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Chapter 8 · Class 11 Mathematics

Probability — Important Questions

34 questions With answers CBSE format

SUMMARY: The chapter on Probability in Class 11 Mathematics introduces the fundamental concepts and principles of probability, including its classical definition and applications.
KEY TOPICS: random experiments, sample space, events, probability of an event, axiomatic approach to probability, mutually exclusive events, exhaustive events, complementary events, conditional probability, independent events

Previous chapter Permutations and Combinations Next chapter Relations and Functions
Multiple Choice Questions (MCQs) 5 questions
Q1 1 Mark

The probability of an impossible event is:

A0
B0.5
C1
D−1
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Correct answer: Option 1 — 0
Q2 1 Mark

The probability of a sure event is:

A0
B0.5
C1
D>1
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Correct answer: Option 3 — 1
Q3 1 Mark

The sample space of tossing 2 coins has size:

A2
B3
C4
D8
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Correct answer: Option 3 — 4
Q4 1 Mark

The probability of getting a head when a fair coin is tossed is:

A0
B1/4
C1/2
D1
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Correct answer: Option 3 — 1/2
Q5 1 Mark

For two mutually exclusive events: P(A ∪ B) equals:

AP(A) + P(B)
BP(A) − P(B)
CP(A) · P(B)
DP(A) / P(B)
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Correct answer: Option 1 — P(A) + P(B)
Short Answer Questions 5 questions
Q6 3 Marks

Define mutually exclusive events with an example.

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Two events are mutually exclusive if they cannot occur simultaneously, i.e. A ∩ B = ∅. Example: in tossing a die, the events 'rolling an even' and 'rolling an odd' are mutually exclusive.
Q7 3 Marks

A die is rolled. Find the probability of getting an even number.

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Sample space S = {1, 2, 3, 4, 5, 6}, |S| = 6. Even outcomes = {2, 4, 6}, count = 3. P(even) = 3/6 = 1/2.
Q8 3 Marks

Two coins are tossed. Find the probability of getting at least one head.

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Sample space {HH, HT, TH, TT}, size 4. Outcomes with at least one H: {HH, HT, TH}, count 3. P = 3/4.
Q9 3 Marks

A card is drawn from a standard deck of 52. Find the probability that it is a king.

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Number of kings = 4; total cards = 52. P(king) = 4/52 = 1/13.
Q10 3 Marks

For events A and B with P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, find P(A ∪ B).

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P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.4 − 0.2 = 0.7.
Long Answer Questions 6 questions
Q11 6 Marks

A fair die is rolled twice. Find the probability that the sum of the numbers is 7.

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Sample space size = 6 × 6 = 36. Outcomes with sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) — count 6. P = 6/36 = 1/6.
Q12 6 Marks

A bag contains 3 red 4 blue and 5 green balls. Two balls are drawn at random without replacement. Find the probability that both are red.

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Total balls = 12. Number of ways to choose 2 reds from 3 = C(3, 2) = 3. Number of ways to choose any 2 from 12 = C(12, 2) = 66. P(both red) = 3/66 = 1/22.
Q13 6 Marks

Two cards are drawn from a deck without replacement. Find the probability that both are aces.

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P(first ace) = 4/52. Given first is ace P(second ace) = 3/51. P(both aces) = (4/52)(3/51) = 12/2652 = 1/221.
Q14 6 Marks

A card is drawn at random from a deck. Find the probability that it is a king or a heart.

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P(king) = 4/52; P(heart) = 13/52; P(king and heart) = P(king of hearts) = 1/52. By inclusion-exclusion: P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.
Q15 6 Marks

Three coins are tossed once. Find the probability of (i) exactly 2 heads (ii) at least 2 heads (iii) all heads.

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Sample space size 2³ = 8. (i) Exactly 2 heads: {HHT, HTH, THH} count 3. P = 3/8. (ii) At least 2 heads: 3 heads + 2 heads = 1 + 3 = 4. P = 4/8 = 1/2. (iii) All heads: {HHH} count 1. P = 1/8.
Q16 6 Marks

Compare classical and empirical (frequentist) definitions of probability with the help of a table.

Assertion–Reason Questions 5 questions
Q17 1 Mark

Assertion (A): The probability of any event lies between 0 and 1 inclusive.

Reason (R): Probability is the ratio of favourable outcomes to total outcomes both of which are non-negative integers with the favourable count not exceeding the total.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q18 1 Mark

Assertion (A): The probability of an impossible event is 0.

Reason (R): An impossible event has no favourable outcomes so its probability is 0/n = 0.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q19 1 Mark

Assertion (A): P(A) + P(A') = 1.

Reason (R): Either A occurs or its complement does — the two are mutually exclusive and exhaustive.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q20 1 Mark

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

Reason (R): Since A and B cannot occur together their intersection is empty so the inclusion-exclusion correction term vanishes.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q21 1 Mark

Assertion (A): For independent events: P(A ∩ B) = P(A) · P(B).

Reason (R): Independence means the occurrence of one event does not affect the probability of the other so probabilities multiply.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Statement-Based Questions 5 questions
Q22 1 Mark

Statement 1: The probability of a sure event is 1.

Statement 2: The probability of an impossible event is 0.

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Correct answer: Option 1 — Both statements are true.
Q23 1 Mark

Statement 1: The probability of getting a head on a fair coin toss is 1/2.

Statement 2: The probability of getting tails on the same toss is also 1/2.

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Correct answer: Option 1 — Both statements are true.
Q24 1 Mark

Statement 1: The probability of rolling a 6 on a fair die is 1/6.

Statement 2: The probability of rolling a number less than 7 on a fair die is 1.

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Correct answer: Option 1 — Both statements are true.
Q25 1 Mark

Statement 1: P(A') = 1 − P(A).

Statement 2: The complement rule is equivalent to P(A) + P(A') = 1.

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Correct answer: Option 1 — Both statements are true.
Q26 1 Mark

Statement 1: For independent events A and B: P(A and B) = P(A) · P(B).

Statement 2: If A and B are independent then P(A | B) = P(A).

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Correct answer: Option 1 — Both statements are true.
Case Study / Passage Questions 3 questions
Q27 3 Marks
Two fair coins are tossed simultaneously. A statistician wants the probability distribution of the number of heads X and to compute E(X) and the probability of at least one head.
  1. The probability of getting at least one head equals:
    A1/4
    B1/2
    C3/4
    D1
  2. The expectation E(X) (mean number of heads) equals:
    A0
    B1/2
    C1
    D2
  3. Compute the variance of X.
Show answersHide answers
1. Option 3 — 3/4
2. Option 3 — 1
3. Sample space {HH HT TH TT} size 4. P(X = 0) = 1/4 P(X = 1) = 2/4 = 1/2 P(X = 2) = 1/4. E(X) = 0(1/4) + 1(1/2) + 2(1/4) = 1. P(at least one H) = 1 − P(no H) = 1 − 1/4 = 3/4.
Q28 3 Marks
A card is drawn at random from a standard deck of 52 cards. A student wants the probability that it is (i) a spade (ii) an ace and (iii) a spade or an ace.
  1. The probability of drawing a spade is:
    A1/4
    B1/13
    C1/52
    D3/13
  2. The probability of drawing a spade or an ace is:
    A4/13
    B16/52
    C1/4
    D17/52
  3. Compute the probability of drawing a king or a heart.
Show answersHide answers
1. Option 1 — 1/4
2. Option 1 — 4/13
3. P(spade) = 13/52 = 1/4. P(ace) = 4/52 = 1/13. P(spade and ace) = P(ace of spades) = 1/52. P(spade or ace) = P(spade) + P(ace) − P(both) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
Q29 3 Marks
Two fair dice are rolled simultaneously. A board-game designer wants the probability that the sum of the numbers is 7 and that the sum is at least 9.
  1. The probability of getting a sum of 7 is:
    A1/6
    B1/9
    C1/12
    D1/36
  2. The probability of getting a sum at least 9 is:
    A1/6
    B5/18
    C1/3
    D7/18
  3. Compute the probability that both dice show the same number.
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1. Option 1 — 1/6
2. Option 2 — 5/18
3. Sample size 36. Outcomes with sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 outcomes; P = 6/36 = 1/6. Outcomes with sum ≥ 9: sums 9, 10, 11, 12 give counts 4, 3, 2, 1 = 10; P = 10/36 = 5/18.
Table-Based Questions 4 questions
Q30 3 Marks

Study the probability distribution:

XP(X)
00.1
10.3
20.4
30.2
  1. The expectation E(X) equals:
    A1.6
    B1.7
    C1.8
    D2.0
  2. P(X = 2) equals:
    A0.1
    B0.2
    C0.3
    D0.4
  3. Compute Var(X) and SD(X).
Show answersHide answers
1. Option 2 — 1.7
2. Option 4 — 0.4
3. Verify: 0.1 + 0.3 + 0.4 + 0.2 = 1.0 ✓. E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7. Variance = E(X²) − E(X)² where E(X²) = 0 + 0.3 + 1.6 + 1.8 = 3.7 so Var = 3.7 − 2.89 = 0.81.
Q31 3 Marks

Study the standard probabilities of common events:

EventP
Coin shows H1/2
Die shows 61/6
Two coins both H1/4
Card is a heart1/4
Die shows even1/2
Spade or club1/2
  1. P(two coins both heads) equals:
    A1/2
    B1/4
    C1/8
    D1/16
  2. P(drawing a heart from a deck) equals:
    A1/4
    B1/13
    C1/52
    D1/2
  3. Compute P(getting at least one head in two tosses).
Show answersHide answers
1. Option 2 — 1/4
2. Option 1 — 1/4
3. For two independent coin tosses P(HH) = (1/2)·(1/2) = 1/4. There are 13 hearts in 52 cards so P(heart) = 13/52 = 1/4. Independent events allow us to multiply probabilities; mutually exclusive events allow us to add.
Q32 6 Marks

For the probability distribution of a discrete random variable X, find (i) the value of k, (ii) E(X), (iii) Var(X), (iv) P(X ≥ 2).

XP(X)
1k
22k
33k
44k
Q33 6 Marks

Two events A and B have the probabilities listed. Compute (i) P(A ∪ B), (ii) P(A | B), (iii) check whether A and B are independent.

EventProbability
P(A)0.5
P(B)0.4
P(A ∩ B)0.2
Picture-Based Questions 1 question
Q34 3 Marks

Study the PMF of two-coin toss and answer:

Probability figure
  1. The number of heads X with the highest probability is:
    A0
    B1
    C2
    D0.5
  2. The expectation E(X) of the distribution equals:
    A0
    B0.5
    C1
    D2
  3. Verify the probability distribution by checking that the probabilities sum to 1.
Show answersHide answers
1. Option 2 — 1
2. Option 3 — 1
3. Sample space {HH, HT, TH, TT}. P(0 heads) = 1/4, P(1 head) = 1/2, P(2 heads) = 1/4. E(X) = 0·(1/4) + 1·(1/2) + 2·(1/4) = 1. Sum of probabilities = 1/4 + 1/2 + 1/4 = 1, satisfying the axiom of total probability.
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