How to Use This Page
This page provides complete, step-by-step NCERT solutions for every exercise in Chapter 14 — Probability. Each solution identifies the sample space, defines the event, counts favorable outcomes, applies the probability formula P(E) = n(E)/n(S), and simplifies. Try each problem before looking at the answer.
Key Concepts (Quick Recap)
- Sample space (S): The set of all possible outcomes
- Event: A subset of the sample space
- Probability: P(E) = n(E)/n(S), where 0 ≤ P(E) ≤ 1
- P(S) = 1 (sure event), P(∅) = 0 (impossible event)
- Complementary events: P(A') = 1 − P(A)
- Mutually exclusive events: A ∩ B = ∅, so P(A ∪ B) = P(A) + P(B)
- General addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- Exhaustive events: A ∪ B = S
- Number of outcomes when n dice are thrown: 6ⁿ
- Number of outcomes when n coins are tossed: 2ⁿ
- Cards in a deck: 52 (13 each of Spades, Hearts, Diamonds, Clubs)
Exercise 14.1
Q1. Write the sample space when a coin is tossed once.
S = {H, T}
n(S) = 2
Q2. Write the sample space when a die is thrown once.
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
Q3. Write the sample space when a coin is tossed twice.
S = {HH, HT, TH, TT}
n(S) = 4
Q4. Write the sample space when a coin is tossed three times.
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
Q5. A die is thrown twice. Write the sample space.
S = {(1,1), (1,2), …, (1,6), (2,1), (2,2), …, (2,6), …, (6,1), …, (6,6)}
n(S) = 6 × 6 = 36
Q6. A coin is tossed and then a die is thrown. Write the sample space.
S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
n(S) = 2 × 6 = 12
Q7. One card is drawn from a well-shuffled deck of 52 cards. Write the sample space.
S = {all 52 cards in the deck}
n(S) = 52
The sample space can be described as:
- Spades: A, 2, 3, …, K (13 cards)
- Hearts: A, 2, 3, …, K (13 cards)
- Diamonds: A, 2, 3, …, K (13 cards)
- Clubs: A, 2, 3, …, K (13 cards)
Exercise 14.2
Q1. A die is thrown once. Find the probability of getting: (i) a prime number, (ii) a number ≥ 3, (iii) a number ≤ 3.
S = {1, 2, 3, 4, 5, 6}, n(S) = 6.
(i) Prime numbers on a die: {2, 3, 5}, n(E) = 3.
P(prime) = 3/6 = 1/2
(ii) Numbers ≥ 3: {3, 4, 5, 6}, n(E) = 4.
P(≥ 3) = 4/6 = 2/3
(iii) Numbers ≤ 3: {1, 2, 3}, n(E) = 3.
P(≤ 3) = 3/6 = 1/2
Q2. A coin is tossed twice. Find the probability of getting: (i) at least one head, (ii) exactly one head, (iii) at most one head.
S = {HH, HT, TH, TT}, n(S) = 4.
(i) At least one head: {HH, HT, TH}, n(E) = 3.
P(at least one head) = 3/4
Or using complement: P(at least one head) = 1 − P(TT) = 1 − 1/4 = 3/4
(ii) Exactly one head: {HT, TH}, n(E) = 2.
P(exactly one head) = 2/4 = 1/2
(iii) At most one head: {HT, TH, TT}, n(E) = 3.
P(at most one head) = 3/4
Q3. A die is thrown once. Find the probability of getting a number between 2 and 6.
Numbers between 2 and 6 (exclusive): {3, 4, 5}, n(E) = 3.
P = 3/6 = 1/2
Q4. Two coins are tossed simultaneously. Find the probability of getting exactly one tail.
S = {HH, HT, TH, TT}, n(S) = 4.
Exactly one tail: {HT, TH}, n(E) = 2.
P = 2/4 = 1/2
Q5. A die is thrown once. Find the probability of getting an even number.
Even numbers: {2, 4, 6}, n(E) = 3.
P(even) = 3/6 = 1/2
Q6. A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is: (i) a king, (ii) a face card, (iii) a red face card.
n(S) = 52.
(i) Kings: 4 (one of each suit).
P(king) = 4/52 = 1/13
(ii) Face cards: 12 (J, Q, K of each suit).
P(face card) = 12/52 = 3/13
(iii) Red face cards: 6 (J, Q, K of Hearts and Diamonds).
P(red face card) = 6/52 = 3/26
Q7. A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is: (i) a black card, (ii) a diamond, (iii) not a diamond.
(i) Black cards: 26 (13 Spades + 13 Clubs).
P(black) = 26/52 = 1/2
(ii) Diamonds: 13.
P(diamond) = 13/52 = 1/4
(iii) Not a diamond: 52 − 13 = 39.
P(not diamond) = 39/52 = 3/4
Or by complement: P(not diamond) = 1 − 1/4 = 3/4.
Q8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random. Find the probability that the ball drawn is: (i) red, (ii) black.
n(S) = 3 + 5 = 8.
(i) P(red) = 3/8
(ii) P(black) = 5/8
Q9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is drawn at random. Find the probability that the marble is: (i) red, (ii) white, (iii) green.
n(S) = 5 + 8 + 4 = 17.
(i) P(red) = 5/17
(ii) P(white) = 8/17
(iii) P(green) = 4/17
Q10. A coin is tossed three times. Find the probability of getting: (i) exactly 2 heads, (ii) at least 2 heads.
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8.
(i) Exactly 2 heads: {HHT, HTH, THH}, n(E) = 3.
P = 3/8
(ii) At least 2 heads: {HHH, HHT, HTH, THH}, n(E) = 4.
P = 4/8 = 1/2
Miscellaneous Exercise
Q1. A box contains 1 red and 3 identical white balls. Two balls are drawn at random one after the other without replacement. Find the probability that at least one ball is white.
Total balls = 4. Drawing 2 without replacement: n(S) = C(4,2) = 6.
At least one white = 1 − P(both red).
Both red: only 1 red ball, so P(both red) = 0 (cannot draw 2 red balls).
P(at least one white) = 1 − 0 = 1
(Since there is only 1 red ball, at least one ball must be white.)
Q2. A bag contains 4 red and 6 black balls. Two balls are drawn at random with replacement. Find the probability that both are red.
P(red on 1st draw) = 4/10 = 2/5.
With replacement, P(red on 2nd draw) = 2/5.
P(both red) = (2/5) × (2/5) = 4/25
Q3. A bag contains 4 red and 6 black balls. Two balls are drawn at random without replacement. Find the probability that both are red.
P(1st red) = 4/10 = 2/5.
P(2nd red given 1st red) = 3/9 = 1/3.
P(both red) = (2/5) × (1/3) = 2/15
Q4. Three coins are tossed once. Find the probability of: (i) 3 heads, (ii) 2 heads, (iii) at least 2 heads, (iv) at most 2 heads.
n(S) = 8.
(i) 3 heads: {HHH}, n(E) = 1. P = 1/8
(ii) 2 heads: {HHT, HTH, THH}, n(E) = 3. P = 3/8
(iii) At least 2 heads: {HHH, HHT, HTH, THH}, n(E) = 4. P = 4/8 = 1/2
(iv) At most 2 heads: {HHT, HTH, THH, HTT, THT, TTH, TTT}, n(E) = 7. P = 7/8
Q5. Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the dice is: (i) 7, (ii) 10, (iii) at least 10.
n(S) = 36.
(i) Sum = 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, n(E) = 6.
P(sum = 7) = 6/36 = 1/6
(ii) Sum = 10: {(4,6), (5,5), (6,4)}, n(E) = 3.
P(sum = 10) = 3/36 = 1/12
(iii) Sum ≥ 10: {(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}, n(E) = 6.
P(sum ≥ 10) = 6/36 = 1/6
Q6. Two dice are thrown. Find the probability that the numbers shown are both even.
Each die shows an even number (2, 4, 6) with probability 3/6 = 1/2.
P(both even) = (1/2) × (1/2) = 1/4
Q7. A committee of 3 persons is to be formed from 2 men and 3 women. Find the probability that the committee has: (i) all women, (ii) at least one man.
n(S) = C(5,3) = 10.
(i) All women: C(3,3) = 1 way.
P(all women) = 1/10
(ii) At least one man = 1 − P(all women) = 1 − 1/10 = 9/10
Q8. If P(A) = 3/8 and P(B) = 1/2 and A and B are mutually exclusive, find P(A ∪ B).
Since A and B are mutually exclusive, A ∩ B = ∅, so P(A ∩ B) = 0.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 3/8 + 1/2 − 0 = 3/8 + 4/8 = 7/8
Internal Links
- [Chapter 14 Guide — Probability](/school/cbse/class/11/maths/cbse-probability)
- [All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
- [Class 11 Maths Subject Guide](/school/cbse/class/11/maths)