NCERT Solutions for Class 12 Maths Chapter 13: Probability (NCERT 2026–27)

These Class 12 Maths Chapter 13 solutions cover Probability from the NCERT textbook (Reprint 2026–27). Every question of Exercise 13.1, 13.2, 13.3 and the Miscellaneous Exercise is solved step by step — conditional probability, the multiplication theorem, independent events and Bayes’ theorem — with each final answer cross-checked against the NCERT answer key, so you can revise the whole chapter quickly and score full marks.

Class: 12 Subject: Mathematics Chapter: 13 — Probability Exercises: 13.1, 13.2, 13.3 + Miscellaneous Part: II Session: 2026–27
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Chapter 13 Overview

Chapter 13, Probability, builds on the axiomatic approach studied earlier and develops four central ideas. It begins with conditional probability — the probability of an event E given that another event F has already occurred — and its properties. It then states the multiplication theorem for the joint occurrence of events, defines independent events, and finally introduces the theorem of total probability and Bayes’ theorem, which lets us reverse a conditional probability to find the probability of a “cause”. The Class 12 Maths Chapter 13 solutions below reproduce every NCERT question verbatim and solve each one step by step, with the final answers matched to the textbook key.

Key Concepts & Definitions

Conditional probability: for events E and F with P(F) ≠ 0, P(E|F) = P(E ∩ F) ÷ P(F). It restricts the sample space to F.

Properties: P(S|F) = P(F|F) = 1;   P(E′|F) = 1 − P(E|F);   P((A∪B)|F) = P(A|F) + P(B|F) − P((A∩B)|F).

Multiplication theorem: P(E ∩ F) = P(E)·P(F|E) = P(F)·P(E|F).

Independent events: E and F are independent if P(E ∩ F) = P(E)·P(F); equivalently P(E|F) = P(E) and P(F|E) = P(F). Independence is about probabilities; mutual exclusion is about the events themselves — they are not the same.

Partition of a sample space: events E1,…,En that are pairwise disjoint, exhaustive (their union is S) and each have positive probability.

Total probability: P(A) = Σ P(Ej)·P(A|Ej) over a partition {Ej}.

Bayes’ theorem: the formula for “reverse” (posterior) probability P(Ei|A) of a hypothesis Ei given that A has occurred.

Important Formulas (Chapter 13)

Conditional probability: P(E|F) = P(E ∩ F) ÷ P(F),  P(F) ≠ 0.

Complement: P(E′|F) = 1 − P(E|F).

Multiplication rule: P(E ∩ F) = P(E)·P(F|E) = P(F)·P(E|F); for three events P(E ∩ F ∩ G) = P(E)·P(F|E)·P(G|EF).

Independence: P(E ∩ F) = P(E)·P(F). Then E&F′, E′&F, E′&F′ are also independent.

At least one of two independent events: P(A ∪ B) = 1 − P(A′)·P(B′).

Total probability: P(A) = P(E1)P(A|E1) + … + P(En)P(A|En).

Bayes’ theorem: P(Ei|A) = P(Ei)P(A|Ei) ÷ Σj P(Ej)P(A|Ej).

Exercise 13.1 — Conditional Probability

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers at the back of the book.

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).

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.

2. 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 = 16/25 (= 0.64).

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)

SOLUTION (i) P(A ∩ B) = P(A)·P(B|A) = 0.8 × 0.4 = 0.32. (ii) P(A|B) = P(A ∩ B) ÷ P(B) = 0.32 ÷ 0.5 = 0.64. (iii) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.8 + 0.5 − 0.32 = 0.98.

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

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

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)

SOLUTION (i) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 6/11 + 5/11 − 7/11 = 4/11. (ii) P(A|B) = P(A ∩ B) ÷ P(B) = (4/11) ÷ (5/11) = 4/5. (iii) P(B|A) = P(A ∩ B) ÷ P(A) = (4/11) ÷ (6/11) = 2/3.

Determine P(E|F) in Exercises 6 to 9.

6. A coin is tossed three times, where (i) E : head on third toss,   F : heads on first two tosses (ii) E : at least two heads,   F : at most two heads (iii) E : at most two tails,   F : at least one tail

SOLUTION Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8. (i) E = {HHH, HTH, THH, TTH}, F = {HHH, HHT}; E ∩ F = {HHH}. P(E|F) = P(E∩F)/P(F) = (1/8)/(2/8) = 1/2. (ii) E = {HHH, HHT, HTH, THH} (at least two heads), F = {HHT, HTH, HTT, THH, THT, TTH, TTT} (at most two heads, i.e. not HHH); E ∩ F = {HHT, HTH, THH}. P(E|F) = (3/8)/(7/8) = 3/7. (iii) E = at most two tails = all except TTT = 7 outcomes; F = at least one tail = all except HHH = 7 outcomes; E ∩ F = all except HHH and TTT = 6 outcomes. P(E|F) = (6/8)/(7/8) = 6/7.

7. Two coins are tossed once, where (i) E : tail appears on one coin,   F : one coin shows head (ii) E : no tail appears,   F : no head appears

SOLUTION S = {HH, HT, TH, TT}, n(S) = 4. (i) E = {HT, TH}, F = {HT, TH}; E ∩ F = {HT, TH}. P(E|F) = (2/4)/(2/4) = 1. (ii) E = no tail = {HH}, F = no head = {TT}; E ∩ F = φ. P(E|F) = 0/(1/4) = 0.

8. A die is thrown three times,   E : 4 appears on the third toss,   F : 6 and 5 appears respectively on first two tosses.

SOLUTION Total outcomes = 63 = 216. F = {(6,5,1), (6,5,2), …, (6,5,6)} has 6 outcomes, so P(F) = 6/216. E ∩ F = {(6,5,4)}, one outcome, so P(E ∩ F) = 1/216. P(E|F) = (1/216) ÷ (6/216) = 1/6.

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

SOLUTION Arrange M, F, S: total 3! = 6 arrangements. F (father in middle) = {MFS, SFM}, so n(F) = 2. In both these, the son is at an end, so E ∩ F = {MFS, SFM}, n(E ∩ F) = 2. P(E|F) = (2/6) ÷ (2/6) = 1.

10. A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

SOLUTION (a) Let black die = 5: outcomes (5,1),…,(5,6) — 6 cases. Sum > 9 needs red ≥ 5: (5,5), (5,6) — 2 favourable. P = 2/6 = 1/3. (b) Red die < 4 means red ∈ {1,2,3}: 18 outcomes. Sum = 8 with red < 4: (6,2) and (5,3) — 2 favourable. P = 2/18 = 1/9.

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 ∩ F)|G)

SOLUTION P(E) = 3/6, P(F) = 2/6, P(G) = 4/6. (i) E ∩ F = {3}, P = 1/6. P(E|F) = (1/6)/(2/6) = 1/2; P(F|E) = (1/6)/(3/6) = 1/3. (ii) E ∩ G = {3,5}, P = 2/6. P(E|G) = (2/6)/(4/6) = 1/2; P(G|E) = (2/6)/(3/6) = 2/3. (iii) E ∪ F = {1,2,3,5}; (E ∪ F) ∩ G = {2,3,5}, P = 3/6. P((E ∪ F)|G) = (3/6)/(4/6) = 3/4. E ∩ F = {3}; (E ∩ F) ∩ G = {3}, P = 1/6. P((E ∩ F)|G) = (1/6)/(4/6) = 1/4.

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?

SOLUTION S = {(b,b), (b,g), (g,b), (g,g)} with the first letter the elder child. Let A = both girls = {(g,g)}. (i) F = youngest (second) is a girl = {(b,g), (g,g)}, n(F) = 2; A ∩ F = {(g,g)}. P(A|F) = (1/4)/(2/4) = 1/2. (ii) F = at least one girl = {(b,g), (g,b), (g,g)}, n(F) = 3; A ∩ F = {(g,g)}. P(A|F) = (1/4)/(3/4) = 1/3.

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?

SOLUTION Total = 300 + 200 + 500 + 400 = 1400. Let A = easy, B = multiple choice. Multiple choice questions = 500 + 400 = 900, so P(B) = 900/1400. Easy and multiple choice = 500, so P(A ∩ B) = 500/1400. P(A|B) = 500/900 = 5/9.

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

SOLUTION Let F = the two numbers are different. Outcomes with equal numbers = 6, so n(F) = 36 − 6 = 30. Let E = sum is 4 with different numbers = {(1,3), (3,1)} (note (2,2) is excluded), n(E ∩ F) = 2. P(E|F) = (2/36) ÷ (30/36) = 2/30 = 1/15.

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’.

SOLUTION If the first throw is 3 or 6 (a multiple of 3) the die is thrown again; for 1,2,4,5 a coin is tossed. The sample space is {(3,1),(3,2),…,(3,6),(6,1),…,(6,6),(1,H),(1,T),(2,H),(2,T),(4,H),(4,T),(5,H),(5,T)}. Let F = ‘at least one die shows a 3’. This requires the die to have been thrown again, and a 3 to appear: F = {(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)}. Let E = ‘the coin shows a tail’. The coin is tossed only when the first throw is not a multiple of 3, so no outcome of E involves a die showing 3. Hence E ∩ F = φ. P(E|F) = 0 ÷ P(F) = 0.

In each of the Exercises 16 and 17 choose the correct answer:

16. If P(A) = 1/2, P(B) = 0, then P(A|B) is (A) 0   (B) 1/2   (C) not defined   (D) 1

SOLUTION P(A|B) = P(A ∩ B) ÷ P(B). Since P(B) = 0, the formula divides by 0, so it is not defined. ∴ correct option is (C) not defined.

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)

SOLUTION P(A|B) = P(A ∩ B)/P(B) and P(B|A) = P(A ∩ B)/P(A). Setting them equal gives P(A ∩ B)/P(B) = P(A ∩ B)/P(A), so P(A) = P(B). ∴ correct option is (D) P(A) = P(B).

Exercise 13.2 — Independent Events

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

SOLUTION For independent events, P(A ∩ B) = P(A)·P(B) = (3/5)(1/5) = 3/25.

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.

SOLUTION There are 26 black cards. P(first black) = 26/52. Given the first is black, P(second black) = 25/51. P(both black) = (26/52)(25/51) = (1/2)(25/51) = 25/102.

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.

SOLUTION The box is approved iff all three drawn oranges are good. Drawing without replacement: P = (12/15)(11/14)(10/13) = 1320 ÷ 2730 = 44/91.

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

SOLUTION P(A) = 1/2, P(B) = 1/6. The coin and die are independent experiments, so P(A ∩ B) = P(head and 3) = 1/12. P(A)·P(B) = (1/2)(1/6) = 1/12 = P(A ∩ B). Hence A and B are independent events.

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?

SOLUTION A = {2,4,6}, P(A) = 3/6 = 1/2. B = {1,2,3} (red), P(B) = 3/6 = 1/2. A ∩ B = {2}, P(A ∩ B) = 1/6. P(A)·P(B) = (1/2)(1/2) = 1/4 ≠ 1/6 = P(A ∩ B). Hence A and B are not independent.

6. Let E and F be events with P(E) = 3/5, P(F) = 3/10 and P(E ∩ F) = 1/5. Are E and F independent?

SOLUTION P(E)·P(F) = (3/5)(3/10) = 9/50. But P(E ∩ F) = 1/5 = 10/50. Since 9/50 ≠ 10/50, E and F are not independent.

7. Given that the events A and B are such that P(A) = 1/2, P(A ∪ B) = 3/5 and P(B) = p. Find p if they are (i) mutually exclusive (ii) independent.

SOLUTION (i) Mutually exclusive: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B). 3/5 = 1/2 + p ⇒ p = 3/5 − 1/2 = 1/10. (ii) Independent: P(A ∩ B) = P(A)·P(B) = p/2. Then 3/5 = 1/2 + p − p/2 = 1/2 + p/2 ⇒ p/2 = 1/10 ⇒ p = 1/5.

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)

SOLUTION (i) P(A ∩ B) = P(A)·P(B) = 0.3 × 0.4 = 0.12. (ii) P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58. (iii) P(A|B) = P(A) = 0.3 (independent). (iv) P(B|A) = P(B) = 0.4 (independent).

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

SOLUTION P(not A and not B) = P(A′ ∩ B′) = P((A ∪ B)′) = 1 − P(A ∪ B). P(A ∪ B) = 1/4 + 1/2 − 1/8 = 2/8 + 4/8 − 1/8 = 5/8. ∴ P(A′ ∩ B′) = 1 − 5/8 = 3/8.

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

SOLUTION P(not A or not B) = P(A′ ∪ B′) = P((A ∩ B)′) = 1 − P(A ∩ B) = 1/4, so P(A ∩ B) = 3/4. P(A)·P(B) = (1/2)(7/12) = 7/24 ≠ 3/4 = P(A ∩ B). Hence A and B are not independent.

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)

SOLUTION (i) P(A and B) = P(A)·P(B) = 0.3 × 0.6 = 0.18. (ii) P(A and not B) = P(A)·P(B′) = 0.3 × 0.4 = 0.12. (iii) P(A or B) = 0.3 + 0.6 − 0.18 = 0.72. (iv) P(neither A nor B) = 1 − P(A or B) = 1 − 0.72 = 0.28.

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

SOLUTION P(odd in one throw) = 3/6 = 1/2, so P(not odd in one throw) = 1/2. P(no odd in all three) = (1/2)3 = 1/8. P(at least one odd) = 1 − 1/8 = 7/8.

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

SOLUTION Total = 18. With replacement, P(red) = 8/18 = 4/9, P(black) = 10/18 = 5/9 each draw. (i) P(both red) = (4/9)(4/9) = 16/81. (ii) P(first black, second red) = (5/9)(4/9) = 20/81. (iii) P(one black, one red) = P(B,R) + P(R,B) = (5/9)(4/9) + (4/9)(5/9) = 20/81 + 20/81 = 40/81.

14. Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved   (ii) exactly one of them solves the problem.

SOLUTION P(A) = 1/2, P(B) = 1/3, so P(A′) = 1/2, P(B′) = 2/3. (i) P(solved) = 1 − P(A′)P(B′) = 1 − (1/2)(2/3) = 1 − 1/3 = 2/3. (ii) P(exactly one) = P(A)P(B′) + P(A′)P(B) = (1/2)(2/3) + (1/2)(1/3) = 2/6 + 1/6 = 3/6 = 1/2.

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? (i) E : ‘the card drawn is a spade’; F : ‘the card drawn is an ace’ (ii) E : ‘the card drawn is black’; F : ‘the card drawn is a king’ (iii) E : ‘the card drawn is a king or queen’; F : ‘the card drawn is a queen or jack’.

SOLUTION (i) P(E) = 13/52 = 1/4, P(F) = 4/52 = 1/13, P(E ∩ F) = 1/52 (ace of spades). P(E)P(F) = (1/4)(1/13) = 1/52 = P(E ∩ F). Independent. (ii) P(E) = 26/52 = 1/2, P(F) = 4/52 = 1/13, P(E ∩ F) = 2/52 = 1/26 (two black kings). P(E)P(F) = (1/2)(1/13) = 1/26 = P(E ∩ F). Independent. (iii) P(E) = 8/52 = 2/13, P(F) = 8/52 = 2/13, E ∩ F = queens = 4/52 = 1/13. P(E)P(F) = (2/13)(2/13) = 4/169 ≠ 1/13. Not independent.

16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

SOLUTION Let H = reads Hindi, E = reads English. P(H) = 0.6, P(E) = 0.4, P(H ∩ E) = 0.2. (a) P(neither) = 1 − P(H ∪ E) = 1 − (0.6 + 0.4 − 0.2) = 1 − 0.8 = 0.2 = 1/5. (b) P(E|H) = P(H ∩ E) ÷ P(H) = 0.2/0.6 = 1/3. (c) P(H|E) = P(H ∩ E) ÷ P(E) = 0.2/0.4 = 1/2.

Choose the correct answer in Exercises 17 and 18.

17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is (A) 0   (B) 1/3   (C) 1/12   (D) 1/36

SOLUTION The only even prime is 2. P(2 on a die) = 1/6, so P(2 on both) = (1/6)(1/6) = 1/36. ∴ correct option is (D) 1/36.

18. Two events A and B will be independent, if (A) A and B are mutually exclusive (B) P(A′B′) = [1 – P(A)] [1 – P(B)] (C) P(A) = P(B) (D) P(A) + P(B) = 1

SOLUTION If A and B are independent then so are A′ and B′, giving P(A′ ∩ B′) = P(A′)P(B′) = [1 − P(A)][1 − P(B)]; this condition is equivalent to independence. ∴ correct option is (B).

Exercise 13.3 — Bayes’ Theorem & Total Probability

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 balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

SOLUTION First draw red (P = 5/10): urn becomes 7 red, 5 black; P(2nd red) = 7/12. Contribution = (1/2)(7/12) = 7/24. First draw black (P = 5/10): urn becomes 5 red, 7 black; P(2nd red) = 5/12. Contribution = (1/2)(5/12) = 5/24. By total probability, P(2nd red) = 7/24 + 5/24 = 12/24 = 1/2.

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.

SOLUTION P(E1) = P(E2) = 1/2. P(R|E1) = 4/8 = 1/2; P(R|E2) = 2/8 = 1/4. By Bayes’: P(E1|R) = [(1/2)(1/2)] ÷ [(1/2)(1/2) + (1/2)(1/4)] = (1/4) ÷ (1/4 + 1/8) = (1/4) ÷ (3/8) = 2/3.

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 a hostlier?

SOLUTION P(H) = 0.6, P(D) = 0.4; P(A|H) = 0.3, P(A|D) = 0.2. P(H|A) = (0.6 × 0.3) ÷ (0.6 × 0.3 + 0.4 × 0.2) = 0.18 ÷ (0.18 + 0.08) = 0.18/0.26 = 9/13.

4. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?

SOLUTION Let K = knows, G = guesses, C = correct. P(K) = 3/4, P(G) = 1/4; P(C|K) = 1, P(C|G) = 1/4. P(K|C) = [(3/4)(1)] ÷ [(3/4)(1) + (1/4)(1/4)] = (3/4) ÷ (3/4 + 1/16) = (3/4) ÷ (13/16) = (12/16)/(13/16) = 12/13.

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 person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

SOLUTION Let D = has disease, Pos = test positive. P(D) = 0.001, P(D′) = 0.999; P(Pos|D) = 0.99, P(Pos|D′) = 0.005. P(D|Pos) = (0.001 × 0.99) ÷ (0.001 × 0.99 + 0.999 × 0.005) = 0.00099 ÷ (0.00099 + 0.004995). = 0.00099 ÷ 0.005985 = 990/5985 = 22/133 (≈ 0.1654).

6. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

SOLUTION P(E1) = P(E2) = P(E3) = 1/3; P(H|E1) = 1, P(H|E2) = 3/4, P(H|E3) = 1/2. P(E1|H) = [(1/3)(1)] ÷ [(1/3)(1 + 3/4 + 1/2)] = 1 ÷ (9/4) = 4/9.

7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

SOLUTION Total = 12000. P(S) = 2000/12000 = 1/6, P(C) = 4000/12000 = 1/3, P(T) = 6000/12000 = 1/2. P(Acc|S) = 0.01, P(Acc|C) = 0.03, P(Acc|T) = 0.15. Denominator = (1/6)(0.01) + (1/3)(0.03) + (1/2)(0.15) = 0.01/6 + 0.03/3 + 0.15/2 = 0.001667 + 0.01 + 0.075 = 0.086667. P(S|Acc) = 0.001667 ÷ 0.086667 = 1/52 = 1/52.

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 it was produced by machine B?

SOLUTION P(A) = 0.6, P(B) = 0.4; P(D|A) = 0.02, P(D|B) = 0.01. P(B|D) = (0.4 × 0.01) ÷ (0.6 × 0.02 + 0.4 × 0.01) = 0.004 ÷ (0.012 + 0.004) = 0.004/0.016 = 1/4.

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.

SOLUTION P(G1) = 0.6, P(G2) = 0.4; P(N|G1) = 0.7, P(N|G2) = 0.3. P(G2|N) = (0.4 × 0.3) ÷ (0.6 × 0.7 + 0.4 × 0.3) = 0.12 ÷ (0.42 + 0.12) = 0.12/0.54 = 2/9.

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?

SOLUTION Let E1 = die shows 1,2,3,4 (P = 4/6 = 2/3); E2 = die shows 5,6 (P = 2/6 = 1/3); A = exactly one head. If E1: one toss, exactly one head means head ⇒ P(A|E1) = 1/2. If E2: three tosses, exactly one head ⇒ P(A|E2) = 3C1(1/2)3 = 3/8. P(E1|A) = [(2/3)(1/2)] ÷ [(2/3)(1/2) + (1/3)(3/8)] = (1/3) ÷ (1/3 + 1/8) = (1/3) ÷ (11/24) = (8/24)/(11/24) = 8/11.

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 it was produced by A?

SOLUTION P(A) = 0.5, P(B) = 0.3, P(C) = 0.2; P(D|A) = 0.01, P(D|B) = 0.05, P(D|C) = 0.07. Denominator = 0.5×0.01 + 0.3×0.05 + 0.2×0.07 = 0.005 + 0.015 + 0.014 = 0.034. P(A|D) = 0.005 ÷ 0.034 = 5/34 = 5/34.

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.

SOLUTION Let E1 = lost card is a diamond (P = 13/52 = 1/4), E2 = lost card is not a diamond (P = 39/52 = 3/4). A = the two drawn cards are diamonds. If E1: 12 diamonds remain among 51 cards: P(A|E1) = (12/51)(11/50) = 132/2550. If E2: 13 diamonds remain among 51 cards: P(A|E2) = (13/51)(12/50) = 156/2550. P(E1|A) = [(1/4)(132/2550)] ÷ [(1/4)(132/2550) + (3/4)(156/2550)] = 132 ÷ (132 + 468) = 132/600 = 11/50.

13. Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is (A) 4/5   (B) 1/2   (C) 1/5   (D) 2/5

SOLUTION Let H = head occurs (P = 1/2), T = tail occurs (P = 1/2); R = A reports head. P(R|H) = 4/5 (truth), P(R|T) = 1/5 (lie). P(H|R) = [(1/2)(4/5)] ÷ [(1/2)(4/5) + (1/2)(1/5)] = (4/10) ÷ (4/10 + 1/10) = 4/5. ∴ correct option is (A) 4/5.

14. If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct? (A) P(A|B) = P(B)/P(A)   (B) P(A|B) < P(A)   (C) P(A|B) ≥ P(A)   (D) None of these

SOLUTION Since A ⊂ B, A ∩ B = A, so P(A|B) = P(A)/P(B). As 0 < P(B) ≤ 1, dividing by P(B) gives P(A|B) ≥ P(A). ∴ correct option is (C) P(A|B) ≥ P(A).

Miscellaneous Exercise on Chapter 13

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

SOLUTION (i) If A ⊂ B, then A ∩ B = A, so P(B|A) = P(A ∩ B)/P(A) = P(A)/P(A) = 1. (ii) If A ∩ B = φ, then P(A ∩ B) = 0, so P(B|A) = 0/P(A) = 0.

2. A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male. (ii) Find the probability that both children are females, if it is known that the elder child is a female.

SOLUTION S = {(b,b), (b,g), (g,b), (g,g)}, first letter = elder child. (i) F = at least one male = {(b,b), (b,g), (g,b)}, n(F) = 3; A = both male = {(b,b)}. P(A|F) = (1/4)/(3/4) = 1/3. (ii) F = elder is female = {(g,b), (g,g)}, n(F) = 2; A = both female = {(g,g)}. P(A|F) = (1/4)/(2/4) = 1/2.

3. Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

SOLUTION P(M) = P(W) = 1/2; P(G|M) = 0.05, P(G|W) = 0.0025. P(M|G) = (0.5 × 0.05) ÷ (0.5 × 0.05 + 0.5 × 0.0025) = 0.05 ÷ (0.05 + 0.0025) = 0.05/0.0525 = 20/21.

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?

SOLUTION Let p = P(right-handed) = 9/10, q = 1/10. ‘At most 6 right-handed’ is the complement of ‘7, 8, 9 or 10 right-handed’. P(at most 6) = 1 − P(7 or more right-handed) = 1 − Σr=710 10Cr (9/10)r(1/10)10−r. ∴ the required probability = 1 − Σr=710 10Cr (9/10)r(1/10)10−r (the NCERT key leaves the answer in this form).

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

SOLUTION A leap year has 366 days = 52 complete weeks + 2 extra days. These two extra days form one of: (Sun,Mon), (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun) — 7 equally likely pairs. The year has 53 Tuesdays if Tuesday is among the two extra days: (Mon,Tue) and (Tue,Wed) — 2 favourable pairs. ∴ required probability = 2/7.

6. Suppose we have four boxes A, B, C and D containing coloured marbles as given below: Box A — Red 1, White 6, Black 3; Box B — Red 6, White 2, Black 2; Box C — Red 8, White 1, Black 1; Box D — Red 0, White 6, Black 4. One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

SOLUTION Each box has 10 marbles and P(box) = 1/4. P(R|A) = 1/10, P(R|B) = 6/10, P(R|C) = 8/10, P(R|D) = 0/10. Denominator = (1/4)(1/10 + 6/10 + 8/10 + 0) = (1/4)(15/10) = 15/40. P(A|R) = (1/4)(1/10) ÷ (15/40) = (1/40)/(15/40) = 1/15. P(B|R) = (1/4)(6/10) ÷ (15/40) = (6/40)/(15/40) = 6/15 = 2/5. P(C|R) = (1/4)(8/10) ÷ (15/40) = (8/40)/(15/40) = 8/15.

7. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

SOLUTION Let E1 = yoga (P = 1/2), E2 = drug (P = 1/2). Base risk = 0.40. P(A|E1) = 0.40 × (1 − 0.30) = 0.40 × 0.70 = 0.28.   P(A|E2) = 0.40 × (1 − 0.25) = 0.40 × 0.75 = 0.30. P(E1|A) = (0.5 × 0.28) ÷ (0.5 × 0.28 + 0.5 × 0.30) = 0.14 ÷ (0.14 + 0.15) = 0.14/0.29 = 14/29.

8. 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 1/2).

SOLUTION A 2×2 determinant has 4 entries, each 0 or 1, so 24 = 16 equally likely matrices. Value = ad − bc. The value is positive (= 1) only when ad = 1 and bc = 0, i.e. a = d = 1 and (b,c) ≠ (1,1). With a = d = 1, (b,c) can be (0,0), (0,1), (1,0) — 3 cases give a positive value. ∴ required probability = 3/16 = 3/16.

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

SOLUTION P(A fails) = 0.2; P(A ∩ B fail) = 0.15. P(B fails alone) = P(B) − P(A ∩ B) = 0.15, so P(B fails) = 0.15 + 0.15 = 0.30. (i) P(A|B failed) = P(A ∩ B) ÷ P(B) = 0.15/0.30 = 0.5. (ii) P(A fails alone) = P(A) − P(A ∩ B) = 0.20 − 0.15 = 0.05.

10. 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.

SOLUTION Let E1 = transferred ball is red (P = 3/7), E2 = transferred ball is black (P = 4/7). After transfer, Bag II has 10 balls. If E1: Bag II = 5 red, 5 black, P(R|E1) = 5/10. If E2: Bag II = 4 red, 6 black, P(R|E2) = 4/10. P(E2|R) = [(4/7)(4/10)] ÷ [(3/7)(5/10) + (4/7)(4/10)] = (16/70) ÷ (15/70 + 16/70) = 16/31 = 16/31.

Choose the correct answer in each of the following:

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

SOLUTION P(B|A) = P(A ∩ B)/P(A) = 1 ⇒ P(A ∩ B) = P(A), which means A is contained in B, i.e. A ⊂ B. ∴ correct option is (A) A ⊂ B.

12. 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)

SOLUTION P(A|B) > P(A) ⇒ P(A ∩ B)/P(B) > P(A) ⇒ P(A ∩ B) > P(A)·P(B). Dividing by P(A) gives P(A ∩ B)/P(A) > P(B), i.e. P(B|A) > P(B). ∴ correct option is (C) P(B|A) > P(B).

13. 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

SOLUTION P(A) + P(B) − P(A ∩ B) = P(A) ⇒ P(B) = P(A ∩ B). Then P(A|B) = P(A ∩ B)/P(B) = P(B)/P(B) = 1. ∴ correct option is (B) P(A|B) = 1.

Common Mistakes to Avoid

Watch out for these

  • Confusing P(E|F) with P(F|E) — always divide P(E ∩ F) by the probability of the given (conditioning) event.
  • Treating independent and mutually exclusive as the same. Independence means P(A ∩ B) = P(A)P(B); mutually exclusive means A ∩ B = φ (so P(A ∩ B) = 0) — two different ideas.
  • Forgetting to change the counts after a draw without replacement (the denominator shrinks; matching balls are removed).
  • In Bayes’ problems, leaving out a hypothesis from the denominator — it must sum P(Ej)P(A|Ej) over all branches.
  • Reading “reduces the risk by 30%” as a probability of 0.30 instead of multiplying the base risk by (1 − 0.30).
  • Using P(A|B) = P(A ∩ B)/P(B) when P(B) = 0 — the conditional probability is then not defined.

Practice MCQs & Assertion–Reason

1. If P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, then P(A|B) is:

(a) 0.4    (b) 0.5    (c) 0.2    (d) 0.8

2. For independent events A and B, P(A ∩ B) equals:

(a) P(A) + P(B)    (b) P(A)·P(B)    (c) P(A) − P(B)    (d) 0

3. A die is thrown twice. The probability of getting an even number both times is:

(a) 1/2    (b) 1/3    (c) 1/4    (d) 1/6

4. If P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, then P(F|E) is:

(a) 1/3    (b) 2/3    (c) 1/2    (d) 2/9

5. Two cards are drawn without replacement from a pack of 52. The probability both are black is:

(a) 1/4    (b) 25/102    (c) 13/51    (d) 1/2

6. A leap year selected at random contains 53 Tuesdays with probability:

(a) 1/7    (b) 2/7    (c) 3/7    (d) 1/53

7. If A and B are independent with P(A) = 0.3, P(B) = 0.6, then P(A ∪ B) is:

(a) 0.9    (b) 0.72    (c) 0.18    (d) 0.28

8. The probability of getting an odd number at least once in three throws of a die is:

(a) 1/8    (b) 1/2    (c) 7/8    (d) 3/8

9. If P(B) = 0, then P(A|B) is:

(a) 0    (b) 1    (c) P(A)    (d) not defined

10. A man speaks truth 3 out of 4 times. He throws a die and reports a six. The probability it is actually a six is:

(a) 3/8    (b) 1/8    (c) 5/8    (d) 1/6

Answer key: 1-(b), 2-(b), 3-(c), 4-(a), 5-(b), 6-(b), 7-(b), 8-(c), 9-(d), 10-(a).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: If A and B are independent, then P(A|B) = P(A).

Reason: For independent events, the occurrence of B does not affect the probability of A.

A-R 2. Assertion: Two mutually exclusive events with non-zero probabilities cannot be independent.

Reason: For mutually exclusive events P(A ∩ B) = 0, whereas independence needs P(A ∩ B) = P(A)·P(B) > 0.

A-R 3. Assertion: For any events E and F with P(F) ≠ 0, P(E′|F) = 1 − P(E|F).

Reason: E and E′ are complementary, and conditional probability is additive over disjoint events given F.

A-R 4. Assertion: If A ⊂ B and P(B) ≠ 0, then P(A|B) ≥ P(A).

Reason: For A ⊂ B, P(A|B) = P(A)/P(B) and 0 < P(B) ≤ 1.

A-R 5. Assertion: If P(B|A) = 1 with P(A) ≠ 0, then B ⊂ A.

Reason: P(B|A) = 1 implies P(A ∩ B) = P(A), i.e. A is contained in B.

Answer key: 1-(A), 2-(A), 3-(A), 4-(A), 5-(D).

Quick Revision Summary

  • Conditional probability: P(E|F) = P(E ∩ F) ÷ P(F), valid only when P(F) ≠ 0.
  • Properties: 0 ≤ P(E|F) ≤ 1, P(E′|F) = 1 − P(E|F), P((E∪F)|G) = P(E|G) + P(F|G) − P((E∩F)|G).
  • Multiplication theorem: P(E ∩ F) = P(E)P(F|E) = P(F)P(E|F).
  • E and F independent ⇔ P(E ∩ F) = P(E)P(F); then E&F′, E′&F, E′&F′ are also independent.
  • P(at least one of two independent events) = 1 − P(A′)P(B′).
  • Total probability: P(A) = Σ P(Ej)P(A|Ej) over a partition {Ej}.
  • Bayes’ theorem: P(Ei|A) = P(Ei)P(A|Ei) ÷ Σj P(Ej)P(A|Ej) — the reverse (posterior) probability of a cause.

How to score full marks in this chapter

Write the formula you are using before substituting (it earns method marks), and for Bayes’ problems first list every hypothesis with its prior P(Ei) and likelihood P(A|Ei) in a small table, then plug into the formula. For “at least one” type questions use the complement 1 − P(none). In without-replacement problems, reduce both the favourable count and the total after each draw. Keep fractions exact and simplify only at the end to avoid rounding errors.

Frequently Asked Questions

What is Class 12 Maths Chapter 13 Probability about?

Chapter 13 covers conditional probability and its properties, the multiplication theorem of probability, independent events, the theorem of total probability and Bayes’ theorem, with applications to drawing balls, cards, defective items and reliability tests.

How many exercises are there in Class 12 Maths Chapter 13?

There are three numbered exercises — Exercise 13.1 (conditional probability), 13.2 (independent events) and 13.3 (Bayes’ theorem and total probability) — plus a Miscellaneous Exercise on Chapter 13. All of them are solved on this page.

What is the difference between independent and mutually exclusive events?

Independent events satisfy P(A ∩ B) = P(A)·P(B) — one event does not affect the other’s probability. Mutually exclusive events cannot occur together, so P(A ∩ B) = 0. Two events with non-zero probabilities cannot be both independent and mutually exclusive.

Are these Class 12 Maths Chapter 13 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every final answer verified against the book’s answer key.

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