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NCERT Solutions · Class 11 Maths · Chapter 14

Chapter 14: Probability (Statistics and Probability)

Q: Which cannot be valid assignments for S = {1, …, 7}? (a) 0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6 (b) 1/7 each (c) 0.1, 0.2, …, 0.7 (d) −0.1, 0.2, 0.3, 0.4, −0.2, 0.1, 0.3 (e) 1/14, 2/14, …, 6/14, 15/14.

(c) not valid; (d) not valid; (e) not valid

Q: A die: P of (i) a prime (ii) ≥ 3 (iii) ≤ 1 (iv) > 6 (v) < 6.

(i) ½; (ii) 2/3; (iii) 1/6; (iv) 0; (v) 5/6

Step-by-step solutions to all exercises of Chapter 14, Probability (NCERT Class 11 Maths, 2026-27): sample spaces, events, mutually exclusive and exhaustive events, axiomatic probability and the Miscellaneous Exercise. All 38 questions are answered, with the key answer highlighted.

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$P (E) = \frac{n ( E )}{n ( S )}$ for equally likely outcomes. $P (not A) = 1 - P (A)$ . $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ ; for mutually exclusive events $P (A \cap B) = 0$ .

Exercise 14.1

A die is rolled; E: "shows 4", F: "shows an even number". Are E and F mutually exclusive?

Solution

E = {4}, F = {2, 4, 6}: E ∩ F = {4} ≠ φ: not mutually exclusive.

not mutually exclusive

A die is thrown. Describe A: < 7, B: > 7, C: multiple of 3, D: < 4, E: even and > 4, F: not less than 3; and find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A − C, D − E, E ∩ F′, F′.

Solution

S = {1, …, 6}: A = S, B = φ, C = {3, 6}, D = {1, 2, 3}, E = {6}, F = {3, 4, 5, 6}. A ∪ B = S; 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′ = {1, 2}, so E ∩ F′ = φ.

A = S; B = φ; C = {3, 6}; D = {1, 2, 3}; E = {6}; F = {3, 4, 5, 6}; S; φ; {3, 6}; {6}; {1, 2, 4, 5}; {1, 2, 3}; {1, 2}

A pair of dice: A: sum > 8, B: 2 occurs on either die, C: sum ≥ 7 and a multiple of 3. Which pairs are mutually exclusive?

Solution

A = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}. B = {(2, 1), …, (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)} (11 points). C: sum 9 or 12 = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}. With a 2 the sum is at most 8, so A ∩ B = φ and B ∩ C = φ; A ∩ C = C ≠ φ. A and B; B and C are mutually exclusive.

A and B; B and C

Three coins: A three heads, B two heads and one tail, C three tails, D head on the first coin. Which are (i) mutually exclusive (ii) simple (iii) compound?

Solution

A = {HHH}, B = {HHT, HTH, THH}, C = {TTT}, D = {HHH, HHT, HTH, HTT}. (i) A and B, A and C, B and C, C and D (A ∩ D, B ∩ D ≠ φ). (ii) simple: A, C. (iii) compound: B, D.

(i) A and B, A and C, B and C, C and D; (ii) A, C; (iii) B, D

Three coins are tossed. Describe (i) two mutually exclusive events (ii) three mutually exclusive and exhaustive events (iii) two events that are not mutually exclusive (iv) two mutually exclusive but not exhaustive events (v) three mutually exclusive but not exhaustive events.

Solution

(Many answers are possible.) (i) "at least two heads" {HHH, HHT, HTH, THH} and "at least two tails" {TTT, TTH, THT, HTT}. (ii) "at most one head", "exactly two heads", "exactly three heads". (iii) "at most two tails" and "exactly two heads" (both contain HHT). (iv) "exactly one head" and "exactly two heads" (TTT, HHH left out). (v) "exactly one tail", "exactly two tails", "exactly three tails" (HHH left out).

See the working above.

Two dice: A even on the first die, B odd on the first die, C sum ≤ 5. Describe (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

C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}. (i) A′ = B = {(x, y): x = 1, 3, 5}. (ii) not B = A. (iii) A ∪ B = S. (iv) A ∩ B = φ. (v) A − C = A without (2, 1), (2, 2), (2, 3), (4, 1): {(2, 4), (2, 5), (2, 6), (4, 2), …, (4, 6), (6, 1), …, (6, 6)} (14 points). (vi) B ∪ C = B ∪ {(2, 1), (2, 2), (2, 3), (4, 1)}. (vii) B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}. (viii) B′ = A, so A ∩ A ∩ C′ = A − C, as in (v).

(i) B; (ii) A; (iii) S; (iv) φ; (v) {(2, 4), (2, 5), (2, 6), (4, 2), …, (4, 6), (6, 1), …, (6, 6)}; (vi) B ∪ {(2, 1), (2, 2), (2, 3), (4, 1)}; (vii) {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}; (viii) A − C

For Q6, true or false: (i) A, B mutually exclusive (ii) A, B mutually exclusive and exhaustive (iii) A = B′ (iv) A, C mutually exclusive (v) A, B′ mutually exclusive (vi) A′, B′, C mutually exclusive and exhaustive.

Solution

(i) T (A ∩ B = φ). (ii) T (also A ∪ B = S). (iii) T. (iv) F (A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)}). (v) F (A ∩ B′ = A ≠ φ). (vi) F (A′ ∩ C = B ∩ C ≠ φ).

(i) T; (ii) T; (iii) T; (iv) F; (v) F; (vi) F

Exercise 14.2

Which cannot be valid assignments for S = {1, …, 7}? (a) 0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6 (b) 1/7 each (c) 0.1, 0.2, …, 0.7 (d) −0.1, 0.2, 0.3, 0.4, −0.2, 0.1, 0.3 (e) 1/14, 2/14, …, 6/14, 15/14.

Solution

(a) all in [0, 1], sum 1.00: valid. (b) sum 1: valid. (c) sum 2.8 ≠ 1: not valid. (d) negative values: not valid. (e) 15/14 > 1 (and sum 36/14): not valid.

A coin is tossed twice. P(at least one tail)?

Solution

S = {HH, HT, TH, TT}; three favourable: ¾.

A die: P of (i) a prime (ii) ≥ 3 (iii) ≤ 1 (iv) > 6 (v) < 6.

Solution

(i) {2, 3, 5}: ½ (ii) {3, 4, 5, 6}: 2/3 (iii) {1}: 1/6 (iv) φ: 0 (v) {1, …, 5}: 5/6.

(i) ½; (ii) 2/3; (iii) 1/6; (iv) 0; (v) 5/6

A card from 52: (a) points in S (b) P(ace of spades) (c) P(i) ace (ii) black card.

Solution

(a) 52 (b) 1/52 (c) (i) 4/52 = 1/13 (ii) 26/52 = ½.

(a) 52; (b) 1/52; (i) 1/13; (ii) ½

A coin marked 1 and 6 and a die are tossed. P(sum (i) 3 (ii) 12)?

Solution

S = {(1, 1), …, (1, 6), (6, 1), …, (6, 6)}: 12 points. (i) only (1, 2): 1/12. (ii) only (6, 6): 1/12.

(i) 1/12; (ii) 1/12

4 men and 6 women on a council; one is chosen. P(woman)?

Solution

6/10 = 3/5.

3/5

A coin is tossed four times; win ₹1 per head, lose ₹1.50 per tail. Find the possible amounts and their probabilities.

Solution

16 outcomes; with k heads the amount is k − 1.5(4 − k): k = 4: gain ₹4, P = 1/16; k = 3: gain ₹1.50, 4/16 = 1/4; k = 2: lose ₹1, 6/16 = 3/8; k = 1: lose ₹3.50, 1/4; k = 0: lose ₹6, 1/16. Five amounts.

gain ₹4; gain ₹1.50; lose ₹1; lose ₹3.50; lose ₹6

Three coins: P of (i) 3 heads (ii) 2 heads (iii) ≥ 2 heads (iv) ≤ 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) ≤ 2 tails.

Solution

n(S) = 8. (i) 1/8 (ii) 3/8 (iii) 4/8 = ½ (iv) 1 − 1/8 = 7/8 (v) 1/8 (vi) 1/8 (vii) 3/8 (viii) 1/8 (ix) 7/8.

(i) 1/8; (ii) 3/8; (iii) ½; (iv) 7/8; (v) 1/8; (vi) 1/8; (vii) 3/8; (viii) 1/8; (ix) 7/8

P(A) = 2/11. Find P(not A).

Solution

1 − 2/11 = 9/11.

9/11

A letter from ASSASSINATION: P(vowel), P(consonant)?

Solution

13 letters: A × 3, I × 2, O × 1 are vowels (6); the rest S × 4, N × 2, T × 1 are consonants (7). 6/13 and 7/13.

6/13 and 7/13

Choose 6 different numbers from 1 to 20; P(they match the 6 fixed numbers)?

Solution

Order does not matter: ²⁰C₆ = 38760 equally likely choices, one favourable: 1/38760.

1/38760

Consistent? (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) A ∩ B ⊂ A needs P(A ∩ B) ≤ P(A), but 0.6 > 0.5: not consistent. (ii) P(A ∩ B) = 0.5 + 0.4 − 0.8 = 0.1, which is ≥ 0 and ≤ both; P(A ∪ B) ≥ P(A): consistent.

not consistent; consistent

Fill in: (i) P(A) = 1/3, P(B) = 1/5, P(A ∩ B) = 1/15, P(A ∪ B) = ? (ii) 0.35, ?, 0.25, 0.6 (iii) 0.5, 0.35, ?, 0.7.

Solution

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

(i) 7/15; (ii) 0.5; (iii) 0.15

P(A) = 3/5, P(B) = 1/5, A, B mutually exclusive: P(A or B)?

Solution

3/5 + 1/5 = 4/5.

4/5

P(E) = ¼, P(F) = ½, P(E and F) = 1/8. Find (i) P(E or F) (ii) P(not E and not F).

Solution

(i) ¼ + ½ − 1/8 = 5/8. (ii) P(E′ ∩ F′) = 1 − P(E ∪ F) = 3/8.

5/8; 3/8

P(not E or not F) = 0.25. Are E and F mutually exclusive?

Solution

E′ ∪ F′ = (E ∩ F)′, so P(E ∩ F) = 1 − 0.25 = 0.75 ≠ 0: not mutually exclusive.

not mutually exclusive

P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16. Find (i) P(not A) (ii) P(not B) (iii) P(A or B).

Solution

(i) 0.58 (ii) 0.52 (iii) 0.42 + 0.48 − 0.16 = 0.74.

(i) 0.58; (ii) 0.52; (iii) 0.74

40% study Maths, 30% Biology, 10% both. P(Maths or Biology)?

Solution

0.4 + 0.3 − 0.1 = 0.6.

0.6

P(pass first) = 0.8, P(pass second) = 0.7, P(at least one) = 0.95. P(both)?

Solution

P(A ∩ B) = 0.8 + 0.7 − 0.95 = 0.55.

0.55

P(pass both English and Hindi) = 0.5, P(neither) = 0.1, P(English) = 0.75. P(Hindi)?

Solution

P(E ∪ H) = 1 − 0.1 = 0.9 = 0.75 + P(H) − 0.5 ⇒ P(H) = 0.65.

P(H) = 0.65

60 students: 30 NCC, 32 NSS, 24 both. P (i) NCC or NSS (ii) neither (iii) NSS but not NCC.

Solution

(i) (30 + 32 − 24)/60 = 38/60 = 19/30. (ii) 22/60 = 11/30. (iii) (32 − 24)/60 = 2/15.

(i) 19/30; (ii) 11/30; (iii) 2/15

Miscellaneous Exercise on Chapter 14

10 red, 20 blue, 30 green marbles; 5 drawn. P (i) all blue (ii) at least one green.

Solution

n(S) = ⁶⁰C₅. (i) ²⁰C₅/⁶⁰C₅. (ii) 1 − P(no green) = 1 − ³⁰C₅/⁶⁰C₅.

(i) ²⁰C₅/⁶⁰C₅; (ii) 1 − ³⁰C₅/⁶⁰C₅

4 cards from 52: P(3 diamonds and 1 spade)?

Solution

¹³C₃ × ¹³C₁/⁵²C₄ (= 286 × 13/270725 = 3718/270725).

¹³C₃ × ¹³C₁/⁵²C₄

A die has faces 1, 1, 2, 2, 2, 3. Find (i) P(2) (ii) P(1 or 3) (iii) P(not 3).

Solution

(i) 3/6 = ½ (ii) (2 + 1)/6 = ½ (iii) 1 − 1/6 = 5/6.

(i) ½; (ii) ½; (iii) 5/6

10000 tickets, 10 prizes. P(no prize) with (a) 1 ticket (b) 2 tickets (c) 10 tickets.

Solution

9990 tickets win nothing. (a) 9990/10000 = 0.999 (b) ⁹⁹⁹⁰C₂/¹⁰⁰⁰⁰C₂ (c) ⁹⁹⁹⁰C₁₀/¹⁰⁰⁰⁰C₁₀.

(a) 0.999; (b) ⁹⁹⁹⁰C₂/¹⁰⁰⁰⁰C₂; (c) ⁹⁹⁹⁰C₁₀/¹⁰⁰⁰⁰C₁₀

100 students form sections of 40 and 60. P(you and your friend are (a) in the same section (b) in different sections)?

Solution

Both in the 40: (40 × 39)/(100 × 99); both in the 60: (60 × 59)/(100 × 99). (a) (1560 + 3540)/9900 = 17/33. (b) 1 − 17/33 = 16/33.

(a) 17/33; (b) 16/33

Three letters go randomly into three addressed envelopes. P(at least one is in its proper envelope)?

Solution

3! = 6 arrangements; only 2 put no letter right (the two cyclic ones). P = 1 − 2/6 = 2/3.

2/3

P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35. Find (i) P(A ∪ B) (ii) P(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′).

Solution

(i) 0.88 (ii) 1 − 0.88 = 0.12 (iii) 0.54 − 0.35 = 0.19 (iv) 0.69 − 0.35 = 0.34.

(i) 0.88; (ii) 0.12; (iii) 0.19; (iv) 0.34

Harish (M, 30), Rohan (M, 33), Sheetal (F, 46), Alis (F, 28), Salim (M, 41): P(spokesperson is male or over 35)?

Solution

Male: Harish, Rohan, Salim; over 35: Sheetal, Salim. Union: 4 people: 4/5.

4/5

4-digit numbers > 5000 from 0, 1, 3, 5, 7: P(divisible by 5) (i) with repetition (ii) without.

Solution

(i) First digit 5 or 7: 2 × 5 × 5 × 5 = 250, less 5000 itself: 249. Divisible by 5 (last 0 or 5): 2 × 5 × 5 × 2 = 100, less 5000: 99. P = 99/249 = 33/83. (ii) Total 2 × 4 × 3 × 2 = 48. First 5: last 0, middle from {1, 3, 7}: 6; first 7: last 0 or 5, then 3 × 2: 12. P = 18/48 = 3/8.

33/83; 3/8

A 4-wheel lock (digits 0–9) opens with one sequence of four different digits. P(right sequence)?

Solution

10 × 9 × 8 × 7 = 5040 sequences: 1/5040.

1/5040

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