Probability — Class 12 MCQs with Answers

Probability — important questions & MCQs with answers (Class 12 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   P(A|B) =

   AP(A)/P(B)BP(A∩B)/P(B)CP(A∪B)/P(B)DP(A)·P(B)
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   Answer: P(A∩B)/P(B)

   Explanation: Conditional probability restricts the sample space to B, so P(A|B) = P(A∩B)/P(B) — dividing by P(B), not by P(A∪B), which double-counts the union.

2. Q2 of 6Easy

   Sample space S:

   ASet of all possible outcomesBSingle outcomeCProbabilityDJust events
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   Answer: Set of all possible outcomes

   Explanation: The sample space S is the complete set of every possible outcome of an experiment — a single outcome is just one sample point, not the whole space.

3. Q3 of 6Medium

   P(getting an even number on a fair die):

   A1/6B1/3C2/3D1/2
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   Answer: 1/2

   Explanation: A fair die has three even faces — 2, 4, 6 — out of six total, so P(even) = 3/6 = 1/2, not 1/3, which would undercount the favourable faces.

4. Q4 of 6Medium

   Assertion (A): If A and B are independent events with P(A) = 0.4 and P(B) = 0.5, then P(A|B) = 0.4. Reason (R): For these events A and B, P(A|B) = P(A)·P(B) = 0.2.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is true but R is false

   Explanation: A is true: for independent events P(A|B) = P(A ∩ B)/P(B) = (0.4 × 0.5)/0.5 = 0.4. R is false: here P(A|B) is 0.4; the product P(A)·P(B) gives P(A ∩ B), not P(A|B).

5. Q5 of 6Hard

   For 3 mutually independent events with probabilities 0.5, 0.4, 0.3, P(all three occur):

   A0.06B0.12C0.05D0.5
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   Answer: 0.06

   Explanation: For mutually independent events, the joint probability is a straight product: P(A∩B∩C) = 0.5 × 0.4 × 0.3 = 0.06 — no overlap subtraction is needed since independence already accounts for it.

6. Q6 of 6Hard

   Assertion (A): Three identical boxes contain two gold coins, two silver coins, and one gold and one silver coin respectively. A box is chosen at random and a coin taken from it at random is gold. The probability that the other coin in that box is also gold is 1/2. Reason (R): If E₁, E₂, E₃ form a partition of the sample space and P(A) ≠ 0, then P(E₁|A) = P(E₁)P(A|E₁) / [P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + P(E₃)P(A|E₃)].

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is false but R is true

   Explanation: A is false: the other coin is gold only if the gold-gold box was chosen. By Bayes' theorem, P(gold-gold box | gold) = (1/3)(1) / [(1/3)(1) + (1/3)(0) + (1/3)(1/2)] = 2/3. R is true: it is Bayes' theorem for three hypotheses.

Probability notes: key concepts (Class 12 Mathematics)

Building on conditional probability P(A|B), the chapter develops the multiplication theorem, independence, the total-probability theorem and Bayes' theorem for reversing conditionals. It then defines a random variable with its distribution, mean and variance, and the binomial distribution for Bernoulli trials.

Conditional probability

P(A|B) = P(A∩B)/P(B) with P(B) > 0; the probability of A given that B has already occurred.

Multiplication theorem

P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B); the chance that both events happen.

Independent events

A and B are independent when P(A∩B) = P(A)·P(B); then P(A|B) = P(A) — B tells nothing about A.

Mutually exclusive events

Events that cannot occur together, so P(A∩B) = 0; distinct from independence, which needs a product.

Addition theorem

P(A∪B) = P(A) + P(B) − P(A∩B); for mutually exclusive events the last term vanishes.

Total probability theorem

For a partition B₁,…,Bₙ, P(A) = Σ P(A|Bᵢ)·P(Bᵢ); decomposes an event across exhaustive cases.

Bayes' theorem

P(Bᵢ|A) = P(A|Bᵢ)P(Bᵢ)/Σ P(A|Bⱼ)P(Bⱼ); reverses a conditional, as in 'infected given a positive test'.

Random variable

A function assigning a real number to each outcome; discrete ones take countable values, continuous ones a whole interval.

Probability distribution

Lists each value of X with its probability; the probabilities are non-negative and sum to 1.

Mean (expected value)

E(X) = Σ xᵢ pᵢ, the long-run average value of the random variable.

Variance & standard deviation

Var(X) = Σ xᵢ²pᵢ − [E(X)]² measures spread; the standard deviation is its square root.

Binomial distribution

n independent trials, each with success probability p: P(X = r) = ⁿCᵣ pʳ q^(n−r), q = 1 − p.

Binomial mean & variance

For binomial (n, p), mean = np, variance = npq and standard deviation = √(npq).