randomly selected item was from machine B, given that it is defective, is given by P (E2|X).

By using Bayes’ theorem, we obtain

969p2ukliW2v2mKto7h7JWkSe7wafwJUqy YG An4oih5r5cfdaUEAZ QMMmCIYNhIjadoygciQoiLflOuIgPUf4uXDj2bp8942kFPOnqRRu9ovjMc9ZUYmluiIXqCvaHiCjVaM
jFaJA644CV6lifeytMr cHs tRcXNElz0Tgs YGDPIAWRz63uRA6ra4Y WE 49oGntjI3UxvbrgiaaeBcqJ9MDlOD9 XZltHecYEFPKZj3pZK3ZJ7 9nyeBGgSgqISDAE8eTW8s

Question 9:

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.

Answer

Let E1 and E2 be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.

P (E1) = Probability that the first group wins the competition = 0.6

P (E2) = Probability that the second group wins the competition = 0.4

P (A|E1) = Probability of introducing a new product if the first group wins = 0.7

P(A|E2)=Probability of introducing a new product if the second group wins = 0.3. The probability that the new product is introduced by these condgroupisgivenby P(E2|A).

By using Bayes’ theorem, we obtain

I0uXaaZIjuOb555KJJmxGP

Question 10:

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?

Answer

Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1, 2, 3, or 4.HvJ8fV1uhReWNO9Kf3JKOXsWeqt H SDj6mRWwbx6orEgBEpwZMYnzTI33zWmUJZSGFiOP0ya eOLsZaxk46NvvKwvegbR0SWXUX4sl X PB8ah6t3Dyn

Let A be 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 or6

P (A|E2) = Probability of getting exactly one head in a single throw of coin if she gets 1,2, 3, or4

The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by P (E2|A).

By using Bayes’ theorem, we obtain

Z auwi2omHc7DrT7fHFjejoKpwCCREJu7NB0zhKZOlqPEbfgbz edn0hR2Dt400RL3Xsc6hpzlCUPe0VwadCwsXG ZNJoz S5FovvvHsYJ WREbNSjPbaQbuQ3jv6PIhb
qWaGiGmu7Id8jUE0J 5t6YXAaIyYul4W9EEA08vOnym FeVNsYbMHejs4hSTiD4lYxGNydC82Bu5DeT ljEHHzAIxQBFlXfMG4oUORC9DV 9aAbrxJpuDp4kIu4 cp8KNHcMVjw

Question 11:

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 was produced by A?

Answer

Let E1, E2, and E3 be the respective events of the time consumed by machines 

A, B, and C for the job.

Let X be the event of producing defective items.WZ8Yg qPi0mxDELoILlJYRCGlL858BcSOo2F53vi nciIewD61Ex4m 2z38AVBoVDe4j0fBcWUGQDu5 3fvEMf29pFYJzBQPJfpms MUYb6c9Nr

fk62ZlSpA1aNwlvY2YfqzedQJ8WsXCxR1YuJgPdSNlJsSUTavWP620KVJ63NlJmyzofNwbTB2l5x1Z2JnfMSSu574Z

The probability that the defective item was produced by A is given by P (E1|A). 

By using Bayes’ theorem, we obtain

Question 12:

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.

Answer

Let E1 and E2 be the respective events of choosing a diamond card and a card which is not diamond.

Let A denote the lost card.

Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

6JiWf8onmt74iOnhZ M2SeFZc

When one diamond card is lost, there are 12 diamond cards out of 51 cards. Two cards can be drawn out of 12 diamond cards in ways.

Similarly, 2 diamond cards can be drawn out of 51 card sin RoU4 xIpZTNVob80CKqtyBclu9Ms uhm vKyavJuEKvpL8JoWOvm0a1mmFrwDxt57Me9Apbc6Uwbt tBZIYfXfoNxsSdO4wAVbPC p28kg7JYORF5rIAkSdkNrDn7XElX1KtqYw ways. The probability of getting two cards, when one diamond card is lost, is given by P (A|E1).
mm7a1C64djyBOJ2LniwGWaN6RpjDj3vSS0RRoeo8Ccu Q2BW0c0 dWz5W4zGw1vdI3kMnAZ3RNo2UApD5mpu8R ekLBKR1XYtSsh4xqkq7LbZaVxUn2ww69ggwyEZEK a pCvKk

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.

Two cards can be drawn out of 13 diamond card sin 5X9vxmcfKfN6 LrRqaRrKKFtfajpWeejeKTLPNUGCcqvYLAxXLOJtkDyG2MOIzmZ1ro66JxI2l7zRDfQ5pu sySrTWPCNvTcIBttNrO27glUgCBw 70GAX9p4cSD4WQuundGL4ways whereas 2 cards can be drawn out of 51 cards in RoU4 xIpZTNVob80CKqtyBclu9Ms uhm vKyavJuEKvpL8JoWOvm0a1mmFrwDxt57Me9Apbc6Uwbt tBZIYfXfoNxsSdO4wAVbPC p28kg7JYORF5rIAkSdkNrDn7XElX1KtqYw ways.

The probability of getting two cards, when one card is lost which is not diamond, is given by P (A|E2).jEI8MEykK3oOO746txX2J5WntrJFBWXZCODH39WdycRrSC6898ZRD6lA3mp5Vi6xSPikM Q54aabvMpxN9d

The probability that the lost card is diamond is given by P (E1|A). By using Bayes’ theorem, we obtain

RFQtKHODoNH q9sDq5DQCGvyELuanua f1mHsB0JEZy7wKC4p2SP xOXWVmbsqdtoFGCRhiQONpIrcaGXPyLxn3qu7ZABa0DuiCYqUaBh3WWJ6Bs3
FIwkKoJGlryb9O0BWmM AxXsGAo90yczLAUuCvuh MpSzHGnKMSzZXzvO qltkZieJegbJYYmjEr4pZMUG bnbkucLY6ylq0f6oCR57tjtKN9YcNvPIPfUO8qeh4WPzvDS0Ovg

*Question 13:

Probability that A speaks truth is . A coin is tossed. Are ports that a head appears. The probability that actually there was head is8lBp2VOq7W bmC2j 2GuUeNSZoz51DmLWuG MnOxpo6XqU T0cUs8cmJByYoGanJ

A.hDtcHGtDCoU10UZwkz55qjpWpmBPJKA5ZwEmPlBPx1sSrEXmiEb7gkk0 pf9TZmv3LUfDuuXJjF05REovGBWmPqGxj KDT5tKxS jQEUtEyZXAvOze96AFgt fOj2H6uCjS0yo

B.fd

C.0Ber6rWjc0Zvmi l8tfzDKlALRvwUaNx VcBcetAqbxSWqn

D.

Answer

Let E1 and E2 be the events such that E1: A speaks truth

E2: A speaks false

Let X be the event that a head appears.B5nfkem eVQp6NPXTJllCM7GDfVublb6IfpJkc8C NCanioP0iVDrg47ar3C80A6b9La7W FEFukJe2XhpWz1yb29za5xYGV9SpZizh

If a coin is tossed, then it may result in either head (H) or tail (T). The probability of getting a head is whether A speaks truth or not.

ZsO0rTcxHeiOeIRFYNKJ5tHkcVNkZRf3rFBDEmQ22 hjJCJdPsMessKW ZLO9t0j3ew7ovSFlf x1UNjI3PBh9gqqYoPOb73qN3Ism0tOMy1ypXu cExgde1JNAYkn0 XcnhDw

The probability that there is actually a head is given by P (E1|X).

guomyZwmDTxZnevt6AlAngE7jh35BByfNfWrpN8ZODC

Therefore, the correct answer is A.

Question 14:

If A and B are two events such that A ⊂ B and P (B) ≠ 0, then which of the following is correct?

A.

B.

C.

D. None of these 

Answer

If A ⊂ B, then A ∩ B = A

⇒ P (A ∩ B) = P (A)

Also, P (A) < P (B)

Consider  …(1)

Consider  …(2)

It is known that, P (B) ≤ 1

ndJX c4blYJpGYh81F q0YT45j6Mt nbdWUANCk3p0s8KJplqH0yQwnCVwV2ET5d6kxcHt6wvsUwltL bsfjr2JZKqBRBkCsb83XEdgDFSuZu3U1WwqlGVIsJTlL1NXTmPhKniw

Thus, from (3), it can be concluded that the relation given in alternative C is correct.

Exercise 13.4

Question 1:

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

(i)

X012
P (X)0.40.40.2

(ii)

X01234
P (X)0.10.50.2− 0.10.3

(iii)

Y−101
P (Y)0.60.10.2
Z3210−1
P (Z)0.30.20.40.10.05

Answer

It is known that the sum of all the probabilities in a probability distribution is one.

Sum of the probabilities = 0.4 + 0.4 + 0.2 =1

Therefore, the given table is a probability distribution of random variables.

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.

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.

Question 2:

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

Answer

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.

∴ 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.

Question 3:

Let X represents 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?

Answer

A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴X (6 H, 0T)

X (5 H, 1 T)UstLG9 vpOb2eNCxYoc JpCMOlZjSBnRRU8U29kLqo9 mum5frigrqBKQ9QTjSoxMSKoXgQb41869Xw7JMpRYx0iMphOOKUkT0gxuDGT KFdhWzavFEDVUdTAwp10ZK16Aui Dw

X (4 H, 2T)

X (3 H, 3T)

X (2 H, 4T)

X (1 H, 5 T)1ya 0E xg8S8Bi1LWC V2xCOF EJastD2aG 40vW IEx9rSXfseGHK fGCLwKYokCS1h4 pL4CbWp3K5aqepTZYbVjDIE

X (0H, 6T)

Thus, the possible values of X are 6, 4, 2, and 0.

Question 4:

Find the probability distribution of number of heads in two tosses of a coin

number of tails in the simultaneous tosses of three coins number of heads in four tosses of a coin 

Answer

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) = 0

Therefore, X can take the value of 0, 1, or 2. It is known that,BTl0D Y6KNN7mb5G1qa4Zj9l7S2jOpkciKOe5Dmwtv4ZwMiw penqAk9aQZJDGgEdWNsZgpaWBEJOSBwr

P (X = 0) = P(TT)

P (X = 1) = P (HT) + P (TH)4fz T5pSnaMNlqsbrQC gL wTVkJ1P3Vko7WB3BVFNOGWD4r5u9JPNqsXGSdU9fyuVm00a36RZEBZRp1FhZXaH5hknn8rlmjGSXRcwsdaqUf1GL2 zTuImK5 R1TCDhiTmPGpOs

P (X = 2) = P(HH)

Thus, the required probability distribution is as follows.

X012

P (X)

Gy9qidz0cteh3yaNCtd s Q5tNVOHplW9yZYU742QHIEvPia2qoO2j1rdXGmV7AEuoAWGTzukU w5cb

hDtcHGtDCoU10UZwkz55qjpWpmBPJKA5ZwEmPlBPx1sSrEXmiEb7gkk0 pf9TZmv3LUfDuuXJjF05REovGBWmPqGxj KDT5tKxS jQEUtEyZXAvOze96AFgt fOj2H6uCjS0yo

Gy9qidz0cteh3yaNCtd s Q5tNVOHplW9yZYU742QHIEvPia2qoO2j1rdXGmV7AEuoAWGTzukU w5cb

When three coins are tossed simultaneously, the sample space is
7uZGUvz7YAgWp3vdP3HsYukzKpkxW5LiJQsiUf9Zxfa346YkZqkQ7xCZVHBXrIXhZcCxlIb2 Lkdef8a96Smy Z9QcNT0vUZGhD6G4Yf48wAQUDNXowYWBBJdqxRZ

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)=

P (X = 1) = P (HHT) + P (HTH) + P(THH)=P (X = 2) = P (HTT) + P (THT) + P (TTH)=

P (X = 3) = P (TTT)=

Thus, the probability distribution is as follows.

X0123

P (X)

6ZR05CKYaN3mWGw CZSkXcZyPY5dTxUHeowGIMA5cqR bPvPVbnAgTSj6j0CJ5lvpE QDzp

eBOkWMZ90fgxRDIcPODlBVnJDoTVuhFs9Ly17xSfLNPreATtJpkVtUyvkbzDReWAP3MaebJA2etD8QZeiDbT71FRq c rW di6Jeh8OhAFZxRzIpKqEDca0kdLS0

eBOkWMZ90fgxRDIcPODlBVnJDoTVuhFs9Ly17xSfLNPreATtJpkVtUyvkbzDReWAP3MaebJA2etD8QZeiDbT71FRq c rW di6Jeh8OhAFZxRzIpKqEDca0kdLS0

6ZR05CKYaN3mWGw CZSkXcZyPY5dTxUHeowGIMA5cqR bPvPVbnAgTSj6j0CJ5lvpE QDzp

When a coin is tossed four times, the sample space is
ADfgHWrhVGYBU0prf6OjE7n53f8e3oDix8t 1wM0NT9CGZnrGyBqp8ogz7yeT2ddOnaiqCV

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 (TTTT) =6a05EqOMz52SWSyAGXgorcIKNDGFMLM8C0FAX0pKG heR5u3g EOXVIhOFGwquC7w v279eYMbpH1eCx gF6L9664HygUZuf8I2wMR RFtkaDNiAirEPDpMCKt774IDabXuby6o

P (X = 1) = P (TTTH) + P (TTHT) + P (THTT) + P (HTTT)

=

P (X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (HTTH) + P (HTHT)

+ P (THTH)

=

P (X = 3) = P (HHHT) + P (HHTH) + P (HTHH) P (THHH)

=

P (X = 4) = P (HHHH)=

Thus, the probability distribution is as follows.

X01234

P (X)

6a05EqOMz52SWSyAGXgorcIKNDGFMLM8C0FAX0pKG heR5u3g EOXVIhOFGwquC7w v279eYMbpH1eCx gF6L9664HygUZuf8I2wMR RFtkaDNiAirEPDpMCKt774IDabXuby6o

Gy9qidz0cteh3yaNCtd s Q5tNVOHplW9yZYU742QHIEvPia2qoO2j1rdXGmV7AEuoAWGTzukU w5cb

eBOkWMZ90fgxRDIcPODlBVnJDoTVuhFs9Ly17xSfLNPreATtJpkVtUyvkbzDReWAP3MaebJA2etD8QZeiDbT71FRq c rW di6Jeh8OhAFZxRzIpKqEDca0kdLS0

Gy9qidz0cteh3yaNCtd s Q5tNVOHplW9yZYU742QHIEvPia2qoO2j1rdXGmV7AEuoAWGTzukU w5cb

6a05EqOMz52SWSyAGXgorcIKNDGFMLM8C0FAX0pKG heR5u3g EOXVIhOFGwquC7w v279eYMbpH1eCx gF6L9664HygUZuf8I2wMR RFtkaDNiAirEPDpMCKt774IDabXuby6o

Question 5:

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than4six appears on at least one die.

Answer

When a die is tossed two times, we obtain (6 × 6) = 36 number of observations. Let X be the random variable, which represents the number of successes.

Here, success refers to the number greater than4.
RTXixrfRvqyvdTtPoZqZ0nN1p04xwK1TOaGYn5N7L6fQ2n8TIQ5TezbRWuGDmLlN7n0Jffimh8UU7HLzd7p

P (X = 0) = P (number less than or equal to 4 on both the tosses) =

P (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)po54XjBYsQnSwypqc2 atbKWlVLQU GY37YEiJp8dPTHiJnmJEbjuBb

P (X = 2) = P (number greater than 4 on both the tosses)
GVjJPmeCd505I4GcK42vBrRj5BF azyNmxfeqfA3GJA x5ibFn94ubAL7Lu7iRFQYemriMirLevdnqMI1p1M 20FcjXdZp8mjemwajAtIaNd4u0hkPK6XcU1GECAblN9je0h6Go

Thus, the probability distribution is as follows.

X112

P (X)

RzEB9BMLMSQ2IGNIn3f6CEdt AY6eNh HdhxPL8lH0zN5F0OVTtYsISyl1r7sXFF0q aPG4 lyh0aDnhg1 94WX3J QYYZgP8E FA5 rVWStXXvIU5dBQpiPRhg40pAltBki3zU

RzEB9BMLMSQ2IGNIn3f6CEdt AY6eNh HdhxPL8lH0zN5F0OVTtYsISyl1r7sXFF0q aPG4 lyh0aDnhg1 94WX3J QYYZgP8E FA5 rVWStXXvIU5dBQpiPRhg40pAltBki3zU

s0YloICZd9ax mUv1sAQ QIc5cjnld21g6 k8GYaQ78ZtgKGlqWgrl4sCTD P c 6DQXNdUD80BGJaItR

(ii) Here, success means six appears on at least one die.

P (Y = 0) = P (six does not appear on any of the dice) P(Y=1)=P (six appears on atleast one of the dice)=

Thus, the required probability distribution is as follows.

Y01

P (Y)


eWfMtJhN3rbH8V x05LQUz3Ja3psvX61X8AX4NmNjV

Question 6:

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.

Answer

It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 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 non-defective and0defective)cEUcv1NOApPGxSQO7OchKGBNK21v08ClTglX3PAtislVMRA9RAe4zPZnYkUvrg et24eXKK335iX3sNX0ONusIK0ojyCkWVd AOGnZn2wjIyk1T

P (X = 1) = P (3 non-defective and1defective)

P (X = 2) = P (2 non-defective and 2defective)

P (X = 3) = P (1 non-defective and 3defective)

P (X = 4) = P (0 non-defective and4defective)

Therefore, the required probability distribution is as follows.

X01234

P (X)

UagNn5IcgAJ5eTFOoXPX78T5gCEcZHJ0U0YIrT 5XcAAa3r bFPezmOJZaFjIy5NE rVOHVutJsYzfFJTSz9ngjDsCvLrWTJL3xvOk

UagNn5IcgAJ5eTFOoXPX78T5gCEcZHJ0U0YIrT 5XcAAa3r bFPezmOJZaFjIy5NE rVOHVutJsYzfFJTSz9ngjDsCvLrWTJL3xvOk

0PnhWG8ZqSX5CXhKMuT JUMxKotKelhFMWv J0W1Sk3uawRc0wFLuqRMMmrAYMYYxDBB5IEKr8wR4mgYgywr0U iWcfSTdCfy8lDLqjlK2dW03bUmgH25 Srgr3wO2FE4sidSk

MOYgkRCYGc8SGp7l314uWXYbaOBCiRibHMnmBy LcYDANmTdU2aefbFQfR 6

ouOihjmB uJx6SeJk90IQbF3Locos nKX1eIToLfVJbdJLHYJxPCATTVyAOdK7A6edcTEDlUlJF776FUCEUQ

Question 7:

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.

Answer

Let the probability of getting a tail in the biased coin be x.

∴P (T) = x

⇒ P (H) = 3x

For a biased coin, P (T) + P (H) = 1
fUnoV cMujWjlLrliEbwKXp9AdHHvIhVinXDjQ7Kzmk9AHV bFsZJeEzsTlOWHLBo3ePvsnYYY0dqcla1lgHPiyO2xPd4cg fZ77w d IiNNJCMDEsPerUTa2WWJLPqe0rTTG2Y

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(notail)=P(H)×P(H)

P (X = 1) = P (one tail) = P (HT) + P (TH)
5Gy3j 1mP0r8o7rA5XdoEZzMdNYJVBBMFa95GSKFOZCHPglZP16tx2 UJ8I5B35O0OsVOTLVSeEk8oVGPyx54vB3vW5raEbUJukCR4xbQ9NsK7 gQp9Lq9D88i9E4NBibbUKAhA

P (X = 2) = P (two tails) = P(TT)

Therefore, the required probability distribution is as follows.

X012

P (X)

Q9F520y1AO6TFANq5 9Hw4e7hL4 6pEJa89LJVgeMtue8KwFUa6qjjllmsdk8JdkpYqg529uw2bJqtV6Yqd 64WqVxxQjJQ8a2EB3R9iB5Wqipx4raQieJ1Z L6aLtWGkHJQBA

eBOkWMZ90fgxRDIcPODlBVnJDoTVuhFs9Ly17xSfLNPreATtJpkVtUyvkbzDReWAP3MaebJA2etD8QZeiDbT71FRq c rW di6Jeh8OhAFZxRzIpKqEDca0kdLS0

6a05EqOMz52SWSyAGXgorcIKNDGFMLM8C0FAX0pKG heR5u3g EOXVIhOFGwquC7w v279eYMbpH1eCx gF6L9664HygUZuf8I2wMR RFtkaDNiAirEPDpMCKt774IDabXuby6o

Question 8:

A random variable X has the following probability distribution.

X01234567
P (X)0k2k2k3kk22k27k2 + k

Determine k

P (X <3)

P (X >6)

P (0 < X < 3)

Answer

(i) It is known that the sum of probabilities of a probability distribution of random variables is one.

k = − 1 is not possible as the probability of an event is never negative.

tAlvwpDDeJcDYJx2FSCJQM UT EYqaNneIEK6Z1Sg0 rcw1OBZpBauTHx6DFAB nnALZd SJ4 0zss8prXWhyWIGUv6

(ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)

LRAsiH1dURJjLP 6nhZen5hG32j34z zHBRqj pG vq5B6lcD2n4m7SeEQimIykzwi 2QI6VG2VXo08U6EfD7kCCD1JUUerJLnl5sLUEEaBo3THEuAjRFpw9216cI1R2YoSpZiQ

(iii) P (X > 6) = P (X = 7)

8zvH2JLgLvVCzV FIQumUXJmZT3fSj15yKOIxN6Z5jb uhTReaXOUxgjKZpGgKeJml7 G NAaM8EMF7dTJa 2AEdMWmUMr3gVu SHLFa7tgfX8LLCzB559ytCnGOThKUaQJ0 Q4

(iv) P (0 < X < 3) = P (X = 1) + P (X = 2)

2AcHyo1c7OIT6f6A1jNUWn7yX3ybu7IxXHzAEtsI7KSUyeySvO4SxURstMoLFUSDtiA6UjS7WMXBG0YAGsUFRqmZtK7Dt8lg sT6ADt2FUNCjy7wlDfZlRkPyl0yXCe7hWifRaE

*Question 9:

The random variable X has probability distribution P(X) of the following form, where k is some number:

(a) Determine the value of k.

(b) Find P(X < 2), P(X ≥ 2), P(X ≥ 2).

Answer

(a) It is known that the sum of probabilities of a probability distribution of random variables is one.

k + 2k + 3k + 0 = 1

⇒ 6k = 1

k =

(b) P(X < 2) = P(X = 0) + P(X = 1)

VRGvGNhOfDGhWN4V 9oz5sjegRzKoKnYrL3lgB9nI1pIYF5uFAwifaxaPtFMhbhXOZDQBWDAEixcMgulzZaY3Bs3XSjfNIkTNyuBmy
INf8T8JnHFOTa13tjYfcEH6LFWMpCJFsChvd5F PLJo8KlBxckgv5jlvwYViTaQQtlHxSQ3QjWijHd6TrCib JBqtf7K1PaiRoZGtX6I0 4p5pZksAj 1M7J67sbmw4TkSXl71E
HfRGgfN2bo4NRo9nvMTD1wH s59zrplukWwW0iNnNutHOTY6OSUWrYQFVg6ikCwr4AgFfR FCYYF8RCPA2Ge1V5jd22nJZk0nkARVYrHivp8XpnQb01ox PUH6wvX4ByZ8cZmw4

Question 10:

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

Answer

Let X denote the success of getting heads. Therefore, the sample space is

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

It can be seen that X can take the value of 0, 1, 2, or 3.
ZNad0Cr0Z7aIoFkAi85jyKlURPM5Ho1uiHrduTFLnTHVqUrwlaJQrVhsy570brAVcB GSqUpYW4tC2jOy3yOkN aG5bcE2O2M3 JKr22e8jIAa5m153bpVYCARI7nm0UI12pcS4

∴P (X = 1) = P (HHT) + P (HTH) + P (THH)
lvBg96ZE44SUcIyMxC7fIlX11S LJhZysUOgiwhCzknTmZmjFYzGoQsKJSRU I5oSnY

∴P(X = 2) = P (HHT) + P (HTH) + P (THH)

lvBg96ZE44SUcIyMxC7fIlX11S LJhZysUOgiwhCzknTmZmjFYzGoQsKJSRU I5oSnY

QdlMItr73ENvj0bqdvTpRSH7jB2JsvivgCqEWPuojpblJ4fKxHtyTKjDyFqILcuRT6V

Therefore, the required probability distribution is as follows.

X0123
P(X)
Y8vdJEiYBjY4hN6gFY07p5ZJQVCLYyNhDnb80N5RYZidjh1vGJ1y92W4z0 7yR3n7 ZN jGMGcBomr 26DvCOnfF2oC3kfrCK60054FMiXkxc8mZ8tHG F7zIu 5lTcTFpBkpXQ

V3BbFMyfk9TrQzBbq71Z2njZj ybMNyz0qPXiQqirpVmPpnGczOSPvoVDPhuJ7sGQOkA9QOXzJRnNG7tXk8zDQW8R znTdjOulFSD2AGCXOKUbwiFu5eema1dnDWyk6SipMCZsM

V3BbFMyfk9TrQzBbq71Z2njZj ybMNyz0qPXiQqirpVmPpnGczOSPvoVDPhuJ7sGQOkA9QOXzJRnNG7tXk8zDQW8R znTdjOulFSD2AGCXOKUbwiFu5eema1dnDWyk6SipMCZsM

Y8vdJEiYBjY4hN6gFY07p5ZJQVCLYyNhDnb80N5RYZidjh1vGJ1y92W4z0 7yR3n7 ZN jGMGcBomr 26DvCOnfF2oC3kfrCK60054FMiXkxc8mZ8tHG F7zIu 5lTcTFpBkpXQ

Mean of X E(X), µ= U5gVL3RdoBHAGMP6nhc5x5OCIhA3Zjzn3KrFAETgzzYYhwRVOvN8iKMg0P8npOaJx9zhAF8J8MxopyxvY9AIy5M hmVNKAc9Oc rqbO9TBpQQTVtLWTlcTNzDCgwURjN6zyefJg

Question 11:

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

Answer

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴P (X = 0) = P (not getting six on any of the dice) =

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)

WEzejAY3zb2Qik5UMr2tgHQcl4KdUpISlQBnli6 VklIpDKFTHjuq3b4JJba7EI0AWR30hMN2jLhXshUGWZUpYHB JgeHNSEzcp4ncjDgHOZ CgjIB58bFCyhvOY85M2vCsaV98

P (X = 2) = P (six on both the dice)=

Therefore, the required probability distribution is as follows.

X012
P(X)
sCLR8cz12SEHAR8cM2KxXCFUiz vu6ajdfe2trC6GwDKALYdKmivf2n6CyFXQpBWTyohRIhoel2vch5gurO0TmcrXE4gw04sZbotuVde4Iix2O47qtHRCB9hVj

xmnKi55yXLrjFagUK0 obaySTR1by4 yAz1yaAzhzGu9R1JAQQWZGO4kU2xRqz YcPr8ywXvuU7wlYBu9 vzSeH8dnc1Hf CKqp2TIPATYhKdrMj0VSs0gqdilRSTAGUkIpkWFo

Then, expectation of X = E(X)= Oe8bt0zKYU14TWXiGqnWdFDBlT8r7yP49BuP1uuVSPsh36L29wvn27WGnTxy9qS OhMjjc7VZFxv

Question 12:

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

Answer

The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X 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).
6Ot6ctZdG zxbwMfgUAfmFJsEpYGaSzU4OofyMMfn1xQd OfTWj1dMQodmQJad2d7eZJS6cv29jNLOP umG4UEovCSweWql

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).
A2W6 ZcSHDht0BQUOPTbHNeU oNktp5eLgPY FnQtrp3nev4e7ze3F raIMTmnxLhcROSH6pteiQwMXy7AL3qPC9j4aI1x0cfE7b5 731cyH743

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

zj kUuBa50DcPbuo57wJm Wv0HUg dNYiw2DUTBRnBFqVfLgJbDXdJSV1RWkGOxx0i31odC3XTPKJUfl0X1Av8zkYygd1UwFCFl B2dLxNpAYjHp0KFGHCR4lJr f4FUiLCeqBQ

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5,2), and (5, 1).
aeDA1heUdNC3jrdA4QrwMgS4 xZx4cJDRkYuxFlJGblNy6SftvCGJMwiKriUH34 U8wVWMBLhIOSD4827jAqAXkG9z5TjypXGov7uNXSNl

For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 4), (6,3), (6, 2), and (6, 1).
D0KTKutLJBjkPWuunq9ZhYHp P4gNFqZgzvQWMZR9SQ d 6N2CxLwnJ 9DjaNmAWxR3uzm7znktvQYGMb5V6T0UaQfkELN7Mqkkp1Ptz9GocBznZ1Q0qJ14kb H zNKV2mC pNs

Therefore, the required probability distribution is as follows.

X23456
P(X)
MfDPlVE16oO1E

NuAJDXwCbVxaUEy5yXgO6 IHdUFr3uAOP jEy7hw V8buaLZm Cxqlx13tklTdM6JheojD6RhF1ASOBS7HiZURkyCs8Qoi1QyJWXiTw UP9r6

ExsDbkz HBJWHxnnS7Vu5rB60LH71PGFhV2wee2oB5dxhAtF75u0BSV4H

BNrPtj8as11G3flbopQ6OaeUxXY6wi bOEYvWrTz3afoRPj2V6O3Qf9HaKQPrDOeo3f64lkxfYMJvoZ ZsuoT5M8zBGiaH1x8syWGo7ay565 HJJb17X9XsKStqv0s VR DDTDo

Cu eZF4NFcmoYLyzq8HeM4Qa35GL swOcDIGxY7ENh5Mb6NQlnRWS4 UT kwX1GPbrz9jKIsz2HH4i9UgeUipp B malHoi
tq2YmnuWX2RCv3Y4SyedpDciAwyRJVySL1ssSUD8XAnC62AAdcujwV2e3Ri2ABsva nres9M0j6BsiGLxOdrrC3G81TRRUUSXBM8Ikgv7XNmBsGmB3mg79sfnTmHquWo 2rDLvI

Question 13:

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

Answer

When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) =

P(X = 3) = P (1, 2) + P(2, 1) =D4HQwEoDOZuaM jeA5XFk9hvEIZpY aH9uXFntTOhBeIJNmLNEBOHhMnzCnMmv15SYRAiABQkpFYLgnRVyNYn4g8UmXUpXPhDzSdPabLvgnxl 99zEDRwy3MufjquSr qewDU24

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) =ceHpICLkZkpcZGLCmejqFosY0sOzKxB3LiAZQAwM6pcZmUvxaJEZXCBgq0vmTa3RdNoe4D s1IvQ054DnpjE S zGGq8UPeysUOlKHW7VGvYbh9GPOTzz7 Ef0WVe7YPLPzZGKA

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) =j5AYiRCKpw6DKX1y1FNDlnPlb6dKpy0OXBNRXaFJfvN3jUzaOPo u9ZfpGsIkRsqPuQmntUjb7E

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) =vKt1ENb36Ts8MOx0gJ zpKoyG3GSS4cOEJlRXXaHz y8JIsDe5EWr39U5PDnKi jnFQ34szMHkIvJMrVghh hbLSQPMqmesENuAQtU GpdRjt2tHnuC 1S3MZ0YuZtO6Ob2m5ww

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3)=

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4)=

P(X = 11) = P(5, 6) + P(6,5)= P(X = 12) = P(6, 6)=

Therefore, the required probability distribution is as follows.

X23456789101112
P(X)

aC7TQLu2vvqwygNcsIJ DEY0RVNTBd5atBLZOf5fsE2C6 Q0KiQZcs3r9TT3finabETTJYw5VBclwAh09c72DGPLiZzW0pKa3cbh1TnNXjC0XKtav aMTXmsbyEj3k9M92qUz4Q

APHw W ZQOc9aEmBumHc8bsve ll bRAfiNOjrg7 siKcJyhUd5VFgoFoMoyxZHsKbUsYtwPpuEWBvLDpztWif37Z DcCTtcLn6zCT9l l pCWU2xbLJon09KFEdmyb6OPgP aU

v7HA9xuGqk4rv2eJXR1nQgKKYlGW0 x3uGKC1cWYXB63tC dM6z3TB5HX q adkDYd6U1Xa8l8kpkivYadCfpPkKa3rWpL7cf3ZbPQJkCLs75SCuHL boE7bNQRlpy0DREWhj0I

vKt1ENb36Ts8MOx0gJ zpKoyG3GSS4cOEJlRXXaHz y8JIsDe5EWr39U5PDnKi jnFQ34szMHkIvJMrVghh hbLSQPMqmesENuAQtU GpdRjt2tHnuC 1S3MZ0YuZtO6Ob2m5ww

5bPPTN5VSaU4XVG1JaLujsKdF jI9bpn3EPryggi3dtXIOb1ghYCpqszSmnT4Fsqj2eo8y0QTnBxrontsa4MBkv1ruYGmdlyRBi2QX XW8LwnwNZB9E2n HNvrAsqtpAdwwWT4

vKt1ENb36Ts8MOx0gJ zpKoyG3GSS4cOEJlRXXaHz y8JIsDe5EWr39U5PDnKi jnFQ34szMHkIvJMrVghh hbLSQPMqmesENuAQtU GpdRjt2tHnuC 1S3MZ0YuZtO6Ob2m5ww

v7HA9xuGqk4rv2eJXR1nQgKKYlGW0 x3uGKC1cWYXB63tC dM6z3TB5HX q adkDYd6U1Xa8l8kpkivYadCfpPkKa3rWpL7cf3ZbPQJkCLs75SCuHL boE7bNQRlpy0DREWhj0I

APHw W ZQOc9aEmBumHc8bsve ll bRAfiNOjrg7 siKcJyhUd5VFgoFoMoyxZHsKbUsYtwPpuEWBvLDpztWif37Z DcCTtcLn6zCT9l l pCWU2xbLJon09KFEdmyb6OPgP aU

aC7TQLu2vvqwygNcsIJ DEY0RVNTBd5atBLZOf5fsE2C6 Q0KiQZcs3r9TT3finabETTJYw5VBclwAh09c72DGPLiZzW0pKa3cbh1TnNXjC0XKtav aMTXmsbyEj3k9M92qUz4Q

Dz7pbT1c8HBhhVJBS

qc84mIAUnxsDTrokabD6SbdnuBmYhEVlddVVaG9 N Q
kMvfvwtWGKSb7Z 1i wc5fTyeC7va85EyWUUrkiHKDLP45Up265tQ ICQNQW15kXqZnuGkeesHoz878weDlC48wsmvNhOf2YFkEVMpnMJoQoGaQ6Icuai3RdyuETb4vfOAO67rc

Question14:

A class has 15 students whose ages are14,17,15,14,21,17,19,20,16,18,20,17, 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.

Answer

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 isMfDPlVE16oO1E

.

The given information can be compiled in the frequency table as follows.

X1415161718192021
f21231231

P(X=14)=,P(X=15)=,P(X=16)=, P(X=16)=,

P(X=18)=,P(X=19)=,P(X=20)=, P(X=21)=

Therefore, the probability distribution of random variable X is as follows.

X1415161718192021
f
NuAJDXwCbVxaUEy5yXgO6 IHdUFr3uAOP jEy7hw V8buaLZm Cxqlx13tklTdM6JheojD6RhF1ASOBS7HiZURkyCs8Qoi1QyJWXiTw UP9r6

MfDPlVE16oO1E

NuAJDXwCbVxaUEy5yXgO6 IHdUFr3uAOP jEy7hw V8buaLZm Cxqlx13tklTdM6JheojD6RhF1ASOBS7HiZURkyCs8Qoi1QyJWXiTw UP9r6

XITI3cEoeWII2 ddFWfk 1fcHskTyEBNRBM9HBVD9443VL

MfDPlVE16oO1E

NuAJDXwCbVxaUEy5yXgO6 IHdUFr3uAOP jEy7hw V8buaLZm Cxqlx13tklTdM6JheojD6RhF1ASOBS7HiZURkyCs8Qoi1QyJWXiTw UP9r6

XITI3cEoeWII2 ddFWfk 1fcHskTyEBNRBM9HBVD9443VL

MfDPlVE16oO1E

Then, mean of X = E(X)
7ROhAmcXXPANLfw94OcklCCDGJ9kDzU CnwckH 5kAunqguOZZdNrl avWtSZrw6hlSfhN6CRmx5WwYN7FKRRbR2KPPMH96ksBP0n dg198BFTpZa8ZhnSQOQ9xWr8HgV7R svE

E(X2) =c a4 sPHR ZXU3ST9XinSYgGyp9SaWckCxOOAPdOuXPhdtGbi9WBAYzsyRzgGjzVuM7ONWaHdF8DCWeUVEM b fi4PVdrwLRqNT4enh3bgVIS52FRfpH6otIDOr8Hi m8OmwsA0

hzK0e5JHk0TgTlWJeogvuCCLC6Sn8FyJj0DGMAAKFppPP21MbSeK75EJEN0ASBpPYv7CnbPJOsai4H7kD54beH0xv 1w01PMcDr6yStcHvo7h9rjLpd6mbCLTusGUrVzgQgu G0

YRLQC3UW36pxL5oepeyR4N9 pJBuVdpg6 JqFa11kbksKs0w7TUwSohQgLfJjOdMq t5lGMan1P0h

*Question 15:

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).

Answer

It is given that P(X = 0) = 30%= mzu1Hs8Gjb HFsBhIc

Therefore, the probability distribution is as follows.

X01
P(X)0.30.7
kl5eYDU7eQiHkc9pDY4iteas8x08bptPY1JjJMjUkMdenTm xXaeWugy4bOcA3tY yJZEmi2Wk8IIrQqFMhMWF FMnIMYrtgUVdaAL hIv p RNrsIowT9JvS1pb0NK8HFZXD0s

jdcRvLZ4gxWsKznAj5FwTRnwflMsaPnsOVdiiGAQlRFwmGYLOZR2PWuNdwyXTeN4ZFOTGl7asmc 5Ia3Q

It is known that, Var (X)=

= 0.7 − (0.7)2

= 0.7 − 0.49

= 0.21

Question 16:

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)  

Answer

Let X be the random variable representing a number on the die. The total number of observations is six.ZtIS38qTZH3msr7kPnO69RdDeVXEXMAFG7upcenRkVsKcSx mzb GWI sAC6H8XMShqI7Uld3suuvfrbuLS6puWPlMeftcON4EXsttDUbGT6mSytfIKv8033YVzcH5RJqspdEu4

Therefore, the probability distribution is as follows.

X125
P(X)
1eG7LYJ7XOuYVLtN tS6so55YKbTyoPkuAfNrvOw GkoChVFLOO8RhOFSvJrsJXQNisvUToWStKaVfOEOl3ezYK28l3 zdwmE7963 ZXvNAKbZD1vPPYeoi 7ZwYTgX1mdK71eY

Cu eZF4NFcmoYLyzq8HeM4Qa35GL swOcDIGxY7ENh5Mb6NQlnRWS4 UT kwX1GPbrz9jKIsz2HH4i9UgeUipp B malHoi

5bPPTN5VSaU4XVG1JaLujsKdF jI9bpn3EPryggi3dtXIOb1ghYCpqszSmnT4Fsqj2eo8y0QTnBxrontsa4MBkv1ruYGmdlyRBi2QX XW8LwnwNZB9E2n HNvrAsqtpAdwwWT4

Mean = E(X) =

z7VfwndYDyBCcFbZSWVpq 0XiG61amjziztHr14XLvV1KQe2arxq1A2 8KQ rU wa8VlRJE21X1Sud7 GEeEeYbUU9PctpDsb89W ZRa7zobIfF2Up

The correct answer is B.

Question 17:

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) (B) (C) (D)

Answer

Let 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 non-ace cards.

∴P(X=0)=P(0aceand2non-acecards)=

P (X = 1) = P (1 ace and 1 non-ace cards)=k n7Yja2rAE7R8 ORqcoHohkWrXagGJRHurjOlKmMrRVWw7Gujczxz51eN8g6mAxJxIX4L 0XRibDrLt

P (X = 2) = P (2 ace and 0 non- ace cards)=

Thus, the probability distribution is as follows.

X012
P(X)
5y9q1QhxA QmNtotBiIuSfimwa9aYLZR8c8cwRAHZqL65 0qdH2QLzr7H SHmDQEG4kZ5hbzdCksPzVUXneK2 kWT


POHb8p85KExyR1I6ZK6vfZiCOxtZRH0tD5FN vfebgszQ0Yq tth 8SU0o2mXHeQ v0zk4B2O7O5UBtKAwGtOTUdG6iGnQi808ibjm 6UIMLHJFXM QsuVCTd5X5 QEa2V50VR8

Then, E(X) =Od8mQjRCUqKuO Ip6SCAPFNgDsB0BbvQxMvqWUPq2Mpg3ZX2ff0niqwlcK3 Nd9Q IlJaaEHbhlj f30sFpFZDDKZ0 WUHaNb PVbzRlX llE5J jdr4ygEeOU3M0V7R6HsJRp8

Therefore, the correct answer is D.

Exercise 13.5

*Question 1:

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? 

Answer

The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is, 7DdWDt3aCsnxV9lH JIi1T5lYuQ8qSze5Cj9ZRUl8zxLzntkFYXUDS zC7NwiVfP sDruKQdXOXJzuTuO zIr0pSxklcxl9oPj7uqs Jqsh7 7aigwCwbbVTHisVh7FQ1SY7Rcc

X has a binomial distribution.

Therefore, P (X = x)= G7nrPbgniprmB3HcgsPAXmC2zdwXPLl9XqFWAy gdfv8geTySabdKDtye7HWzR3hFM0rt duHbd7KThHglYEnfXjSzuUATTDs2
t7qtN

P (5 successes) = P (X =5)

P(at least 5 successes) = P(X ≥5)

ebnVcS2XmG3kpX9raS6FBkRmAeBCtOCg3tYSjPL5SMmYR2cRDTAbssJ7rwUDqziBc0RMA584t9iguH iM0DOj7RhGdDGkJ9HxjcKvUhdmEeDhhxwFbquPENzV7CXtKSV 5lHCU4

P (at most 5 successes) = P(X ≤5)
GwzqiPwf9L7LWup7fdlQrZdR 8 aOF3y2ki9XWDZwHZpXskaKueKM3ph1nmJllc9J2imUU5XMufJHTkycC5Q5of59wXuV7MdEWAav5hjX DAEOrSfzXlzpAk11CFCVT0fJHwYR0

Question 2:

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

Answer

The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is
8i0LBTLzhmpwqozYFy6D3CpMNVcprIHIwrnp9vhiaKPqYz9x0g4PZU csakbKGvbksyGIZ jJ8qrHmQosgDsKXBKIKhE2eB3qy706cCYtWb4SZcAb7KUD1b0t3ammkWTpM3GNS8ht0Z7w oKtCSCsg4acB3noUGTKG0X lgHoGTQd1qQHGNh7nAC63zWdRczIa0QWYcaEe1AkgGrHUTIS9go4bDR7Vh6Lu9YjFh61gKwkLm9QTV

Clearly, X has the binomial distribution with n = 4,Y mV5AVV4uoPLuVM5vF6 y63GpZezcVy7buMZjEPCm4yV8hfUUDXMHhdRUtmCDWPcN4fjv9RNWcol4ZNznD5dD
o dLHxbOQJ9O012N4on JTSKLkH3r4QyJ1GcQLu f05lIJgBlGuWMBs9RW6elPIcAFT9yFOxqcR0BbQdO0bHT4Mg1

PXVsJCmUWpHMnsLqQy7JJ8usaCj1Jr8WjN6Q

∴ P (2 successes) = P (X = 2)

Question 3:

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?

Answer

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.hnci6YiqTLpr8 rTXbQO4fU EMzkL47JheEULkUDAK Y4ZhQYroIZTqSrBPbvRZfMZ5d76lNHRrPDMxZ mQKw4SIwk3uqx1hm8npoqe1N31O5vVaZbPbyN TQwNM4kvX YdWg

X has a binomial distribution with n =10andP(X = x)=

P (not more than 1 defective item) = P (X ≤ 1)

izi9HYLQUIZJOf I6Sndn8 2r3 tvddPAr2uFQ9GZgGs86BaHXegZmb Et7oLhlkW9njuMshpk3x gnsTUqezsCWzqU7wyaH82sdFZ2UXDd3ulgIZq PVK1Yj6eQxuhpgW Xts
8AxpogpwlSLpr3tYTD bz23o6mE5oz iouRCL4b1tfnADnBBgeMnxrKErj0ySWACf XG381ypcEFXfX B8IjpxFv73r

Question 4:

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? 

Answer

Let X represent 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.

X has a binomial distribution with n = 5and nE39awauTc6dreRrWOk0DAxkHCRRb7Ujq5YAu2IxYmafXOdBN17pOpPDnz7afoNUQIT5dIEpRxZvvTtyyq 2eCHbKZ9v2ZglJR05r CBVXyng4vYmnOWnZn KI3HCPc9d fimC0

0IQxuWsoyLBgerBUgRP7eP8fPFp7H8GoYnpLDAFDgC9oOVkz umHQ9PuBHopIdY6SJdL1BFOYISNveeZNSMi4vwCnflHQLTN7geDjoAd0iPEu Nl BlIQCvyTYOX8DLBdzbBVjc

P (all five cards are spades) = P(X =5)

njt30wUSOj Kqf4sztJyS66KL4Q2KUHuWH1zDgi YW4 H8d6Hrsn5fCiw4j3 GOROuf2MiJmiqL11d6ajtV7HoOXYI7WjsCKubSSCqD1E9FsHiHyMQNZ99mAGke7AZRBTXbMTXQ

P (only 3 cards are spades) = P(X =3)
pHhDcIUsaHH2eMXjnNgjuujRvHoKR54J4HKg8KKevfGZc26ZLwQF0xOiTPKcOsiSBySTYpvhgzRy1BxL8uFP4Xn7uTiZUUii3RYNy kArUd cZaEYHZrLpfOO6R6ave z3FcaCI

P (none is a spade) = P(X =0)
Dk9 NBrGGQKNM4 8r32GrtebD5i1OlXmEInVaEUOcxPd9kzlfXFuhV4CgX amsvzz5KffExJWmzMREfYa5nEOQov0qBvKPSMEStQklPw FO3apdzjrZKgZoyp8Gdx2POj 1iUu4

Question 5:

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.

Answer

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.

It is given that, p = 0.05

tttqUu5kp4ZeDlhqzDevLQkGIEQDJRqDI2FDCMW6zJRQOEpETaW94TlhsSMTeyvcBYmp0fENtCKpywMdaWCjr0YHJ9t5h1kTtc4keACEaZt1VX8hGjq9ubF2PGxa0Di w55 4UI

X has a binomial distribution with n = 5 and p = 0.05
bdHLu lQq4yg3DIa

P (none) = P(X =0)
kiFIIvIfzoeLq2rDnCnakhqdLzr00lqmFoZl4O1e04yKhZvIrUyXyhbsUcCqznryeM9TsTJnJeIANKd W4PtgbHX6krqqf0OEJiYBUhAjLh0Yj44AteRnCn77YPEBalQqJLEHeY

P (not more than one) = P(X ≤1)
5x1 kUEG vumwI8vFi2ZQ7Yb7IxozowcXD0tbHc0PsqkrEE8qdevp7AObYviQY8YH9Lw csEa2rAnzjwXobrCZpbsMuNXV860Gsvf633hN38XamAKkrzVI8rvNlqho2h4Bh2Atw

P (more than 1) = P(X >1)
kZ7hnJO1IggB c5fbSMUzm ioFBhM2FuOvN1mTUMgW EJ2wGyK3moqdr55s7TOefIFPtFVCM4svd34Z0FNAm6mWzkTbIVl 2dYfyzWSrbll0A1loBgw4sUYGX3hgq4B cgD NUU

P (at least one) = P(X ≥1)

Question 6:

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?

Answer

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 iAbJc7oaS24IL38jpS4Yhmg 4jHwiPD

P (none marked with 0) = P (X = 0)
2lURvzk3ksa6LzytSveWyfE ftHZN7yeBpU6oGMfDL0f2wnDYi87SU7hFGc0HygF82AQb5OZFTjwD6RjjU SabPzFSC234wgatpsAfGzG8 MKeoL2T2udndPyJ XU4mXia74DA

Question 7:

In an examination, 20 questions of true-false type are asked. Suppose a student tosses affair 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 atleast 12 questions correctly.

Answer

Let X represent the number of correctly answered questions out of 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 are Bernoulli trials.

p =
FfsrZpVAt792j3UoBdTugF999HfDxcX7yKp7v27gUHsBW S1QDfmsuhy2seKBbdERJLTQKp1ogiadq4QKIuIPW8PsM2ndSASpgwZGd6IDts4g 0O18B5cLIs3H6Kwf2NJzw4Z5M

X has a binomial distribution with n = 20 and p=

P (at least 12 questions answered correctly) = P(X ≥ 12)
zuEpDBrx5uQJsIhI j07QKQOg7Wu8mwR4ycwhUsXAplsxzI8I3bjNETq1qakky4NkMJ0j7tQqRCfD1 aHB9djlgxPOWCyVO7a40 I jE7PrNylNHtn

Question 8:

Suppose X has a binomial distribution . Show that X=3 is the most likely outcome.

Answer

X is the random variable whose binomial distribution is

It can be seen that P(X=x) will be maximum, if xuksebcleCRGQmAgWkdt0o7vu8YtO4 yiyX5dVnpkBQMuT0g6D6yjPCkYzbG1ftQXTpFsyeFvNJnCpnCNe 3rHbABSULF will be maximum.
a slerjjHNzdpB2H5KaZ0o0Qw1FgE4t9nKQv4FbRVE51wJNIIAJvZPzIXZKbV0 aS6XOdEaEcdKOrlGZu1YavYu4oP9wCGdhin5mc0E07sZt vIe99WAtzViKkD34AbuVp8jsYo

The value of ciWwnkoNWEVrmYw3EMWlz0dqoUCrhLDdXvKn3QeiICaooIndyqjgxYGKuEFVAx czXw8wO2 Y8Cct4xL97q9dodD33sXWxsQW2WuL2JBFZOiKrPJSV7Llh kpLtJCPkb dWDfKA is maximum. Therefore, for x=3, P(X=x) is maximum. Thus, X = 3 is the most likely outcome.

Question 9:

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?

Answer

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, piZvFUcpONdAkemM2QW Id6O79lPvdQxmVFsUcPH m93cxxGYzOdDSiHXc3F3kv2BN4TOKUau 8OjKtFGY47EPXImznFhtVi8BOJvcJXw976yi5LCyBChxyHlLvo
b y8FRlNR85H hmHJSWJkuCJQnJxEKTySOlyye6YJeeGY9urT6HKHa

Clearly, X has a binomial distribution with n = 5 and p

G7N awL8FLim fp3uU7SgUB4iTYYRmx5QH8FlBOT9nh4AKSxfabj8HmPUFTZvU4fezDxML6hsBye6L4zNjCEcNbA9KKOd3i1eW w3t9MRBqAh1I0z28WXOJTtlIcgnXuvS2PSEA

P (guessing more than 4 correct answers) = P(X ≥ 4)

Question 10:

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

Answer

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, X has a binomial distribution with n = 50 and

e0Be 0K3rQCTLMVpe52QCLgEhWiiDyRMPYVQv SHKQul0tEpSL6WP0rlD 0p1J9eVi84dHL2PPsfvVZ43q4euwHGRdGqJ6A3sHWdGDipsYi1q GE2L8pqJCN Xk3umv7JN JvBQ

P (winning at least once) = P (X ≥1)

P (winning exactly once) = P(X =1)

GxFX hHCXOUSZ2H2bTbIgn3m1TUfcNxAt v3LffcBxkauIFZAEL7JIxOl5Pec2ZdtoFTDNY9emBM53jqxSFi xmYuYWELkFBO4blf8pp7wcC V09EvloHH5cCH0uRPB6lrL8fQc

P (at least twice) = P(X ≥ 2)

*Question 11:

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

Answer

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

ht0Z7w oKtCSCsg4acB3noUGTKG0X lgHoGTQd1qQHGNh7nAC63zWdRczIa0QWYcaEe1AkgGrHUTIS9go4bDR7Vh6Lu9YjFh61gKwkLm9QTV

Clearly, X has the probability distribution with n = 7 and p

Xz6ly8PHH9WTj5GhUHFVBnymg3HnsRr JE7lQAfwYQ3HBd DEZ3HTv18WwSxrsXD7 SNUb6z2DLeORh7 DZRCv

P (getting 5 exactly twice) = P(X = 2)

Question 12:

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

Answer

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

ht0Z7w oKtCSCsg4acB3noUGTKG0X lgHoGTQd1qQHGNh7nAC63zWdRczIa0QWYcaEe1AkgGrHUTIS9go4bDR7Vh6Lu9YjFh61gKwkLm9QTV

Clearly, X has a binomial distribution with n = 6
HGYQabKgd1qsC14hMKJyTJjz44tUlFPFcT f LWz8WB0vlDygEpO4Fl7qEEY7LwDyEmoNlULERyckvtCjnn4zYLrHLWaFZEW0JmCMwfGpfU1j3fY0bRJkFDR1w8HUhdG MiZ3tw

P (at most 2 sixes) = P(X ≤ 2)

xWXqw3cR0kaVi37dJL8OACyWazHCs0oJL5SMdFVkTRiWzOrPjRBphcTRSTvexBYw5BTjKp4wUbRmIxWoLDpLEdcBn8D7JVyZ M8jZ0cn2r2A6AIv6 5 i5dVFbTF6DiWXgymNDk

Question 13:

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?

Answer

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.
XVN4NmmswsTi l8sYK0WzJ7 ZuDMjmmcgtlrOILOeoP48kzJSfL0E75ECop6iDQ1iJcAK6PYftAUlVNgSeVQ0y9LAe5VhBc3wY5gxsARx5HQVHCkhCNyPHcxBi2HS9B3LWfBr k

Clearly, X has a binomial distribution with n = 12 and p = 10% =

4sVi4hrRfkBSKoLuqacIvXDtanKRJZ331WjNyKks9il2SAbwrj0YcRpzSxOmU3OTVX uOZCTSzX0MDyGIeuF2 R6rW0cD0H rrNZdmaZMddyp8RibVYWPWvfB3zWlVWu9n2jV94

P (selecting 9 defective articles) =DB0eN1EX14O6YhVyHyU2HJ6RTwdGX o7R6sLKBZ3b9xmAVr1

Question 14:

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

(A) 10−1

(B)

(C)

(D)

Answer

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

Clearly, X has a binomial distribution with n = 5 and

P (none of the bulbs is defective) = P(X = 0)

pkwB0GbLVOh4ttGyQGF2BMAENAKKr7juTO7eEn2JJuCrSc1HzcVid5mlgcZ
XKImobrnEl91f0 fR3qznvMyiKmW kYT08T1DyV6PIG4qC5yQoCigCjWQ1sUcnpn2EsmRHNFYvuqrmUacpEjFsuLvB9s4xwrRgyL9nDlRnWRxg7nms

The correct answer is C.

Question 15:

The probability that a student is not a swimmer is . Then the probability that out of five students, four are swimmers is

(A) (B)

(C) (D) None of these 

Answer

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 6eXgYJABBA3Vuca6qOL3RnQasYXmTbUNKgQow NMXKbkSF90IeFoAhnPMBqmMTqZolyfydEVgiAXzMBqC8urMMRcWGyWsQe6bzV NSPYSTvsJuYCFD1YTXd4kYnfjgCUHQ7 ES4

Clearly, X has a binomial distribution with n = 5 and

l2JYP44otyQZz12J LxyFb1mzX5e9eRNp9kkQNKRtXPgt1OumH1pPkj5F3UGaNbenx A7Nk45ZDvGtFzWL64Uj Uy15kuP1MPU8FjsEz1HhBlxEMnrsHuTY 4mjL9iCP r6a3IQ

P (four students are swimmers) = P(X=4)Therefore, the correct answer is A.

Miscellaneous Solutions

Question 1:

A and B are two events such that P (A) ≠ 0. Find P (B|A), if

(i) A is a subset of B (ii) A ∩ B = Φ 

Answer

It is given that, P (A) ≠ 0

(i) A is a subset of B.

r2IuYYY0XCDnb0FvYZ

(ii)

JgZgOCNa3TG7eA Q2NTnD9lTPJZMrwi8xw1aEohMcJ9nhI6u6kfz H6P8hDHj57REFvnFYRdFF C0gcX2Pan1H715GwbHGL YpN zE9Nr3gmW8xP88JsUumGo7uGxYChL

Question 2:

A couple has two children,

(i) Find the probability that both children are males, if it is known that atleast 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.

Answer

If a couple has two children, then the sample space is S = {(b, b), (b, g), (g, b), (g, g)}

Let E and F respectively denote the events that both children are males and atleast one of the children is a male.

LmovrVroenWcXi3YFZ N4Dn5CQx6nv4pwehC8hAKa0Cb7Y35PoxcVORCcTpKmODEp CnwdDqgPoCngKbMynrPx3 wkheeiwQqCAg6eNXIaeSjOekTBMXPxPPygwTaDdkb kOyDM
4 esyLhVo3yy8IPMOc

Let A and B respectively denote the events that both children are females and the elder child is a female.

uw wrJ4Kd2YykKtD3gRROIQajeAxoAQa8a5fNWqJQTfKq2db8DphmiJUJF33gBtUw6myuVB5ryPsid Z1IWJyA6vuuKfXKe7cwOj2VqkOpCWJVnXcLFH5yrKqDyF0Q IC1BQyi8

Question 3:

Suppose that 5% of men and 0.25% of women have grey hair. A 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. 

Answer

It is given that 5% of men and 0.25% of women have grey hair. Therefore, percentage of people with grey hair = (5 + 0.25) % = 5.25%

∴ Probability that the selected haired person is a male

Question 4:

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?

Answer

A person can be either right-handed or left-handed. It is given that 90% of the people are right-handed.SLtbOR6l97ZZDlhHY4jj6onH1UVmP doqm28qlAr2ae0xhiyoQ3IkTmCve9feAzm eCWeiIMSs2jZ29 dHv276Ql 593yd5oOkths9yReK EP 0HyDdBqG24Q5UuMDPsqOrsy9E

Using binomial distribution, the probability that more than 6 people are right-handed is given by,

kMoAaOZ5nQA9M3TWZLZVTwDCbzVoW3f9sApk2AEeyGj E

Therefore, the probability that at most 6 people are right-handed

= 1 − P (more than 6 are right-handed)
8p9BrjUX5tk6el2dL4ZT1UW73WhV81Cx40MJWJv5iMMFSwhcHLFCzL74zGs ji6 jd1dTPjrBphSsFQNzjJE6nVjhVtn rjQ2NAGP8OYufdeyeKFttfF CsVrsWobcbGPh6 dq8

Question 5:

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 (i) that 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. 

Answer

Total number of balls in the urn =25 Balls bearing mark ‘X’ =10

Balls bearing mark ‘Y’ =15

p = P (ball bearing mark ‘X’) =

q = P (ball bearing mark ‘Y’) =

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials. Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

Clearly, Z has a binomial distribution with n = 6 and p = .

∴ P (Z = z)=

P (all will bear ‘X’ mark) = P (Z = 0)=

P (not more than 2 bear ‘Y’ mark) = P (Z ≤2)

= P (Z = 0) + P (Z = 1) + P (Z = 2)
8CpMfD9AyGEcKf LEt36hxso1AhUHC9TJPiNMrwDWECjMnYFoXR9pSaU0h5ChEyEmo6anOk9H5 ZBG2ZcKSVh0ou5CxQA7i0Nwe1on4QGJkquCNhuxLJdGs7Hfkes09L MD FMQ

P (at least one ball bears ‘Y’ mark) = P (Z ≥ 1) = 1 − P (Z =0)
GairBojcvp4YZBjxQqr GwwcUBE7nng5DMXHO88LIasmZIiFqkcOt4Ftq096ozF16G4hBxxobf2sVFxhT ou9zxT0n4TC4 7 oUd 3 sxQSojEUPfgji Iz2ptL WE1uY2IbVr0

P (equal number of balls with ‘X’ mark and ‘Y’ mark) = P (Z =3)
BCgwe5aFgDV3pukh9SDkkHGgSzaCrNlMxvfA rGRdTuKX4bWzRYexbLTsrhEmcenBvi9aiTWuy LxChLqLLyYprj1jvUQlkqmSCE6L

=

Question 6:

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

Answer

Let p and q respectively be the probabilities that the player will clear and knock down the hurdle.kRuUnamSK y80Jsc5Yr2XTsjNBdAdBgWnwceYtMdLfVWeZ5olxGE79dyZ6yxZgowno1XB5pg8vE4uIsOzQAHtOTwE5SDvc0Kzqn g5yW9CQUZjpJO W469CECp TL8db8lDb7NY

Let X be the random variable that represents the number of times the player will knock down the hurdle.

Therefore, by binomial distribution, we obtain

P (X = x)= YfX aMGhDt7WlDR77Oum8Sx GkK6S yFs5K7wc4hsoHPY4IFG8dlma6

P (player knocking down less than 2 hurdles) = P (X < 2)

= P (X = 0) + P (X = 1)

=6rFFjJA8sGN6aIFGMucuRtseOjnTuUR hCphu1TGC5q MPHZ h5FCwNT8OALW1uj8MmKEF 6 Hvifzr33YnrZzv9kZBCS

j62RYsfifb1aCEN3J DkLZZKZqcOpA59tIH4bv GC6zfJ73yiqU Et9E zViY7W4ElipHNhzslRsbuHZYJK5Iy XK07MrUqt911eboqO7fLEZVc1j4 ukf4Pzgavu4TPrSYZrsc

Question 7:

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.

Answer

The probability of getting a six in a throw of die is and not getting a six is . Let

The probability that the 2 sixes come in the first five throws of the die is

T0703 wTtINq 4LfbjdGbIwhP5Iqq4uuTptDNtlaGNndpid21d oK841KIEI RNW1wmOYt2qK4oEITnyKwgAezLtfRTgAynwU46G9E ham4pO8MiMY5DY4eVW8gJm8oahd5ArzE

∴ Probability that third six comes in the sixth throw =lNcrRml60Sq7HoogTwYSDpsiuYKmoqjqLNDplYqPDkEElpIHf8xTv4Q1nrolBtqSboLqIbClbu rlf5De auki3rcH1cN8fen5zmF6r5SMyzO1abNGDam8BSH5bWz zk9P87c N3tWeWcKcdqwdorUC

Question 8:

If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays? 

Answer

In a leap year, there are 366 days i.e., 52 weeks and 2 days. In 52 weeks, there are 52 Tuesdays.

Therefore, the probability that the leap year will contain 53 Tuesdays is equal to 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

Total number of cases = 7 Favourable cases = 2

∴Probability that a leap year will have 53 Tuesdays =

Question 9:

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.

Answer

The probability of success is twice the probability of failure. Let the probability of failure be x.

∴ Probability of success = 2xtKgmt2O

Let p = and q =

Let X be the random variable that represents the number of successes in six trials. By binomial distribution, we obtain

P (X = x) = YfX aMGhDt7WlDR77Oum8Sx GkK6S yFs5K7wc4hsoHPY4IFG8dlma6

Probability of at least 4 successes = P (X ≥ 4)

= P (X = 4) + P (X = 5) + P (X = 6)
UwGm8eneFJtWalXQ0SbIYLuPmNZkygSJGiTW2ciqHRmgkHtMsGGHfnMAveYL0YhYXRWb7szMTo9VQ8LN04IBqnNbi2PyQUDhlocWh4B 2IIVFL7S2lGkA6QUuoz G10eOWvG44U

Question 10:

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

Answer

Let the man toss the coin n times. The n tosses are n Bernoulli trials.

Probability (p) of getting a head at the toss of a coin is .

p = q =

It is given that,

P (getting at least one head) > P (x ≥ 1) > 0.9YzG gkIkMRGl6RCfq XzEQskIX8HQX3st6s9PVz RyEKVWad BUUh22rwvQvzwcLl6vlPRB1xr2q44VMSfVnZTSyqIKiMFhI9ki7GOFMEo3x TN099pgzIvFvzozmcSVWDCCaw

∴ 1 − P (x = 0) > 0.9

The minimum value of n that satisfies the given inequality is 4. Thus, the man should toss the coin 4 or more than 4times.

Question 11:

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.

Answer

In a throw of a die, the probability of getting a six is and the probability of not getting a 6 is

Three cases can occur.

If he gets a six in the first throw, then the required probability is .

Amount he will receive = Re 1

If he does not get a six in the first throw and gets a six in the second throw, then

Probability =

Amount he will receive = −Re 1 + Re 1 = 0

If he does not get a six in the first two throws and gets a six in the third throw, then probability =

 Amount he will receive = −Re 1 − Re 1 + Re 1 = −1

Expected value he can win

n70hbdS9xIpB4QQc7Ng2xAZrlcag YQAH8KQf4iqH0ANO tIoTv4IFPqYFkH0 4uRGFj72gspRYcQsRgkqJr4uS m4CPUTLk XRXJYlzi93ZYgBKXfMJ7kStzfnX5DL2K2o080A

Question 12:

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

BoxMarble colour
RedWhiteBlack
A163
B622
C811
D064

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? 

Answer

Let R be the event of drawing the red marble.

Let EA, EB, and EC respectively denote the events of selecting the box A, B, and C. Total number of marbles = 40

Number of red marbles = 15
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Probability of drawing the red marble from box A is given by P (EA|R).
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Probability that the red marble is from box B is P(EB|R).
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Probability that the red marble is from box C is P(EC|R).
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*Question 13:

Assume that the chances of the patient having a heart attack are 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?

Answer

Let A, E1, and E2 respectively denote the events that a person has a heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.dwoqURNAMQX8Dcryt9x5HlevwU7tZq WW mcf8Qu32LsLdBkcnQlEmfahw134TXGoNxyPbgsf0rZ3YHjv1ejD0gpI8Q rrBghlcIMC1xpBKV eAxPaMZyzxtkbYbIyHJGE P3IcWyTpAbSmX9h ckmYXNpkoAhPy9ePK2W9d0k0eipQsGRLCk2ha0 dZB3Ha2CphMdEWR4TWYoZx6lN13 HXHben4PRTA3rt4M64JPU1j3lUVw wH4ZzkrrvYz3C5SAIL4ls sHHas

Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by P (E1|A).oab7fM58oml2w6 U42n4czvy9S9AEjV6NDwG88OYTzELaOwIDgg PIEVpgExMWklDNsUmtjslSv 8A H62jKAjn baPm5bvS5wZrz4z8RC U3nsLdkkH9zgjN2rXMWmcA678vTs

Question 14:

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  .

Answer

The total number of determinants of second order with each element being 0 or 1 is (2)4= 16
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The value of determinant is positive in the following   cases.

∴Required probability=

Question 15:

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

P(A fails| B has failed) (ii) P(A fails alone)

Answer

Let the event in which A fails and B fails be denoted by EA and EB. P (EA) = 0.2

P (EA∴ EB) = 0.15

P (B fails alone) = P (EB) − P (EA∴ EB)

∴ 0.15 = P (EB) − 0.15

∴ P (EB) = 0.3

(i)KDexo7lOpFzmAt91u Ao Vr 1kvySv9TEjuWpWKUPqtGzOUuKJtcaSLj4zsqYL eif7wLuz6uin4kFxmvNnfxj7DGL etUJbN S2 RbEumN4E46mcIsKKjIqv18RoavMViWFfV4

P (A fails alone) = P (EA) − P (EA∴EB)

= 0.2 − 0.15

= 0.05

Question 16:

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. 

Answer

LetE1andE2respectivelydenotetheeventsthataredballistransferredfrombagItoII and a black ball is transferred from bag I to II.

Let A be the event that the ball drawn is red. When a red ball is transferred from bag I to II,

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When a black ball is transferred from bag I to II,
7TF3rgljJiQWeHn9EqTeXKbBKQ624Ckqf7o16dTa2 9jVY8fmzQlMCSDzoKPZTg3 DozeMR5s4aZ3Xyl3LOuh6KWciAqCfy6fJWgixnWme sGY1ROTvclSTgpimr8RAmzDD5OY

Question 17:

If A and B are two events such that P (A) ≠ 0 and P(B|A) = 1, then.

(A) A C B

(B) B C A

(C) B =Φ

(D) A = Φ 

Answer

P (A) ≠ 0and LQlABPziyejzgjqV9hk9r16Kbtcc44f9j2TyPCdLdjuvVc6hey9FSbmcSS0bNcN5uj1DJuvyOHUWlbU pB2I2UemPWWUHWDjFWH1iG3o WoerOwOk tLiAJsX9lh6wEcNjGtk9k

Thus, the correct answer is A.

Question 18:

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) 

Answer

RlCrbhpYUZzAUkw4hh9RuSTbrZRy2mLokvg3H7wS3VrZk liPFdPiQSHKJoVR0L8SIpEoDPBAfS37eEIs8Zxaln0rCwuop 1XuGs 63OI yCrpDMTqwrG 0 XeNI6XuBhpTPWuQ

Thus, the correct answer is C.

Question 19:

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

Answer
ILralM0ibDzhDe4KXM1OmsRbQY Z7xb4mUgNzZzYXa8BSVkgW nofn24BNBNBSoXfeqqJN

Thus, the correct answer is B.

NCERT Solutions For Class 12 Math Chapter 13 – Probability

We address random experiments, sample space, and other events related with the many experiments in the traditional approach to probability. We are more familiar with the word ‘chance’ than the word ‘probability’ in our everyday lives. Because mathematics is all about measuring things, probability theory basically measures the possibilities of occurrences occurring or not occurring. In probability, there are various sorts of events.

Topics to study in Probability Class 12 Math

Section no.Topics
13.1Introduction
13.2Conditional Probability
13.2.1Properties of Conditional Probability
13.3Multiplication Theorem of Probability
13.4Independent Events
13.5Bayes’ Theorem
13.5.1Partition of a sample space
13.5.2Theorem of total probability
13.6Random Variables and its Probability Distributions
13.6.1Probability distribution of a random variable
13.6.213.6.2 Mean of a random variable
13.6.3Variance of a random variable
13.7Bernoulli Trials and Binomial Distribution
13.7.1Bernoulli trials
13.7.2Binomial Distribution

Weightage of Probability class 12 Math in CBSE Exam

ChaptersMarks
Probability7 Marks

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Related Links

NCERT Solution for Class XIIth Maths Chapter 1 Relations and FunctionNCERT Solution for Class XIIth Maths Chapter 4 Determinants
NCERT Solution for Class XIIth Maths Chapter 2 Inverse TrigonometryNCERT Solution for Class XIIth Maths Chapter 5 Continuity and Differentiability
NCERT Solution for Class XIIth Maths Chapter 3 MatricesNCERT Solution for Class XIIth Maths Chapter 6 Applications of Derivatives

Conclusion NCERT Solution for Class 12 Mathematics Chapter 13, “Probability,” is an important resource for students who want to develop a strong understanding of probability theory. The solutions provide a clear and concise explanation of various concepts and principles of probability, including conditional probability, Bayes’ theorem, random variables, probability distributions, and mathematical expectation. The step-by-step solutions and solved examples help students to improve their problem-solving skills and gain confidence in tackling complex probability problems. Additionally, the solutions are designed in accordance with the latest CBSE syllabus, making them useful for students preparing for board exams or other competitive exams like JEE and NEET. Overall, NCERT Solution for Class 12 Mathematics Chapter 13 is an excellent study material that can help students to excel in mathematics and build a solid foundation in probability theory.

Who invented probability?

The mathematical methods of probability arose in the 1560s investigations of Gerolamo Cardano (not published for another 100 years), and then in the correspondence of Pierre de Fermat and Blaise Pascal (1654) on topics like the fair division of the stake in an intermittent game of chance.

What is probability in real life?

Weather forecasting is perhaps the most common real-life use of probability. Weather forecasters use probability to determine the likelihood of rain, snow, clouds, and other precipitation on a given day in a specific location.

What are the 3 types of probability?

There are three major types of probabilities:

  • Theoretical Probability.
  • Experimental Probability.
  • Axiomatic Probability.

What are the four rules of probability?

What are the 4 Laws of Probability?

  • Addition rule: P(A or B) = P(A) + P(B) – P(A and B)
  • Multiplication rule: P(A and B) = P(A) . P(B/A)
  • The sum of the probabilities of all possible outcomes = 1.
  • Complementary law:

What is axiomatic probability?

If candidate A wins, candidate B will be unable to win the election. The third axiom of probability asserts that if A and B are mutually incompatible events, then P (A1 ∪ A2) = P (A1) + P (A2).

What is the basic probability?

A probability is a number that expresses the chance or likelihood of an event occurring. Probabilities can be stated as proportions ranging from 0 to 1, as well as percentages ranging from 0% to 100%.

How do you solve for probability?

Subtract the total number of events from the total number of possible outcomes. This will offer us the chance that a single event will occur. When you roll a 3 on a die, the number of events is 1 (since each die only has one 3), and the number of outcomes is 6.

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