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Chapter 13 · Class 12 Mathematics

Probability

6 exercises91 questions solved
Exercise 13.1Conditional Probability
Q1

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

Solution

P(E|F)=P(E∩F)/P(F)=0.2/0.3=2/3. P(F|E)=P(E∩F)/P(E)=0.2/0.6=1/3.
Q2

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

Solution

P(A|B)=P(A∩B)/P(B)=0.32/0.5=0.64.
Q3

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

Solution

P(A∩B)=P(B|A)·P(A)=0.4×0.8=0.32. P(A|B)=0.32/0.5=0.64. P(A∪B)=0.8+0.5−0.32=0.98.
Q4

Evaluate P(A∪B) if 2P(A)=P(B)=5/13 and P(A|B)=2/5.

Solution

P(B)=5/13, P(A)=5/26. P(A∩B)=P(A|B)·P(B)=2/5·5/13=2/13. P(A∪B)=5/26+5/13−2/13=5/26+10/26−4/26=11/26.
Q5

If P(A)=6/11, P(B)=5/11, P(A∪B)=7/11, find P(B|A), P(A|B), P(A∩B).

Solution

P(A∩B)=P(A)+P(B)−P(A∪B)=6/11+5/11−7/11=4/11. P(B|A)=(4/11)/(6/11)=2/3. P(A|B)=(4/11)/(5/11)=4/5.
Q6

A coin is tossed three times. E: at least 2 heads. F: at most 2 heads. Find P(E|F).

Solution

S={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}. E={HHH,HHT,HTH,THH}. F={HHT,HTH,THH,HTT,THT,TTH,TTT}. E∩F={HHT,HTH,THH}. P(E|F)=(3/8)/(7/8)=3/7.
Q7

Two coins are tossed once, where 1st coin has head on both sides and 2nd coin has tail on both sides. A coin is chosen at random and tossed once. Find P(head).

Solution

P(choose C1)=1/2, P(H|C1)=1. P(choose C2)=1/2, P(H|C2)=0. P(H)=1/2·1+1/2·0=1/2.
Q8

A die is thrown three times. E: 4 appears on the third throw. F: 6 and 5 appear on first and second throws. P(E|F)=?

Solution

F={(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}, P(F)=6/216=1/36. E∩F={(6,5,4)}, P(E∩F)=1/216. P(E|F)=(1/216)/(6/216)=1/6.
Q9

Mother, father, son line up at random. A: son on one end. B: father in middle. P(A|B)=?

Solution

Sample space: 3!=6. B: father in middle → FFM arrangements with father in middle: son-Father-mother or mother-Father-son → 2 arrangements. A∩B: son on end AND father in middle → both: {S,F,M} and {M,F,S} → 2. P(A|B)=2/2=1.
Q10

A black and a red die are rolled. Condition: sum of numbers on the dice is 4. P(number on black die = 3 | condition)=?

Solution

Condition: sum=4. Outcomes: (1,3),(2,2),(3,1). Black die=3: (3,1). P=1/3.
Q11

A fair die is rolled. Consider events A={1,3,5}, B={2,3}, and C={2,3,4,5}. Find (i) P(A|B) (ii) P(B|C) (iii) P(A∩B|C) (iv) P(A∪B|C).

Solution

(i) A∩B={3}. P(A|B)=(1/6)/(2/6)=1/2. (ii) B∩C={2,3}. P(B|C)=(2/6)/(4/6)=1/2. (iii) A∩B={3}, A∩B∩C={3}. P(A∩B|C)=(1/6)/(4/6)=1/4. (iv) A∪B={1,2,3,5}. (A∪B)∩C={2,3,5}. P(A∪B|C)=(3/6)/(4/6)=3/4.
Q12

Assume 1 in 10,000 people are infected with a rare disease. Test gives 99% accurate results. Find the probability that a randomly selected person who tested positive actually has the disease.

Solution

P(D)=0.0001, P(D')=0.9999. P(+|D)=0.99, P(+|D')=0.01. P(D|+)=P(+|D)P(D)/[P(+|D)P(D)+P(+|D')P(D')] = 0.99×0.0001/(0.99×0.0001+0.01×0.9999) = 0.0000099/(0.0000099+0.009999)≈0.0000099/0.01≈0.0099≈0.98%.
Q13

An instructor has a question bank of 300 easy True/False Qs, 200 difficult T/F Qs, 500 easy MCQs, and 400 difficult MCQs. If a question is selected at random, find P(easy|T/F).

Solution

Total T/F=500. Easy T/F=300. P(easy|T/F)=300/500=3/5.
Q14

Given P(A)=1/2, P(B)=1/3, P(A∩B)=1/4. Find P(A'|B').

Solution

P(A'∩B')=1−P(A∪B)=1−(1/2+1/3−1/4)=1−7/12=5/12. P(B')=1−1/3=2/3. P(A'|B')=(5/12)/(2/3)=5/8.
Q15

If A and B are events such that P(A|B)>P(A), then show P(B|A)>P(B).

Solution

P(A|B)>P(A) → P(A∩B)/P(B)>P(A) → P(A∩B)>P(A)P(B). P(B|A)=P(A∩B)/P(A)>P(A)P(B)/P(A)=P(B). ✓
Q16

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

Solution

P(B|A)=1 → P(A∩B)/P(A)=1 → P(A∩B)=P(A). Since P(A∩B)≤P(A) always, equality means every element of A is in B → A⊂B. ✓
Q17

If P(A|B)=P(A), show A and B are independent.

Solution

P(A|B)=P(A) → P(A∩B)/P(B)=P(A) → P(A∩B)=P(A)P(B) → A and B are independent. ✓
Exercise 13.2Multiplication Theorem on Probability
Q1

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

Solution

P(A∩B)=P(A)·P(B)=3/5·1/5=3/25.
Q2

Two cards are drawn at random without replacement from a pack of 52 cards. What is the probability that both are black?

Solution

P(1st black)=26/52=1/2. P(2nd black|1st black)=25/51. P(both black)=1/2·25/51=25/102.
Q3

A box of oranges is inspected by examining three randomly selected oranges. If all three oranges are good, the box is approved. If there are 15 oranges and 12 are good, find probability box is approved.

Solution

P=C(12,3)/C(15,3)=(12·11·10)/(15·14·13)=1320/2730=44/91.
Q4

A fair coin and an unbiased die are tossed. Let A={head appears on coin} and B={3 on die}. Check if A and B are independent.

Solution

P(A)=1/2, P(B)=1/6. P(A∩B)=1/12=P(A)·P(B). ✓ Independent.
Q5

A die is thrown once. E='the number appearing is a multiple of 3'. F='the number appearing is even'. Are E and F independent?

Solution

E={3,6}, P(E)=1/3. F={2,4,6}, P(F)=1/2. E∩F={6}, P(E∩F)=1/6=P(E)·P(F). ✓ Independent.
Q6

Let E₁ and E₂ be the events: a person has a job, and a person has a house. P(E₁)=0.7, P(E₂)=0.5, P(E₁∩E₂)=0.3. Find P(E₁'|E₂').

Solution

P(E₁'∩E₂')=1−P(E₁∪E₂)=1−(0.7+0.5−0.3)=0.1. P(E₂')=0.5. P(E₁'|E₂')=0.1/0.5=0.2.
Q7

Given two independent events A and B, P(A)=0.3, P(B)=0.6. Find P(A∩B), P(A∩B'), P(A'∩B), P(A'∩B'), P(A∪B).

Solution

P(A∩B)=0.18. P(A∩B')=0.3×0.4=0.12. P(A'∩B)=0.7×0.6=0.42. P(A'∩B')=0.7×0.4=0.28. P(A∪B)=0.3+0.6−0.18=0.72.
Q8

Let A and B be independent events, P(A)=0.3, P(A∪B)=0.6. Find P(B).

Solution

P(A∪B)=P(A)+P(B)−P(A)P(B)=0.6. 0.3+P(B)−0.3P(B)=0.6. 0.7P(B)=0.3. P(B)=3/7.
Q9

If A and B are two independent events then show P(A'∪B')=1−P(A)P(B).

Solution

P(A'∪B')=1−P(A∩B)=1−P(A)P(B) (since A,B independent). ✓
Q10

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

Solution

P(A'∪B')=P((A∩B)')=1−P(A∩B)=1/4→P(A∩B)=3/4. P(A)P(B)=7/24≠3/4. Not independent.
Q11

Given P(A)=0.6, P(B)=0.3 and P(A∩B)=0.2. P(A') and P(A|B) and check if independent.

Solution

P(A')=0.4. P(A|B)=0.2/0.3=2/3≠P(A)=0.6. Not independent.
Q12

A and B are two independent events. P(both A and B)=1/6, P(neither A nor B)=1/3. Find P(A) and P(B).

Solution

P(A)P(B)=1/6. (1−P(A))(1−P(B))=1/3. Let a=P(A),b=P(B): ab=1/6, 1−a−b+ab=1/3→a+b=1−1/3+1/6=1/2+1/6=5/6... wait: 1−a−b+1/6=1/3→a+b=1+1/6−1/3=5/6. So a+b=5/6 and ab=1/6. Quadratic: t²−5t/6+1/6=0→6t²−5t+1=0→(2t−1)(3t−1)=0→t=1/2 or t=1/3. P(A)=1/2,P(B)=1/3 or vice versa.
Q13

An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted, and it is returned with 5 additional balls of the same colour. A ball is then drawn again. What is the probability that second ball drawn is red?

Solution

P(R₁)=1/2. P(R₂|R₁)=10/15=2/3. P(B₁)=1/2. P(R₂|B₁)=5/15=1/3. P(R₂)=P(R₁)P(R₂|R₁)+P(B₁)P(R₂|B₁)=1/2·2/3+1/2·1/3=1/3+1/6=1/2.
Q14

In a factory, machine A produces 30% of items and machine B produces 70% of items. 5% of A's items are defective and 1% of B's items are defective. An item is chosen at random. Find P(defective).

Solution

P(D)=P(A)P(D|A)+P(B)P(D|B)=0.3×0.05+0.7×0.01=0.015+0.007=0.022.
Q15

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find probability of (i) both red (ii) first red, second black (iii) one red, one black.

Solution

(i) P=(8/18)²=64/324=16/81. (ii) P=8/18·10/18=80/324=20/81. (iii) P=2·(8/18·10/18)=160/324=40/81.
Q16

Probability of solving a problem by A is 1/2 and by B is 1/3. Find P(problem is solved).

Solution

P(not solved)=P(A')(P(B')=1/2·2/3=1/3. P(solved)=1−1/3=2/3.
Q17

One card is drawn at random from a well shuffled deck. In which of the following are the events A and B independent? (i) A: the card is a spade, B: the card is an ace.

Solution

P(A)=13/52=1/4. P(B)=4/52=1/13. P(A∩B)=1/52=P(A)P(B)=1/52. ✓ Independent.
Q18

In Ex 17, is A: 'black card', B: 'king' independent?

Solution

P(A)=26/52=1/2. P(B)=4/52=1/13. P(A∩B)=2/52=1/26=1/2·1/13. ✓ Independent.
Exercise 13.3Bayes' Theorem
Q1

An urn contains 5 red and 5 black balls. A ball is drawn at random; if red, 2 red balls are added; if black, 2 black balls are added. A second ball is drawn. Find P(first ball is red | second ball is red).

Solution

P(R₁)=1/2, P(R₂|R₁)=7/12, P(R₂|B₁)=5/12. P(R₁|R₂)=P(R₁)P(R₂|R₁)/P(R₂). P(R₂)=1/2·7/12+1/2·5/12=12/24=1/2. P(R₁|R₂)=(1/2·7/12)/(1/2)=7/12.
Q2

A bag I contains 3 red and 4 black balls; bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find P(it was from bag I).

Solution

P(B₁)=P(B₂)=1/2. P(R|B₁)=3/7. P(R|B₂)=5/11. P(R)=1/2·3/7+1/2·5/11=3/14+5/22=33/154+35/154=68/154=34/77. P(B₁|R)=(1/2·3/7)/(34/77)=(3/14)·(77/34)=3·77/(14·34)=231/476=33/68.
Q3

Of the students in a class, it is known that 30% have 100% attendance and 70% students are irregular. Prev year results show 70% of regular students passed and 10% of irregular students passed. If a student passed, find P(he is regular).

Solution

P(A)=0.3, P(B)=0.7. P(Pass|A)=0.7, P(Pass|B)=0.1. P(A|Pass)=0.3×0.7/(0.3×0.7+0.7×0.1)=0.21/(0.21+0.07)=0.21/0.28=3/4.
Q4

In a factory, machines X, Y, Z produce 25%, 35%, 40% of total items respectively. 5% from X, 4% from Y, 2% from Z are defective. An item is found defective. Find P(it came from machine X).

Solution

P(X)=0.25, P(Y)=0.35, P(Z)=0.40. P(D|X)=0.05, P(D|Y)=0.04, P(D|Z)=0.02. P(D)=0.25×0.05+0.35×0.04+0.40×0.02=0.0125+0.014+0.008=0.0345. P(X|D)=0.0125/0.0345=125/345=25/69.
Q5

A laboratory blood test is 99% effective in detecting a disease when the patient has it. However, the test gives false positive for 0.5% of healthy people. In a given population, 0.1% have the disease. A person chosen at random tests positive. What is P(he has the disease)?

Solution

P(D)=0.001, P(H)=0.999. P(+|D)=0.99, P(+|H)=0.005. P(D|+)=(0.99×0.001)/(0.99×0.001+0.005×0.999)=(0.00099)/(0.00099+0.004995)=0.00099/0.005985=990/5985=22/133.
Q6

There are three boxes: box 1 has 1 white, 2 red, 3 black; box 2 has 2 white, 3 red, 1 black; box 3 has 3 white, 1 red, 2 black. A box is chosen at random and a ball drawn. Find P(ball is red). If ball is red, find P(it came from box 1).

Solution

P(R|B₁)=2/6=1/3, P(R|B₂)=3/6=1/2, P(R|B₃)=1/6. P(R)=1/3(1/3+1/2+1/6)=1/3·1=1/3. P(B₁|R)=(1/3·1/3)/(1/3)=1/3.
Q7

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. P(accident)=0.01, 0.03, 0.15 respectively. One of the insured persons meets an accident. P(he is a scooter driver)?

Solution

Total=12000. P(S)=2/12=1/6, P(C)=4/12=1/3, P(T)=6/12=1/2. P(A|S)=0.01, P(A|C)=0.03, P(A|T)=0.15. P(A)=1/6·0.01+1/3·0.03+1/2·0.15=0.01/6+0.03/3+0.15/2=(1+6+45)/600=52/600. P(S|A)=(1/6·0.01)/(52/600)=(1/600)/(52/600)=1/52.
Q8

A factory has two machines A and B. Past data shows machine A produced 60% of items and B produced 40%. Also 2% of A's items and 1% of B's items were defective. All items put in store and one drawn at random is found defective. P(it came from machine B)?

Solution

P(A)=0.6, P(B)=0.4. P(D|A)=0.02, P(D|B)=0.01. P(D)=0.6·0.02+0.4·0.01=0.012+0.004=0.016. P(B|D)=0.4·0.01/0.016=0.004/0.016=1/4.
Q9

Two groups are equally likely to be chosen. Group I has 2 boys and 1 girl; Group II has 3 boys and 3 girls. A group is selected and 1 person is chosen. If the person is a girl, what is P(she is from Group II)?

Solution

P(I)=P(II)=1/2. P(G|I)=1/3. P(G|II)=3/6=1/2. P(II|G)=(1/2·1/2)/(1/2·1/3+1/2·1/2)=(1/4)/(1/6+1/4)=(1/4)/(5/12)=3/5.
Q10

Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin 3 times and notes number of tails. If she gets 3,4,5,6, she tosses coin once and notes head/tail. If she obtained exactly one tail, what is P(she threw 3,4,5,6)?

Solution

P(A: 1 or 2)=2/6=1/3. P(B: 3−6)=4/6=2/3. P(1 tail|A)=C(3,1)/8=3/8. P(1 tail|B)=P(tail on 1 toss)=1/2. P(B|1 tail)=(2/3·1/2)/((1/3·3/8)+(2/3·1/2))=(1/3)/(1/8+1/3)=(1/3)/(11/24)=8/11.
Q11

A manufacturer has 3 machines. Machine I works 1/2 the time, machine II works 1/4 time, machine III works 1/4 time. Machine I produces 1 defective item per 20; machine II: 1 per 25; machine III: 1 per 40. An item is defective. What is the probability it was produced on machine I?

Solution

P(I)=1/2,P(II)=1/4,P(III)=1/4. P(D|I)=1/20, P(D|II)=1/25, P(D|III)=1/40. P(D)=1/2·1/20+1/4·1/25+1/4·1/40=1/40+1/100+1/160. LCM=800: =20/800+8/800+5/800=33/800. P(I|D)=(1/40)/(33/800)=(20/800)/(33/800)=20/33.
Q12

A card from a pack of 52 cards is lost. From the remaining 51 cards, 2 are drawn and found to be both diamonds. Find P(lost card is diamond).

Solution

P(Diamond lost)=13/52=1/4. P(Other lost)=39/52=3/4. P(2 diamonds from 51|diamond lost)=C(12,2)/C(51,2)=66/1275. P(2 diamonds from 51|other lost)=C(13,2)/C(51,2)=78/1275. P(D lost|2D drawn)=(1/4·66/1275)/((1/4·66/1275)+(3/4·78/1275))=(66/4)/((66/4)+(234/4))=(66)/(300)=11/50.
Q13

Probability that A speaks truth is 4/5 and probability that B speaks truth is 3/4. They independently toss a coin and report same outcome (both say head). Find probability they are right.

Solution

P(A right, B right)=(4/5)(3/4)=3/5. P(A wrong, B wrong)=(1/5)(1/4)=1/20. Both say head: either both right or both wrong. P(right|both say head)=(3/5)/((3/5)+(1/20))=(12/20)/((12+1)/20)=12/13.
Q14

Bag I: 3 red, 4 black balls. Bag II: 4 red, 2 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. If the ball drawn is red, find P(transferred ball was red).

Solution

P(R transferred)=3/7. P(B transferred)=4/7. After transfer: Bag II has 7 or 8 balls. If R transferred: 5 red, 2 black (7 balls). P(R|R trans)=5/7. If B transferred: 4 red, 3 black (7 balls). P(R|B trans)=4/7. P(R drawn)=3/7·5/7+4/7·4/7=15/49+16/49=31/49. P(R trans|R drawn)=(3/7·5/7)/(31/49)=15/31.
Exercise 13.4Random Variables and their Probability Distributions
Q1

State which of the following are not the probability mass functions of a random variable: (i) X: 0,1,2; P(X): 0.4, 0.4, 0.2 (ii) X: 0,1,2,3; P(X): 0.1, 0.5, 0.2, −0.1

Solution

(i) Sum=1, all P≥0 ✓ → is a PMF. (ii) P(X=3)=−0.1<0 ✗ → NOT a PMF.
Q2

An urn contains 10 black and 5 white balls. Two balls are drawn with replacement. Let X=number of white balls drawn. Find E(X).

Solution

X~Binomial(n=2, p=1/3). E(X)=np=2/3.
Q3

A fair coin is tossed four times. Let X denote the number of tails. Find the PMF and mean of X.

Solution

X~B(4,1/2). P(X=k)=C(4,k)/16. P(0)=1/16,P(1)=4/16,P(2)=6/16,P(3)=4/16,P(4)=1/16. E(X)=4×1/2=2. Var(X)=4×1/2×1/2=1.
Q4

Find the probability distribution of: number of heads in 3 tosses of a fair coin.

Solution

X: 0,1,2,3. P(X=k)=C(3,k)/8. P(0)=1/8, P(1)=3/8, P(2)=3/8, P(3)=1/8. E(X)=3/2.
Q5

Find the probability distribution of number of sixes in 2 throws of a die.

Solution

p=1/6, q=5/6, n=2. P(X=0)=25/36, P(X=1)=10/36, P(X=2)=1/36. E(X)=2/6=1/3.
Q6

Two dice are thrown simultaneously. X = absolute difference of the two numbers. Find E(X).

Solution

P(X=0)=6/36=1/6. P(X=1)=10/36. P(X=2)=8/36. P(X=3)=6/36. P(X=4)=4/36. P(X=5)=2/36. E(X)=0+10/36+16/36+18/36+16/36+10/36=70/36=35/18.
Q7

A coin is biased so P(head)=0.6. If thrown 4 times, find the probability distribution of the number of heads.

Solution

X~B(4,0.6). P(X=k)=C(4,k)(0.6)^k(0.4)^(4−k). P(0)=0.0256, P(1)=0.1536, P(2)=0.3456, P(3)=0.3456, P(4)=0.1296.
Q8

From a lot of 10 items containing 3 defectives, 4 are chosen at random. Let X be the number of defective items chosen. Find E(X).

Solution

X follows Hypergeometric. E(X)=n·K/N=4·3/10=6/5=1.2.
Q9

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

Solution

E(X)=7. E(X²)=∑x²P(x). Var(X)=E(X²)−(E(X))²=329/6−49=35/6≈5.83. SD=√(35/6).
Q10

A class has 15 students whose heights are measured. The heights in cm are recorded. Find E(X) and Var(X) from the distribution.

Solution

This needs actual data. For a general distribution: E(X)=∑xP(x), Var(X)=∑(x−E(X))²P(x)=∑x²P(x)−(E(X))².
Q11

In a meeting, 70% of 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).

Solution

P(X=1)=0.7, P(X=0)=0.3. E(X)=0.7. E(X²)=0.7. Var(X)=0.7−0.49=0.21.
Q12

The mean of the number obtained on throwing a die having written 1 on 3 faces, 2 on 2 faces and 5 on 1 face. Find E(X) and Var(X).

Solution

P(X=1)=3/6=1/2, P(X=2)=2/6=1/3, P(X=5)=1/6. E(X)=1/2+2/3+5/6=3/6+4/6+5/6=2. E(X²)=1/2+4/3+25/6=3/6+8/6+25/6=6. Var(X)=6−4=2.
Q13

Two numbers are selected at random (without replacement) from first 6 positive integers. Let X denote the larger of the two numbers. Find E(X).

Solution

Total pairs=C(6,2)=15. P(X=2)=1/15, P(X=3)=2/15, P(X=4)=3/15, P(X=5)=4/15, P(X=6)=5/15. E(X)=(2+6+12+20+30)/15=70/15=14/3.
Q14

A random variable X has the following probability distribution: X: 0,1,2,3,4,5,6,7; P(X): 0,k,2k,2k,3k,k²,2k²,7k²+k. Find k and evaluate P(X<6), P(X≥6), P(0<X<5).

Solution

Sum=0+k+2k+2k+3k+k²+2k²+7k²+k=10k²+9k=1. 10k²+9k−1=0. (10k−1)(k+1)=0. k=1/10. P(X<6)=0+1/10+2/10+2/10+3/10+1/100=8/10+1/100=81/100. P(X≥6)=2/100+7/100+1/10=2/100+7/100+10/100=19/100. P(0<X<5)=1/10+2/10+2/10+3/10=8/10=4/5.
Q15

The random variable X can take only values 0,1,2. Given P(X=0)=P(X=1)=p and E(X²)=E(X). Find p.

Solution

P(X=2)=1−2p. E(X)=0+p+2(1−2p)=2−3p. E(X²)=0+p+4(1−2p)=4−7p. E(X²)=E(X): 4−7p=2−3p → 4p=2 → p=1/2... Check P(X=2)=1−1=0. E(X)=2−3/2=1/2. E(X²)=4−7/2=1/2. ✓
Q16

Find the mean number of heads in 3 tosses of a fair coin.

Solution

X~B(3,1/2). E(X)=np=3/2.
Q17

Two dice are thrown simultaneously. Find the probability distribution of X = sum obtained. Also find E(X) and Var(X).

Solution

P(X=k) for k=2 to 12 as standard. E(X)=7 (by symmetry). Var(X)=∑(x−7)²P(x)=35/6.
Exercise 13.5Binomial Distribution
Q1

An experiment succeeds twice as often as it fails. Find P(at least 5 successes in 6 trials).

Solution

p=2/3, q=1/3. P(X≥5)=P(5)+P(6)=C(6,5)(2/3)⁵(1/3)+C(6,6)(2/3)⁶=6·32/3⁶·1/3+64/3⁶=192/729+64/729=256/729.
Q2

A pair of dice is thrown 4 times. P(getting a doublet exactly 2 times)=?

Solution

P(doublet)=6/36=1/6, q=5/6. P(X=2)=C(4,2)(1/6)²(5/6)²=6·25/1296=150/1296=25/216.
Q3

There are 5% defective items in a large bulk. Probability that a sample of 10 items has not more than 1 defective item.

Solution

p=0.05, q=0.95, n=10. P(X≤1)=P(0)+P(1)=(0.95)¹⁰+10(0.05)(0.95)⁹=0.95⁹(0.95+0.5)=0.95⁹·1.45... = (0.95)⁹·1.45. (0.95)⁹≈0.6302. P≈0.9138.
Q4

Five cards are drawn from a well-shuffled pack of 52 cards. What is the probability that all 5 cards are spades?

Solution

This is hypergeometric but using binomial approx: p=1/4. P(X=5)=C(5,5)(1/4)⁵=(1/4)⁵=1/1024. (Exact: C(13,5)/C(52,5)=1287/2598960=33/66640.)
Q5

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find probability (in a sample of 5) that (i) none fuses (ii) not more than 1 fuses.

Solution

p=0.05, q=0.95, n=5. (i) P(X=0)=(0.95)⁵≈0.7738. (ii) P(X≤1)=(0.95)⁵+5(0.05)(0.95)⁴=(0.95)⁴(0.95+0.25)=(0.95)⁴·1.2≈0.8145·1.2≈0.9774.
Q6

A bag consists of 10 balls of which 4 are red and 6 are black. A ball is drawn at random and its colour is noted and it is replaced. This is done 5 times. Find P(at least one red ball).

Solution

p=4/10=2/5. P(at least one red)=1−P(0 red)=1−(3/5)⁵=1−243/3125=2882/3125.
Q7

On a multiple choice examination with 3 possible answers for each of 5 questions, what is P(a candidate gets 4 or more correct answers just by guessing?)

Solution

p=1/3, q=2/3, n=5. P(X≥4)=P(4)+P(5)=C(5,4)(1/3)⁴(2/3)+C(5,5)(1/3)⁵=5·2/3⁵+1/3⁵=11/243.
Q8

A person buys a lottery ticket in 50 lotteries, in each of which chance of winning is 1/100. Find P(winning at least once).

Solution

P(no win)=(99/100)⁵⁰. P(winning≥1)=1−(99/100)⁵⁰=1−(0.99)⁵⁰≈1−0.605=0.395.
Q9

Find the probability that in 10 throws of a fair die, a score of 5 appears exactly twice.

Solution

p=1/6, n=10, k=2. P(X=2)=C(10,2)(1/6)²(5/6)⁸=45·(5/6)⁸/36=45·390625/(36·1679616)=45·390625/60466176.
Q10

If getting 5 or 6 in a throw of die is 'success', find P(at least 2 successes in 7 throws).

Solution

p=2/6=1/3, q=2/3, n=7. P(X≥2)=1−P(0)−P(1)=1−(2/3)⁷−7(1/3)(2/3)⁶=1−128/2187−448/2187=1−576/2187=1611/2187.
Q11

In an examination, 20 questions of T/F type are asked. A student tosses a fair coin to determine his answer. Find P(he answers at least 12 questions correctly).

Solution

p=1/2, n=20. P(X≥12)=∑_{k=12}^{20}C(20,k)/2²⁰. = (C(20,12)+C(20,13)+...+C(20,20))/2²⁰ = (125970+77520+38760+15504+4845+1140+190+20+1)/1048576 = 264210/1048576 = 16513/65536.
Q12

Suppose X has a binomial distribution B(6,1/2). Show that X=3 is the most likely outcome.

Solution

P(X=k)=C(6,k)/64. P(3)=20/64=5/16. Checking all: P(0)=P(6)=1/64, P(1)=P(5)=6/64, P(2)=P(4)=15/64. Largest is P(3)=20/64. ✓
Q13

On a multiple choice test with 5 alternatives for each of 4 questions, find P(student guesses all correctly).

Solution

p=1/5, n=4. P(X=4)=(1/5)⁴=1/625.
Q14

A box of 100 mangoes contains 10 that are bad. Find P(sample of 5 contains at most 1 bad mango). Use binomial with p=0.1.

Solution

p=0.1, q=0.9, n=5. P(X≤1)=P(0)+P(1)=(0.9)⁵+5(0.1)(0.9)⁴=0.59049+5·0.1·0.6561=0.59049+0.32805=0.91854.
Q15

The mean and variance of a binomial distribution are 4 and 2 respectively. Find P(X≥1).

Solution

np=4, npq=2→q=1/2→p=1/2→n=8. P(X≥1)=1−P(0)=1−(1/2)⁸=255/256.
Exercise MiscMiscellaneous Exercise
Q1

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=∅.

Solution

(i) A⊂B→A∩B=A→P(B|A)=P(A)/P(A)=1. (ii) A∩B=∅→P(B|A)=0/P(A)=0.
Q2

A couple has 2 children. Find P(both children are boys if it is known that (i) at least one is boy (ii) elder child is boy).

Solution

S={BB,BG,GB,GG}. (i) A={BB,BG,GB}, B={BB}. P(B|A)=(1/4)/(3/4)=1/3. (ii) A={BB,BG}={elder is boy}, B={BB}. P(B|A)=(1/4)/(2/4)=1/2.
Q3

Suppose that 5% of men and 0.25% of women have grey hair. A grey-haired person is selected at random. What is P(person is male)? Assume equal M/F population.

Solution

P(M)=P(F)=0.5. P(G|M)=0.05, P(G|F)=0.0025. P(M|G)=(0.5·0.05)/(0.5·0.05+0.5·0.0025)=0.025/(0.025+0.00125)=0.025/0.02625=20/21.
Q4

Suppose a family has 2 children. What is probability that both are girls if it is known that (i) at least one is girl (ii) older child is girl?

Solution

(i) P(GG|at least one G)=(1/4)/(3/4)=1/3. (ii) P(GG|elder G)=(1/4)/(2/4)=1/2.
Q5

An urn contains 25 balls of which 10 are white and the rest have red colour. Six balls are drawn and the probability that the selection has exactly two white balls is P. Find P.

Solution

P=C(10,2)·C(15,4)/C(25,6)=45·1365/177100=61425/177100=45·1365/177100. Simplify: 45·1365=61425. 61425/177100≈0.347.
Q6

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper, and 20% read both. A student is selected at random. (i) Find P(she reads neither Hindi nor English) (ii) if she reads Hindi, P(she reads English newspaper) (iii) P(reads English | reads at least one of two).

Solution

(i) P(H∪E)=0.6+0.4−0.2=0.8. P(neither)=0.2. (ii) P(E|H)=0.2/0.6=1/3. (iii) P(E|H∪E)=0.4/0.8=1/2.
Q7

The probability that at least one of A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, find P(A')+P(B').

Solution

P(A∪B)=0.6. P(A)+P(B)−P(A∩B)=0.6. P(A)+P(B)=0.8. P(A')+P(B')=2−0.8=1.2.
Q8

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

Solution

P(A'∩B')=P((A∪B)')=1−P(A∪B)=1−(1/4+1/2−1/8)=1−5/8=3/8.
Q9

Events A and B are independent. P(A)=1/3, P(B)=1/2. Find P(neither A nor B).

Solution

P(A'∩B')=P(A')P(B')=(2/3)(1/2)=1/3.
Q10

Given P(A)=0.3, P(B)=0.4, P(A∩B)=0.5 (note: P(A∩B)≤min(P(A),P(B)), so this is invalid)... correct: P(A)=0.6, P(B)=0.3, P(A|B)=0.4. Find P(A∩B), P(A'|B'), P(A∪B).

Solution

P(A∩B)=P(A|B)P(B)=0.4×0.3=0.12. P(B')=0.7. P(A'∩B')=1−P(A∪B)=1−(0.6+0.3−0.12)=1−0.78=0.22. P(A'|B')=0.22/0.7=22/70=11/35. P(A∪B)=0.78.
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