NCERT Solutions for Class 11 Maths Chapter 14: Probability (NCERT 2026–27)

These Class 11 Maths Chapter 14 solutions cover Probability from the NCERT textbook (Reprint 2026–27). Every question of Exercise 14.1, Exercise 14.2 and the Miscellaneous Exercise is reproduced verbatim and solved step by step using the axiomatic approach — events and sample spaces, mutually exclusive and exhaustive events, equally likely outcomes, and the addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Class: 11 Subject: Mathematics Chapter: 14 – Probability Exercises: 14.1, 14.2, Miscellaneous Session: 2026–27
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Chapter 14 Overview

Chapter 14, Probability, develops the axiomatic approach to probability introduced by Kolmogorov. It begins by defining an event as a subset of the sample space and classifies events as impossible, sure, simple, compound, complementary, mutually exclusive and exhaustive. It then states the three probability axioms, derives key results such as P(φ) = 0, the addition theorem P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and the complement rule P(not A) = 1 − P(A), and applies them to coins, dice, cards, committees and counting problems using permutations and combinations. The Class 11 Maths Chapter 14 solutions below work through every exercise question in order.

Key Concepts & Definitions

Event: any subset E of a sample space S.

Impossible & sure event: the empty set φ is the impossible event; the whole sample space S is the sure event.

Simple (elementary) event: an event with exactly one sample point; a compound event has more than one.

Complementary event: ‘not A’ = A′ = S − A.

Event ‘A or B’ = A ∪ B;  event ‘A and B’ = A ∩ B;  event ‘A but not B’ = A − B = A ∩ B′.

Mutually exclusive events: A and B with A ∩ B = φ (they cannot occur together).

Exhaustive events: events E1, E2, …, En with E1 ∪ E2 ∪ … ∪ En = S.

Equally likely outcomes: outcomes each having the same probability.

Important Formulas (Chapter 14)

Axioms: P(E) ≥ 0;   P(S) = 1;   if A, B mutually exclusive then P(A ∪ B) = P(A) + P(B).

Equally likely outcomes: P(A) = n(A) / n(S) = (favourable outcomes) / (total outcomes).

Addition theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Complement rule: P(not A) = P(A′) = 1 − P(A).

De Morgan: P(A′ ∩ B′) = P((A ∪ B)′) = 1 − P(A ∪ B).

‘A but not B’: P(A ∩ B′) = P(A) − P(A ∩ B).

Exercise 14.1 Solutions

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

1. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?

SOLUTION S = {1, 2, 3, 4, 5, 6}, E = {4}, F = {2, 4, 6}. E ∩ F = {4} ≠ φ, so 4 is common to both events. ∴ E and F are not mutually exclusive.

2. A die is thrown. Describe the following events: (i) A: a number less than 7   (ii) B: a number greater than 7 (iii) C: a multiple of 3   (iv) D: a number less than 4 (v) E: an even number greater than 4   (vi) F: a number not less than 3 Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′.

SOLUTION S = {1, 2, 3, 4, 5, 6}. (i) A = {1, 2, 3, 4, 5, 6}   (ii) B = φ   (iii) C = {3, 6} (iv) D = {1, 2, 3}   (v) E = {6}   (vi) F = {3, 4, 5, 6} A ∪ B = {1, 2, 3, 4, 5, 6};   A ∩ B = φ. B ∪ C = {3, 6};   E ∩ F = {6};   D ∩ E = φ. A – C = {1, 2, 4, 5};   D – E = {1, 2, 3}. F′ = S − F = {1, 2};   E ∩ F′ = {6} ∩ {1, 2} = φ.

3. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than 8, B: 2 occurs on either die, C: the sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive?

SOLUTION A (sum > 8, i.e. 9, 10, 11, 12) = {(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}. B (a 2 on either die) = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)}. C (sum ≥ 7 and a multiple of 3, i.e. sum = 9 or 12) = {(3,6), (6,3), (5,4), (4,5), (6,6)}. A ∩ B = φ (no element of A contains a 2), so A and B are mutually exclusive. B ∩ C = φ (no element of C contains a 2), so B and C are mutually exclusive. A ∩ C = {(3,6), (6,3), (5,4), (4,5), (6,6)} ≠ φ, so A and C are not mutually exclusive. ∴ (A, B) and (B, C) are mutually exclusive.

4. Three coins are tossed once. Let A denote the event ‘three heads show’, B denote the event “two heads and one tail show”, C denote the event “three tails show” and D denote the event ‘a head shows on the first coin’. Which events are (i) mutually exclusive?   (ii) simple?   (iii) Compound?

SOLUTION S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. A = {HHH}; B = {HHT, HTH, THH}; C = {TTT}; D = {HHH, HHT, HTH, HTT}. (i) Checking pairs for disjointness: A ∩ B = φ, A ∩ C = φ, B ∩ C = φ, C ∩ D = φ. So the mutually exclusive pairs are A and B; A and C; B and C; C and D. (ii) Simple events (one sample point): A and C. (iii) Compound events (more than one sample point): B and D.

5. Three coins are tossed. Describe (i) Two events which are mutually exclusive. (ii) Three events which are mutually exclusive and exhaustive. (iii) Two events, which are not mutually exclusive. (iv) Two events which are mutually exclusive but not exhaustive. (v) Three events which are mutually exclusive but not exhaustive.

SOLUTION Take S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. (Other valid answers exist.) (i) A: “getting at least two heads” and B: “getting at least two tails” — A ∩ B = φ. (ii) “Getting no heads”, “getting exactly one head” and “getting at least two heads” — pairwise disjoint and their union is S. (iii) “Getting at most two tails” and “getting exactly two tails” — the second is a subset of the first, so they overlap. (iv) “Getting exactly one head” and “getting exactly two heads” — disjoint, but their union is not S. (v) “Getting exactly one tail”, “getting exactly two tails” and “getting exactly three tails” — pairwise disjoint, but their union misses HHH, so not exhaustive. Note: several other sets of events are equally acceptable.

6. Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice ≤ 5. Describe the events (i) A′   (ii) not B   (iii) A or B   (iv) A and B   (v) A but not C   (vi) B or C   (vii) B and C   (viii) A ∩ B′ ∩ C′

SOLUTION A (first die even) = {(2,1)…(2,6), (4,1)…(4,6), (6,1)…(6,6)} (18 outcomes). B (first die odd) = {(1,1)…(1,6), (3,1)…(3,6), (5,1)…(5,6)} (18 outcomes). C (sum ≤ 5) = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}. (i) A′ = B (all outcomes with first die odd). (ii) not B = B′ = A (all outcomes with first die even). (iii) A or B = A ∪ B = S (every outcome). (iv) A and B = A ∩ B = φ (first die cannot be both even and odd). (v) A but not C = A − C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. (vi) B or C = B ∪ C = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1)…(3,6), (5,1)…(5,6)} together with the C-outcomes having an even first die: (2,1), (2,2), (2,3), (4,1). (vii) B and C = B ∩ C = {(1,1), (1,2), (1,3), (1,4), (3,1), (3,2)}. (viii) A ∩ B′ ∩ C′ = A ∩ A ∩ C′ = A − C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} (same as (v)).

7. Refer to question 6 above, state true or false: (give reason for your answer) (i) A and B are mutually exclusive (ii) A and B are mutually exclusive and exhaustive (iii) A = B′ (iv) A and C are mutually exclusive (v) A and B′ are mutually exclusive. (vi) A′, B′, C are mutually exclusive and exhaustive.

SOLUTION (i) True. A ∩ B = φ (first die cannot be both even and odd). (ii) True. A ∩ B = φ and A ∪ B = S, so they are mutually exclusive and exhaustive. (iii) True. B′ = A, hence A = B′. (iv) False. A ∩ C = {(2,1), (2,2), (2,3), (4,1)} ≠ φ, so they can occur together. (v) False. B′ = A, so A ∩ B′ = A ≠ φ; they are not mutually exclusive. (vi) False. A′ = B and B′ = A, so A′ ∩ B′ = B ∩ A = φ, but A′ ∩ C = B ∩ C ≠ φ (e.g. (1,1)), so they are not pairwise mutually exclusive.

Exercise 14.2 Solutions

1. Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = {ω1, ω2, ω3, ω4, ω5, ω6, ω7} (a) 0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6 (b) 1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7 (c) 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7 (d) −0.1, 0.2, 0.3, 0.4, −0.2, 0.1, 0.3 (e) 1/14, 2/14, 3/14, 4/14, 5/14, 6/14, 15/14

SOLUTION A valid assignment needs each probability in [0, 1] and the total equal to 1. (a) Sum = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1, all in [0, 1] → Valid (Yes). (b) Sum = 7 × (1/7) = 1, all in [0, 1] → Valid (Yes). (c) Sum = 0.1 + 0.2 + … + 0.7 = 2.8 ≠ 1 → Not valid (No). (d) Two values are negative (−0.1, −0.2) → Not valid (No). (e) 15/14 > 1 (and sum = 36/14 ≠ 1) → Not valid (No). ∴ (c), (d) and (e) cannot be valid assignments.

2. A coin is tossed twice, what is the probability that atleast one tail occurs?

SOLUTION S = {HH, HT, TH, TT}, n(S) = 4. “At least one tail” = {HT, TH, TT}, so favourable = 3. ∴ P = 3/4 = 3/4.

3. A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear, (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear, (v) A number less than 6 will appear.

SOLUTION S = {1, 2, 3, 4, 5, 6}, n(S) = 6. (i) Primes {2, 3, 5} → 3/6 = 1/2. (ii) {3, 4, 5, 6} → 4/6 = 2/3. (iii) {1} → 1/6. (iv) None → 0/6 = 0. (v) {1, 2, 3, 4, 5} → 5/6.

4. A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) black card.

SOLUTION (a) n(S) = 52. (b) There is exactly one ace of spades → 1/52. (c)(i) 4 aces → 4/52 = 1/13. (c)(ii) 26 black cards (clubs + spades) → 26/52 = 1/2.

5. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.

SOLUTION Coin shows 1 or 6; die shows 1–6. Sample space has 2 × 6 = 12 equally likely outcomes (c, d), c ∈ {1, 6}. (i) Sum = 3: only (1, 2). Favourable = 1 → 1/12. (ii) Sum = 12: only (6, 6). Favourable = 1 → 1/12.

6. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

SOLUTION Total members = 4 + 6 = 10; women = 6. ∴ P(woman) = 6/10 = 3/5.

7. A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

SOLUTION In 4 tosses, n(S) = 24 = 16. If there are h heads and (4 − h) tails, the net amount = h(1) − (4 − h)(1.50) = 2.5h − 6. h = 4: 2.5(4) − 6 = Rs 4.00 gain; ways = 4C4 = 1 → P = 1/16. h = 3: 2.5(3) − 6 = Rs 1.50 gain; ways = 4C3 = 4 → P = 4/16 = 1/4. h = 2: 2.5(2) − 6 = Re 1.00 loss; ways = 4C2 = 6 → P = 6/16 = 3/8. h = 1: 2.5(1) − 6 = Rs 3.50 loss; ways = 4C1 = 4 → P = 4/16 = 1/4. h = 0: 2.5(0) − 6 = Rs 6.00 loss; ways = 4C0 = 1 → P = 1/16. ∴ there are 5 different amounts with the probabilities above.

8. Three coins are tossed once. Find the probability of getting (i) 3 heads   (ii) 2 heads   (iii) atleast 2 heads (iv) atmost 2 heads   (v) no head   (vi) 3 tails (vii) exactly two tails   (viii) no tail   (ix) atmost two tails

SOLUTION S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8. (i) 3 heads {HHH} → 1/8. (ii) exactly 2 heads {HHT, HTH, THH} → 3/8 = 3/8. (iii) at least 2 heads {HHT, HTH, THH, HHH} → 4/8 = 1/2. (iv) at most 2 heads = all except HHH → 7/8 = 7/8. (v) no head {TTT} → 1/8. (vi) 3 tails {TTT} → 1/8. (vii) exactly two tails {HTT, THT, TTH} → 3/8. (viii) no tail {HHH} → 1/8. (ix) at most two tails = all except TTT → 7/8 = 7/8.

9. If 2/11 is the probability of an event, what is the probability of the event ‘not A’.

SOLUTION P(not A) = 1 − P(A) = 1 − 2/11 = 9/11.

10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii) a consonant.

SOLUTION ASSASSINATION has 13 letters: A, S, S, A, S, S, I, N, A, T, I, O, N. Vowels: A, A, A, I, I, O = 6; consonants: S, S, S, S, N, T, N = 7. (i) P(vowel) = 6/13 = 6/13. (ii) P(consonant) = 7/13 = 7/13.

11. In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint order of the numbers is not important.]

SOLUTION Total ways to choose 6 numbers from 20 (order not important) = 20C6. 20C6 = (20 × 19 × 18 × 17 × 16 × 15)/(6!) = 38760. Only one combination wins → P(win) = 1/38760.

12. Check whether the following probabilities P(A) and P(B) are consistently defined (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 (ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

SOLUTION (i) We need P(A ∩ B) ≤ P(A) and ≤ P(B). Here P(A ∩ B) = 0.6 > P(A) = 0.5, which is impossible → not consistently defined (No). (ii) Then P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.8 = 0.1 ≥ 0, and 0.1 ≤ both 0.5 and 0.4 → consistently defined (Yes).

13. Fill in the blanks in following table: (i) P(A) = 1/3, P(B) = 1/5, P(A ∩ B) = 1/15, P(A ∪ B) = … (ii) P(A) = 0.35, P(B) = …, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6 (iii) P(A) = 0.5, P(B) = 0.35, P(A ∩ B) = …, P(A ∪ B) = 0.7

SOLUTION Use P(A ∪ B) = P(A) + P(B) − P(A ∩ B). (i) P(A ∪ B) = 1/3 + 1/5 − 1/15 = 5/15 + 3/15 − 1/15 = 7/15 → 7/15. (ii) P(B) = P(A ∪ B) − P(A) + P(A ∩ B) = 0.6 − 0.35 + 0.25 = 0.5. (iii) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.35 − 0.7 = 0.15.

14. Given P(A) = 3/5 and P(B) = 1/5. Find P(A or B), if A and B are mutually exclusive events.

SOLUTION For mutually exclusive events, P(A or B) = P(A) + P(B). = 3/5 + 1/5 = 4/5.

15. If E and F are events such that P(E) = 1/4, P(F) = 1/2 and P(E and F) = 1/8, find (i) P(E or F), (ii) P(not E and not F).

SOLUTION (i) P(E or F) = P(E) + P(F) − P(E ∩ F) = 1/4 + 1/2 − 1/8 = 2/8 + 4/8 − 1/8 = 5/8 → 5/8. (ii) P(not E and not F) = P(E′ ∩ F′) = 1 − P(E ∪ F) = 1 − 5/8 = 3/8.

16. Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.

SOLUTION P(E′ ∪ F′) = 0.25. By De Morgan, E′ ∪ F′ = (E ∩ F)′, so P((E ∩ F)′) = 0.25. ∴ P(E ∩ F) = 1 − 0.25 = 0.75 ≠ 0. Since P(E ∩ F) ≠ 0, E and F can occur together → they are not mutually exclusive (No).

17. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B).

SOLUTION (i) P(not A) = 1 − 0.42 = 0.58. (ii) P(not B) = 1 − 0.48 = 0.52. (iii) P(A or B) = 0.42 + 0.48 − 0.16 = 0.74.

18. In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

SOLUTION P(M) = 0.40, P(B) = 0.30, P(M ∩ B) = 0.10. P(M or B) = 0.40 + 0.30 − 0.10 = 0.6.

19. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both?

SOLUTION Let A, B be passing the first and second; P(A) = 0.8, P(B) = 0.7, P(A ∪ B) = 0.95. P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 + 0.7 − 0.95 = 0.55.

20. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

SOLUTION Let E, H be passing English and Hindi. P(E ∩ H) = 0.5, P(E′ ∩ H′) = 0.1, P(E) = 0.75. P(E ∪ H) = 1 − P(E′ ∩ H′) = 1 − 0.1 = 0.9. P(H) = P(E ∪ H) − P(E) + P(E ∩ H) = 0.9 − 0.75 + 0.5 = 0.65.

21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS. (ii) The student has opted neither NCC nor NSS. (iii) The student has opted NSS but not NCC.

SOLUTION n(S) = 60. P(NCC) = 30/60, P(NSS) = 32/60, P(NCC ∩ NSS) = 24/60. (i) P(NCC or NSS) = 30/60 + 32/60 − 24/60 = 38/60 = 19/30. (ii) P(neither) = 1 − 19/30 = 11/30 → 11/30. (iii) P(NSS but not NCC) = P(NSS) − P(NCC ∩ NSS) = 32/60 − 24/60 = 8/60 = 2/15.

Miscellaneous Exercise on Chapter 14 Solutions

1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) atleast one will be green?

SOLUTION Total marbles = 10 + 20 + 30 = 60; 5 drawn in 60C5 ways. (i) All 5 blue: choose from 20 blue → P = 20C5 / 60C5. (ii) P(at least one green) = 1 − P(no green) = 1 − (30C5 / 60C5), the “no green” case choosing all 5 from the 30 non-green marbles.

2. 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

SOLUTION Total ways to draw 4 cards = 52C4. 3 diamonds from 13 and 1 spade from 13 → favourable = 13C3 × 13C1. ∴ P = (13C3 × 13C1) / 52C4.

3. A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine (i) P(2)   (ii) P(1 or 3)   (iii) P(not 3)

SOLUTION 6 faces total: two show 1, three show 2, one shows 3. (i) P(2) = 3/6 = 1/2. (ii) P(1 or 3) = (2 + 1)/6 = 3/6 = 1/2. (iii) P(not 3) = 1 − P(3) = 1 − 1/6 = 5/6.

4. In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.

SOLUTION There are 10 winning tickets and 9990 non-winning tickets out of 10000. (a) One ticket: P(no prize) = 9990/10000 = 999/1000. (b) Two tickets: P(both non-prize) = 9990C2 / 10000C2. (c) Ten tickets: P(all ten non-prize) = 9990C10 / 10000C10.

5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (a) you both enter the same section? (b) you both enter the different sections?

SOLUTION Total ways to split 100 students into sections of 40 and 60 = 100C40 (the rest form the other section). (a) Both in the 40-section: remaining 38 chosen from 98 → 98C38; both in the 60-section: remaining 58 chosen from 98 → 98C58. P(same section) = (98C38 + 98C58) / 100C40 = 17/33. (b) P(different sections) = 1 − 17/33 = 16/33.

6. Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

SOLUTION Total arrangements of 3 letters into 3 envelopes = 3! = 6. The number of derangements (no letter correct) of 3 objects = 2 (the arrangements with all wrong). So P(none correct) = 2/6 = 1/3, hence P(at least one correct) = 1 − 1/3 = 2/3.

7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B) (ii) P(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′)

SOLUTION (i) P(A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88. (ii) P(A′ ∩ B′) = 1 − P(A ∪ B) = 1 − 0.88 = 0.12. (iii) P(A ∩ B′) = P(A) − P(A ∩ B) = 0.54 − 0.35 = 0.19. (iv) P(B ∩ A′) = P(B) − P(A ∩ B) = 0.69 − 0.35 = 0.34.

8. From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows: 1. Harish – M – 30   2. Rohan – M – 33   3. Sheetal – F – 46   4. Alis – F – 28   5. Salim – M – 41 A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

SOLUTION n(S) = 5. Let M = male, O = over 35 years. Males = {Harish, Rohan, Salim} → P(M) = 3/5; over 35 = {Sheetal (46), Salim (41)} → P(O) = 2/5. Male and over 35 = {Salim} → P(M ∩ O) = 1/5. P(male or over 35) = 3/5 + 2/5 − 1/5 = 4/5 → 4/5.

9. If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed?

SOLUTION (i) Repetition allowed. For a number > 5000, the first digit is 5 or 7 (2 choices); the other three digits are any of the 5 → 2 × 5 × 5 × 5 = 250 numbers. But 5000 itself is not > 5000; subtracting the single number 5000 gives the count 250 − 1 = 249. Divisible by 5 means last digit 0 or 5 (2 choices): 2 × 5 × 5 × 2 = 100, again excluding 5000 gives 100 − 1 = 99. ∴ P = 99/249 = 33/83. (ii) No repetition. First digit 5 or 7 (2 ways); remaining 3 places from the other 4 digits = 4 × 3 × 2 = 24; total = 2 × 24 = 48 numbers > 5000. Divisible by 5 (last digit 0 or 5): if last is 0, first is 5 or 7 (2), middle two from remaining 3 = 3 × 2 = 6, giving 2 × 6 = 12; if last is 5, first must be 7 (1), middle two from remaining 3 (0,1,3) = 3 × 2 = 6, giving 6. Total favourable = 12 + 6 = 18. ∴ P = 18/48 = 3/8.

10. The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

SOLUTION Number of 4-digit sequences with no repeats = 10 × 9 × 8 × 7 = 5040. Only one sequence opens the lock → P = 1/5040.

Common Mistakes to Avoid

Watch out for these

  • Calling events mutually exclusive without checking that A ∩ B = φ — one shared outcome breaks it.
  • Forgetting that a valid probability assignment needs both 0 ≤ P(ωi) ≤ 1 and a total of 1.
  • Using P(A ∪ B) = P(A) + P(B) for events that are not mutually exclusive — you must subtract P(A ∩ B).
  • Mixing up “A but not B”: P(A ∩ B′) = P(A) − P(A ∩ B), not P(A) − P(B).
  • Forgetting De Morgan: P(A′ ∩ B′) = 1 − P(A ∪ B), and P(A′ ∪ B′) = 1 − P(A ∩ B).
  • In counting problems, using permutations where order does not matter (use combinations) — e.g. selecting a committee or lottery numbers.

Practice MCQs & Assertion–Reason

1. An event is defined as:

(a) the sample space    (b) any subset of the sample space    (c) a single outcome only    (d) the empty set only

2. For any event E, which assignment is valid?

(a) P(E) = −0.2    (b) P(E) = 1.4    (c) P(E) = 0.6    (d) P(S) = 0

3. A coin is tossed twice. The probability of at least one tail is:

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

4. A die is thrown. The probability of getting a prime number is:

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

5. If P(A) = 0.3, P(B) = 0.4 and P(A ∩ B) = 0.1, then P(A ∪ B) is:

(a) 0.5    (b) 0.6    (c) 0.7    (d) 0.8

6. If P(A) = 2/11, then P(not A) is:

(a) 2/11    (b) 9/11    (c) 11/9    (d) 1

7. One card is drawn from a pack of 52. The probability that it is a black card is:

(a) 1/13    (b) 1/4    (c) 1/2    (d) 1/26

8. A letter is chosen at random from ‘ASSASSINATION’. The probability that it is a vowel is:

(a) 6/13    (b) 7/13    (c) 1/2    (d) 5/13

9. Two events A and B are mutually exclusive when:

(a) A ∪ B = S    (b) A ∩ B = φ    (c) A = B    (d) P(A) = P(B)

10. A number lock has 4 wheels (0–9) and opens with 4 digits with no repeats. The probability of opening it in one try is:

(a) 1/10000    (b) 1/5040    (c) 1/4    (d) 1/24

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

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: For mutually exclusive events A and B, P(A ∪ B) = P(A) + P(B).

Reason: If A and B are mutually exclusive, then A ∩ B = φ and P(A ∩ B) = 0.

A-R 2. Assertion: For any event A, P(not A) = 1 − P(A).

Reason: A and A′ are mutually exclusive and exhaustive, so P(A) + P(A′) = 1.

A-R 3. Assertion: The assignment of probabilities 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7 to seven outcomes is valid.

Reason: Each assigned probability lies between 0 and 1.

A-R 4. Assertion: If P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.6, the probabilities are consistently defined.

Reason: P(A ∩ B) must be less than or equal to both P(A) and P(B).

A-R 5. Assertion: If P(E′ ∪ F′) = 0.25, then E and F are not mutually exclusive.

Reason: P(E′ ∪ F′) = 1 − P(E ∩ F), so P(E ∩ F) = 0.75 ≠ 0.

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

Quick Revision Summary

  • An event is any subset of the sample space; φ is impossible and S is the sure event.
  • Axioms: P(E) ≥ 0, P(S) = 1, and for mutually exclusive A, B, P(A ∪ B) = P(A) + P(B).
  • For equally likely outcomes, P(A) = n(A)/n(S).
  • Addition theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
  • Complement: P(not A) = 1 − P(A);   ‘A but not B’: P(A ∩ B′) = P(A) − P(A ∩ B).
  • De Morgan: P(A′ ∩ B′) = 1 − P(A ∪ B); P(A′ ∪ B′) = 1 − P(A ∩ B).
  • A valid probability assignment needs each value in [0, 1] and a total of 1.

How to score full marks in this chapter

List the sample space clearly and state n(S) before counting favourable outcomes. For “or” problems, always check whether events are mutually exclusive before deciding to subtract P(A ∩ B). Convert worded conditions (“at least”, “at most”, “neither”) into set operations and use the complement rule to shorten long cases. In selection problems decide combinations versus permutations by asking whether order matters, and leave answers as exact fractions unless a decimal is given.

Frequently Asked Questions

What is Class 11 Maths Chapter 14 Probability about?

Chapter 14 develops the axiomatic approach to probability: events as subsets of a sample space; impossible, sure, simple, compound, complementary, mutually exclusive and exhaustive events; the probability axioms; and the addition and complement rules applied to coins, dice, cards and counting problems.

How many exercises are there in Class 11 Maths Chapter 14?

There are two main exercises — Exercise 14.1 and Exercise 14.2 — plus a Miscellaneous Exercise on Chapter 14. Every question of all three is solved on this page.

What is the addition theorem of probability?

For any two events A and B, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). When A and B are mutually exclusive, A ∩ B = φ, so it reduces to P(A ∪ B) = P(A) + P(B).

Are these Class 11 Maths Chapter 14 solutions free?

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

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