Question 1
Given P(E)=0.6, P(F)=0.3 and P(E∩F)=0.2, find P(E|F) and P(F|E).
Steps only
P(E|F)=P(E∩F)/P(F)=0.2/0.3.
P(F|E)=P(E∩F)/P(E)=0.2/0.6.
Answer
P(E|F)=2/3 and P(F|E)=1/3.
CLASS 12 · MATHEMATICS PART II · CHAPTER 13
Probability
Worked solutions for conditional probability, independence, Bayes’ theorem, random variables and the chapter’s four numbered problem sets.
61 worked answersAll 4 exercise sets
What you will learn
- Calculate conditional probabilities from restricted sample spaces.
- Use multiplication and independence rules for events.
- Apply total probability and Bayes’ theorem to reverse conditional information.
61 worked answers · short steps first, explanations when you need them
Exercise 13.1 — Conditional probability
Textbook pages 413–415
Question 2
Find P(A|B), if P(B)=0.5 and P(A∩B)=0.32.
Steps only
P(A|B)=P(A∩B)/P(B)=0.32/0.5.
Answer
P(A|B)=0.64.
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), and (iii) P(A∪B).
Steps only
P(A∩B)=P(A)P(B|A)=0.8×0.4=0.32.
P(A|B)=0.32/0.5=0.64.
P(A∪B)=P(A)+P(B)−P(A∩B)=0.8+0.5−0.32.
Answer
- (i) 0.32,
- (ii) 0.64,
- (iii) 0.98.
Question 4
Find P(A∪B), if 2P(A)=P(B)=5/13 and P(A|B)=2/5.
Steps only
P(B)=5/13 and 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=11/26.
Answer
P(A∪B)=11/26.
Question 5
If P(A)=6/11, P(B)=5/11 and P(A∪B)=7/11, find (i) P(A∩B), (ii) P(A|B), (iii) P(B|A).
Steps only
P(A∩B)=P(A)+P(B)−P(A∪B)=(6+5−7)/11=4/11.
Divide the intersection by P(B) for P(A|B), and by P(A) for P(B|A).
Answer
- (i) 4/11,
- (ii) 4/5,
- (iii) 2/3.
Question 6
A fair coin is tossed three times. Find P(E|F) for: (i) E=tail on third toss, F=heads on first two tosses; (ii) E=at least two heads, F=at most two heads; (iii) E=at most two tails, F=at least one tail.
Steps only
(i) Given HH on the first two tosses, the third toss is equally likely to be H or T, so the chance of T is 1/2.
(ii) F excludes HHH, leaving 7 outcomes; E∩F is HHT, HTH and THH, so 3/7.
(iii) F excludes HHH; E excludes TTT. Their intersection has 6 of the 7 outcomes in F.
Answer
- (i) 1/2,
- (ii) 3/7,
- (iii) 6/7.
Question 7
Two fair coins are tossed. Find P(E|F) for (i) E=exactly one tail, F=at least one head; (ii) E=no tail, F=no head.
Steps only
(i) F={HH,HT,TH}; E∩F={HT,TH}, so 2/3.
(ii) E={HH} and F={TT}; the events cannot occur together.
Answer
- (i) 2/3,
- (ii) 0.
Question 8
A die is thrown three times. E=4 on the third toss; F=6 and then 5 on the first two tosses. Find P(E|F).
Steps only
Once the first two results are fixed as 6 and 5, the third roll is still a fair die roll.
Only one of its six equally likely faces is 4.
Answer
P(E|F)=1/6.
Question 9
Mother, father and son line up at random. E=son at one end; F=father in the middle. Find P(E|F).
Steps only
When father is fixed in the middle, mother and son occupy the two ends.
In both possible orders, the son is at an end.
Answer
P(E|F)=1.
Question 10
A black and a red die are rolled. Find (a) P(sum>9 | black die=5), (b) P(sum=8 | red die<4).
Steps only
(a) With black=5, red must be 5 or 6: 2 of 6 outcomes.
(b) Red can be 1, 2 or 3; a sum of 8 occurs only for red=2 and black=6: 1 of 3 outcomes.
Answer
(a) 1/3, (b) 1/3.
Question 11
A fair die is rolled with E={1,3,5}, F={2,3}, G={2,3,4,5}. Find (i) P(E|F), P(F|E); (ii) P(E|G), P(G|E); (iii) P((E∪F)|G), P((E∩F)|G).
Steps only
(i) E∩F={3}; divide by |F|=2 or |E|=3.
(ii) E∩G={3,5}; G∩E has 2 of 4 outcomes, while G has 2 of the 3 outcomes in E.
(iii) E∪F contains all of G; E∩F={3}, one outcome among the four in G.
Answer
- (i) 1/2, 1/3;
- (ii) 1/2, 2/3;
- (iii) 1, 1/4.
Question 12
Each child is equally likely to be a boy or girl. Find the probability both are girls given (i) the youngest is a girl, (ii) at least one is a girl.
Steps only
Use ordered outcomes (elder, youngest): BB, BG, GB, GG.
(i) With youngest a girl, the possibilities are BG and GG.
(ii) With at least one girl, the possibilities are BG, GB and GG.
Answer
- (i) 1/2,
- (ii) 1/3.
Question 13
A question bank has 300 easy and 200 difficult true/false questions, and 500 easy and 400 difficult multiple-choice questions. Given that a selected question is multiple choice, find the probability it is easy.
Steps only
There are 500+400=900 multiple-choice questions.
Of those, 500 are easy.
Answer
The probability is 500/900=5/9.
Question 14
Given that the numbers on two dice are different, find the probability their sum is 4.
Steps only
There are 36 ordered outcomes; 6 are doubles, leaving 30 different-number outcomes.
The sum 4 with different numbers occurs as (1,3) or (3,1).
Answer
The probability is 2/30=1/15.
Question 15
Roll a die. If it is a multiple of 3, roll again; otherwise toss a coin. Find the probability the coin shows tail given that at least one die shows 3.
Steps only
For at least one die to show 3, the first roll must be 3 or 6; if it is 6, the second roll must be 3.
Whenever a die roll occurs in these cases, the coin is not tossed.
Answer
The conditional probability is 0.
Question 16
If P(A)=1/2 and P(B)=0, what is P(A|B)?
Steps only
P(A|B)=P(A∩B)/P(B), but the denominator is zero.
Answer
It is not defined (choice C).
Question 17
If P(A|B)=P(B|A), which conclusion must follow?
Steps only
When P(A∩B)>0, equality implies P(A)=P(B).
But if A and B are disjoint and both have positive probability, both conditional probabilities are 0 even when P(A)≠P(B).
Answer
As printed, none of the listed conclusions is guaranteed. Choice (D), P(A)=P(B), follows only if P(A∩B)>0 is additionally assumed.
Exercise 13.2 — Multiplication rule and independence
Textbook pages 421–423
Question 1
If P(A)=3/5 and P(B)=1/5 and A,B are independent, find P(A∩B).
Steps only
For independent events, P(A∩B)=P(A)P(B).
Answer
P(A∩B)=(3/5)(1/5)=3/25.
Question 2
Two cards are drawn without replacement from a standard 52-card deck. Find the probability both are black.
Steps only
There are 26 black cards. P(first black)=26/52. After one black card is removed, P(second black)=25/51.
Answer
The probability is (26/52)(25/51)=25/102.
Question 3
A box has 12 good and 3 bad oranges. Three are drawn without replacement. Find the probability all are good.
Steps only
P(all good)=(12/15)(11/14)(10/13).
Equivalently, choose 3 of the 12 good oranges out of all ways to choose 3 of 15.
Answer
The probability is 44/91.
Question 4
A fair coin and an unbiased die are tossed. A=head on the coin; B=3 on the die. Are A and B independent?
Steps only
P(A)=1/2, P(B)=1/6 and P(A∩B)=1/12.
P(A)P(B)=(1/2)(1/6)=1/12.
Answer
Yes, A and B are independent.
Question 5
A die has red faces 1,2,3 and green faces 4,5,6. A=even; B=red. Are A and B independent?
Steps only
P(A)=3/6=1/2 and P(B)=3/6=1/2.
A∩B={2}, so P(A∩B)=1/6, while P(A)P(B)=1/4.
Answer
No. The events are not independent.
Question 6
P(E)=3/5, P(F)=3/10 and P(E∩F)=1/5. Are E and F independent?
Steps only
If independent, the intersection would be (3/5)(3/10)=9/50.
But 1/5=10/50, which is different.
Answer
No, E and F are not independent.
Question 7
P(A)=1/2, P(A∪B)=3/5 and P(B)=p. Find p if (i) A,B are mutually exclusive, (ii) A,B are independent.
Steps only
(i) Mutually exclusive: 3/5=1/2+p, so p=1/10.
(ii) Independent: 3/5=1/2+p−(1/2)p, so p=1/5.
Answer
- (i) p=1/10;
- (ii) p=1/5.
Question 8
A and B are independent, P(A)=0.3, P(B)=0.4. Find (i) P(A∩B), (ii) P(A∪B), (iii) P(A|B), (iv) P(B|A).
Steps only
The intersection is 0.3×0.4=0.12.
The union is 0.3+0.4−0.12=0.58.
For independent events, conditioning on the other event leaves the probability unchanged.
Answer
- (i) 0.12,
- (ii) 0.58,
- (iii) 0.3,
- (iv) 0.4.
Question 9
P(A)=1/4, P(B)=1/2 and P(A∩B)=1/8. Find P(A′∩B′).
Steps only
P(A∪B)=1/4+1/2−1/8=5/8.
Neither occurs is the complement of A∪B.
Answer
P(A′∩B′)=1−5/8=3/8.
Question 10
P(A)=1/2, P(B)=7/12 and P(A′∪B′)=1/4. Are A and B independent?
Steps only
By De Morgan’s law, A′∪B′=(A∩B)′, so P(A∩B)=3/4.
An intersection cannot have probability greater than P(A)=1/2.
Answer
These data are inconsistent, so no such events exist and independence cannot be assessed.
Question 11
A and B are independent, with P(A)=0.3 and P(B)=0.6. Find (i) P(A∩B), (ii) P(A∩B′), (iii) P(A∪B), (iv) P(A′∩B′).
Steps only
P(A∩B)=0.3×0.6=0.18.
P(A∩B′)=0.3×0.4=0.12.
P(A∪B)=0.3+0.6−0.18=0.72.
P(A′∩B′)=(0.7)(0.4)=0.28.
Answer
- (i) 0.18,
- (ii) 0.12,
- (iii) 0.72,
- (iv) 0.28.
Question 12
A die is tossed three times. Find the probability of getting an odd number at least once.
Steps only
The complement is getting an even number on all three tosses, with probability (1/2)^3.
Answer
The probability is 1−1/8=7/8.
Question 13
A box has 10 black and 8 red balls. Two balls are drawn with replacement. Find (i) both red, (ii) first black and second red, (iii) one black and one red.
Steps only
P(red)=8/18=4/9 and P(black)=10/18=5/9; replacement makes the draws independent.
(i) (4/9)^2. (ii) (5/9)(4/9). (iii) Either order: 2(5/9)(4/9).
Answer
- (i) 16/81,
- (ii) 20/81,
- (iii) 40/81.
Question 14
A and B independently solve a problem with probabilities 1/2 and 1/3. Find (i) probability it is solved, (ii) probability exactly one solves it.
Steps only
(i) P(at least one)=1−P(neither)=1−(1/2)(2/3)=2/3.
(ii) P(A only)+P(B only)=(1/2)(2/3)+(1/2)(1/3)=1/2.
Answer
- (i) 2/3,
- (ii) 1/2.
Question 15
For each pair, decide whether E and F are independent: (i) E=spade, F=ace; (ii) E=black, F=king; (iii) E=king or queen, F=queen or jack.
Steps only
(i) P(E∩F)=1/52=(1/4)(1/13), so independent.
(ii) P(E∩F)=2/52=1/26=(1/2)(1/13), so independent.
(iii) P(E)=P(F)=8/52=2/13, but P(E∩F)=4/52=1/13, not 4/169.
Answer
- (i) Independent;
- (ii) independent;
- (iii) not independent.
Question 16
In a hostel, 60% read Hindi, 40% English and 20% both. Find (a) probability neither, (b) P(English|Hindi), (c) P(Hindi|English).
Steps only
P(H∪E)=0.6+0.4−0.2=0.8, so neither=0.2.
P(E|H)=0.2/0.6=1/3.
P(H|E)=0.2/0.4=1/2.
Answer
(a) 0.2, (b) 1/3, (c) 1/2.
Question 17
A pair of dice is rolled. Find the probability of obtaining an even prime number on each die.
Steps only
The only even prime is 2.
Both dice must show 2, one outcome among 36 equally likely pairs.
Answer
The probability is 1/36 (choice D).
Question 18
Which condition is equivalent to independence of A and B?
Steps only
If A and B are independent, their complements are independent too.
Thus P(A′∩B′)=P(A′)P(B′)=[1−P(A)][1−P(B)].
Answer
Choice (B).
Exercise 13.3 — Bayes’ theorem
Textbook pages 431–432
Question 1
An urn has 5 red and 5 black balls. Draw one, replace it, then add two balls of the drawn colour. Find the probability the next ball is red.
Steps only
If the first ball is red (probability 1/2), the urn has 7 red out of 12.
If the first ball is black (probability 1/2), it has 5 red out of 12.
Total: (1/2)(7/12)+(1/2)(5/12).
Answer
The probability is 1/2.
Question 2
Bag I has 4 red, 4 black; Bag II has 2 red, 6 black. Choose a bag equally likely, then draw red. Find P(Bag I|red).
Steps only
The red likelihoods are 1/2 from Bag I and 1/4 from Bag II.
Bayes: [(1/2)(1/2)]/[(1/2)(1/2)+(1/2)(1/4)].
Answer
P(Bag I|red)=2/3.
Question 3
60% of students live in a hostel; 40% are day scholars. A grades occur for 30% of hostel students and 20% of day scholars. Given a student has an A, find the probability they are a hostel student.
Steps only
P(hostel and A)=0.6×0.3=0.18.
P(A)=0.18+(0.4×0.2)=0.26.
Divide 0.18 by 0.26.
Answer
P(hostel|A)=9/13.
Question 4
A student knows an answer with probability 3/4; otherwise guesses. A guess is correct with probability 1/4. Given the answer was correct, find the probability the student knew it.
Steps only
P(known and correct)=3/4.
P(correct)=3/4+(1/4)(1/4)=13/16.
The required probability is (3/4)/(13/16).
Answer
The probability is 12/13.
Question 5
A test detects 99% of diseased people and gives a false positive for 0.5% of healthy people. Disease prevalence is 0.1%. Given a positive result, find the probability the person is diseased.
Steps only
True-positive probability in the population: 0.001×0.99=0.00099.
False-positive probability: 0.999×0.005=0.004995.
Posterior probability=0.00099/(0.00099+0.004995).
Answer
The probability is 22/133≈0.1654, or about 16.5%.
Question 6
One of three coins is chosen equally: a two-headed coin, a coin with P(H)=0.75, or a fair coin. It shows heads. Find the probability it was the two-headed coin.
Steps only
The prior probabilities are each 1/3; head likelihoods are 1, 3/4 and 1/2.
Posterior for the two-headed coin is 1/(1+3/4+1/2).
Answer
The probability is 4/9.
Question 7
An insurer covers 2,000 scooter, 4,000 car and 6,000 truck drivers. Accident probabilities are 0.01, 0.03 and 0.15. Given an accident, find the probability the driver used a scooter.
Steps only
Expected accident counts are 20 scooters, 120 cars and 900 trucks.
Total accidents represented: 1,040.
Answer
P(scooter|accident)=20/1040=1/52.
Question 8
Machine A makes 60% of items, B makes 40%; defect rates are 2% and 1%. Given an item is defective, find the probability it came from B.
Steps only
P(B and defective)=0.4×0.01=0.004.
P(defective)=0.6×0.02+0.4×0.01=0.016.
Answer
P(B|defective)=0.004/0.016=1/4.
Question 9
Two groups win with probabilities 0.6 and 0.4. Their probabilities of introducing a product are 0.7 and 0.3. Given a product was introduced, find the probability the second group won.
Steps only
Second group and product: 0.4×0.3=0.12.
Product overall: 0.6×0.7+0.4×0.3=0.54.
Answer
The probability is 0.12/0.54=2/9.
Question 10
A girl rolls a die. On 5 or 6 she tosses a coin three times; on 1–4 she tosses once. Given exactly one head, find the probability the die showed 1–4.
Steps only
P(1–4 and one head)= (4/6)(1/2)=1/3.
P(5–6 and exactly one head)= (2/6)(3/8)=1/8.
Divide 1/3 by 1/3+1/8.
Answer
The probability is 8/11.
Question 11
Operators A, B, C make 50%, 30%, 20% of items, with defect rates 1%, 5%, 7%. Given a defective item, find the probability A made it.
Steps only
A’s defective contribution is 0.5×0.01=0.005.
All-defect probability is 0.005+0.3×0.05+0.2×0.07=0.034.
Answer
P(A|defective)=0.005/0.034=5/34.
Question 12
One card is lost from a 52-card deck. Two of the remaining cards drawn are both diamonds. Find the probability the lost card was a diamond.
Steps only
If a diamond was lost, 12 diamonds remain; the chance of drawing two is C(12,2)/C(51,2).
If a non-diamond was lost, 13 diamonds remain; likelihood is C(13,2)/C(51,2).
Weight by prior probabilities 1/4 and 3/4; the common denominator cancels: (1/4)66 : (3/4)78 = 11:39.
Answer
The probability the lost card was a diamond is 11/50.
Question 13
A reports truth with probability 4/5. A coin is tossed and A reports heads. Find the probability heads actually occurred.
Steps only
P(actual head and report head)=(1/2)(4/5)=2/5.
P(report head)=(1/2)(4/5)+(1/2)(1/5)=1/2.
Answer
The required probability is (2/5)/(1/2)=4/5.
Miscellaneous Exercise on Chapter 13
Textbook pages 435–437
Question 1
P(A)≠0. Find P(B|A) if (i) A⊆B, (ii) A∩B=∅.
Steps only
(i) Whenever A occurs, B occurs, so P(A∩B)=P(A).
(ii) If A and B are disjoint, P(A∩B)=0.
Answer
- (i) 1,
- (ii) 0.
Question 2
For two children, find (i) P(both boys | at least one boy), (ii) P(both girls | elder is a girl).
Steps only
Ordered outcomes are BB, BG, GB, GG.
(i) At least one boy leaves BB, BG, GB, so 1/3.
(ii) Elder girl leaves GB and GG, so 1/2.
Answer
- (i) 1/3,
- (ii) 1/2.
Question 3
5% of men and 0.25% of women have grey hair; there are equal numbers of men and women. Given grey hair, find the probability the person is male.
Steps only
Male-and-grey weight: (1/2)(0.05)=0.025.
Female-and-grey weight: (1/2)(0.0025)=0.00125.
The posterior is 0.025/(0.02625).
Answer
The probability is 20/21.
Question 4
90% of people are right-handed. In a random sample of 10, find the probability at most 6 are right-handed.
Steps only
Let X be binomial with n=10 and p=0.9.
P(X≤6)=Σ[k=0 to 6] C(10,k)(0.9)^k(0.1)^(10−k).
Answer
P(X≤6)=0.0127951984 (approximately).
Question 5
A leap year is selected at random. Find the probability it contains 53 Tuesdays.
Steps only
A leap year has 366=52×7+2 days.
The two extra consecutive weekdays include Tuesday if the year starts on Monday or Tuesday.
Answer
The probability is 2/7.
Question 6
Choose one of four boxes equally. Their red counts are A:1 of 10, B:6 of 10, C:8 of 10, D:0 of 10. Given a red marble, find the probability it came from A, B or C.
Steps only
The common prior for each box is 1/4. Red likelihoods are 1/10, 6/10, 8/10 and 0.
Their likelihoods sum to 15/10. Divide each likelihood by this sum.
Answer
P(A|red)=1/15, P(B|red)=2/5, P(C|red)=8/15.
Question 7
Initial heart-attack risk is 40%. Meditation reduces the risk by 30%; a drug reduces it by 25%. Each option is equally likely. Given a heart attack, find the probability meditation was chosen.
Steps only
Risk after meditation is 0.4×0.7=0.28; after the drug it is 0.4×0.75=0.30.
Equal priors cancel, so posterior meditation is 0.28/(0.28+0.30).
Answer
The probability is 14/29.
Question 8
Each entry of a 2×2 matrix is independently 0 or 1 with equal chance. Find the probability its determinant is positive.
Steps only
For determinant ad−bc to be positive, ad must be 1 and bc must be 0.
This requires a=d=1, while (b,c) can be (0,0),(0,1), or (1,0). There are 3 favourable matrices among 16.
Answer
The probability is 3/16.
Question 9
P(A fails)=0.2, P(B fails alone)=0.15 and P(A and B fail)=0.15. Find (i) P(A fails|B fails), (ii) P(A fails alone).
Steps only
P(B fails)=0.15+0.15=0.30.
(i) 0.15/0.30=1/2.
(ii) P(A fails but B does not)=0.20−0.15=0.05.
Answer
- (i) 1/2,
- (ii) 0.05.
Question 10
Bag I has 3 red, 4 black; Bag II has 4 red, 5 black. A ball moves from I to II, then a red is drawn from II. Given the draw is red, find the probability the transferred ball was black.
Steps only
P(black transfer)=4/7; then P(red draw)=4/10=2/5.
P(red transfer)=3/7; then P(red draw)=5/10=1/2.
Bayes gives [(4/7)(2/5)]/[(4/7)(2/5)+(3/7)(1/2)].
Answer
The probability is 16/31.
Question 11
If P(A)≠0 and P(B|A)=1, which set relation must hold?
Steps only
P(B|A)=1 means P(A∩B)=P(A), so no part of A has positive probability outside B.
Answer
A⊆B (choice A, up to probability-zero outcomes).
Question 12
If P(A|B)>P(A), which statement follows?
Steps only
P(A∩B)/P(B)>P(A), so P(A∩B)>P(A)P(B).
Divide by P(A)>0 to get P(B|A)>P(B).
Answer
Choice (C): P(B|A)>P(B).
Question 13
If P(A)+P(B)−P(A∩B)=P(A), which conditional probability equals 1?
Steps only
The union formula simplifies to P(B)=P(A∩B), so B is contained in A up to zero-probability outcomes.
Answer
P(A|B)=1 when P(B)>0 (choice B).