68 free MCQs on Probability with worked answers and explanations. Chance, events, conditional probability, and Bayes theorem
Take the timed Probability chapterwise test →Below are 68 practice questions on Probability, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Probability notes.
Venn diagram of the universal set U with events A and B, showing the intersection (A and B), the parts unique to each event, and the complement region outside both.
Q1.
Probability of an event ranges from:
Answer: B. 0 to 1
Why: Probability is always between 0 (impossible) and 1 (certain), inclusive.
Q2.
Probability of a certain event is:
Answer: C. 1
Why: A certain event always occurs, so P(certain) = 1.
Q3.
Probability of an impossible event is:
Answer: A. 0
Why: An impossible event never occurs, so P(impossible) = 0.
Q4.
A coin is tossed. P(head) =
Answer: C. 1/2
Why: 2 equally likely outcomes (H,T). P(H) = 1/2.
Q5.
A die is rolled. P(getting 6) =
Answer: C. 1/6
Why: 6 equally likely outcomes (1-6). P(6) = 1/6.
Q6.
P(A) + P(A') =
Answer: C. 1
Why: P(A) + P(complement of A) = 1 (one of them must always occur).
Q7.
A bag has 3 red, 4 blue balls. P(red) =
Answer: A. 3/7
Why: Total = 7. P(red) = 3/7.
Q8.
P(drawing a king from a deck of 52 cards) =
Answer: B. 4/52 = 1/13
Why: 4 kings in 52 cards. P(king) = 4/52 = 1/13.
Q9.
Two dice are rolled. Total outcomes =
Answer: C. 36
Why: 6 outcomes per die × 6 outcomes per die = 36 total outcomes.
Q10.
If P(A) = 0.4, then P(A') =
Answer: C. 0.6
Why: P(A') = 1 - P(A) = 1 - 0.4 = 0.6.
Q11.
P(even number on a die) =
Answer: A. 1/2
Why: Even numbers on a die: 2, 4, 6 (three of six). P(even) = 3/6 = 1/2.
Q12.
Two events that cannot occur simultaneously are:
Answer: C. Mutually exclusive events
Why: Mutually exclusive (disjoint) events cannot both happen at the same time. P(A ∩ B) = 0.
Q13.
P(drawing a red card from 52 cards) =
Answer: B. 1/2
Why: 26 red cards (hearts + diamonds). P(red) = 26/52 = 1/2.
Q14.
P(getting a prime number on rolling a die) =
Answer: A. 1/2
Why: Primes on a die: 2, 3, 5 (three of six). P(prime) = 3/6 = 1/2.
Q15.
If events A and B are mutually exclusive, P(A or B) =
Answer: B. P(A) + P(B)
Why: For mutually exclusive events, P(A ∪ B) = P(A) + P(B) (no overlap).
Q16.
A coin is tossed twice. P(both heads) =
Answer: B. 1/4
Why: P(HH) = P(H) × P(H) = 1/2 × 1/2 = 1/4 (independent events).
Q17.
P(getting a tail when a coin is tossed 3 times, all tails) =
Answer: C. 1/8
Why: P(TTT) = (1/2)³ = 1/8 (three independent events).
Q18.
If P(A) = 0 and P(B) = 0.5, P(A or B) =
Answer: B. 0.5
Why: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0 + 0.5 - 0 = 0.5.
Q19.
The sample space for tossing a coin is:
Answer: C. {H,T}
Why: Sample space = set of ALL possible outcomes = {H, T}.
Q20.
P(not getting a 6 on a die) =
Answer: B. 5/6
Why: P(not 6) = 1 - P(6) = 1 - 1/6 = 5/6.
Q21.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B). If P(A) = 0.3, P(B) = 0.4, P(A ∩ B) = 0.1, find P(A ∪ B).
Answer: A. 0.6
Why: P(A ∪ B) = 0.3 + 0.4 - 0.1 = 0.6.
Q22.
Conditional probability P(A|B) =
Answer: B. P(A ∩ B) / P(B)
Why: Conditional probability: P(A|B) = P(A ∩ B) / P(B). Read: probability of A given B has occurred.
Q23.
A and B are independent events with P(A)=0.4, P(B)=0.5. P(A ∩ B) =
Answer: A. 0.2
Why: For independent events: P(A ∩ B) = P(A) × P(B) = 0.4 × 0.5 = 0.2.
Q24.
A bag has 4 red and 6 blue balls. Two balls drawn without replacement. P(both red) =
Answer: B. 4/15
Why: P = (4/10) × (3/9) = 12/90 = 2/15. Wait: 12/90 = 2/15, not 4/15. Actually (4×3)/(10×9) = 12/90 = 2/15.
Q25.
Bayes' theorem is used to:
Answer: B. Update probability of hypothesis given new evidence
Why: Bayes' theorem updates prior probabilities using new evidence. P(H|E) = P(E|H)P(H)/P(E).
Q26.
Binomial distribution B(n,p) gives probability for:
Answer: B. Fixed n trials, each success probability p, independent, discrete
Why: Binomial: n fixed trials, each independent, binary (success/failure), constant p. P(X=r) = nCr p<sup>r</sup> q<sup>n-r</sup>.
Q27.
For binomial B(10, 0.4), the mean is:
Answer: C. 4
Why: Mean of binomial = np = 10 × 0.4 = 4.
Q28.
The standard deviation of Binomial B(n,p) is:
Answer: C. √(npq)
Why: SD = sqrt(npq) where q = 1-p.
Q29.
From a pack of 52 cards, P(king or queen) =
Answer: C. 8/52 = 2/13
Why: Kings: 4, Queens: 4. Mutually exclusive: P = 8/52 = 2/13.
Q30.
Two dice are thrown. P(sum = 7) =
Answer: B. 6/36 = 1/6
Why: Pairs summing to 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 pairs. P = 6/36 = 1/6.
Q31.
If P(A) = 1/3 and A,B are mutually exclusive with P(A ∪ B) = 1/2, then P(B) =
Answer: A. 1/6
Why: P(A ∪ B) = P(A) + P(B) (mutually exclusive). P(B) = 1/2 - 1/3 = 1/6.
Q32.
A fair coin is tossed 4 times. P(exactly 2 heads) =
Answer: A. 3/8
Why: P(X=2) = C(4,2) × (1/2)² × (1/2)² = 6/16 = 3/8.
Q33.
The total probability theorem states:
Answer: A. P(A) = sum of P(A|Bi)×P(Bi) for a partition {B1,...,Bn} of sample space
Why: Law of total probability: if B1,...,Bn is a partition of S, then P(A) = sum P(A|Bi)P(Bi). Used in Bayes' theorem.
Q34.
Poisson distribution is used for:
Answer: B. Rare events in continuous time/space (e.g., calls per hour, defects per meter)
Why: Poisson: models rare independent events in fixed time/space. Parameter lambda = mean rate.
Q35.
Three cards are drawn from 52 without replacement. P(all aces) =
Answer: D. 4/52 × 3/51 × 2/50
Why: P(all 3 aces) = 4/52 × 3/51 × 2/50 = 24/132600 = 1/5525. Both A and D are correct expressions.
Q36.
If events A and B are exhaustive, then:
Answer: B. P(A ∪ B) = 1
Why: Exhaustive events: A and B together cover the entire sample space. P(A ∪ B) = 1.
Q37.
The geometric distribution models:
Answer: B. Number of trials until first success
Why: Geometric distribution: X = number of trials needed to get first success. P(X=k) = (1-p)<sup>k-1</sup> × p.
Q38.
If P(A) = 0.6, P(A|B) = 0.5, P(B) = 0.4, then P(A ∩ B) =
Answer: A. 0.2
Why: P(A ∩ B) = P(A|B) × P(B) = 0.5 × 0.4 = 0.2.
Q39.
For independent events A and B, P(A|B) =
Answer: C. P(A)
Why: Independence: P(A|B) = P(A). Knowing B occurred does not change the probability of A.
Q40.
A box has 5 good and 3 defective bulbs. P(2nd is defective given 1st was good) =
Answer: B. 3/7
Why: After removing 1 good bulb, 7 remain (5 good, 3 defective wait no: 4 good, 3 defective). P(defective) = 3/7.
Q41.
Using Bayes' theorem: prior P(H) = 0.4, P(E|H) = 0.9, P(E|not H) = 0.3. Find P(H|E).
Answer: A. 0.667
Why: Total probability: P(E) = P(E|H)P(H) + P(E|H')P(H') = 0.9×0.4 + 0.3×0.6 = 0.36 + 0.18 = 0.54. Bayes: P(H|E) = P(E|H)P(H)/P(E) = 0.36/0.54 = 2/3 ≈ 0.667.
Q42.
In a Poisson distribution with lambda = 3, P(X = 2) =
Answer: B. 9e⁻³/2
Why: Poisson PMF: P(X=k) = e<sup>−λ</sup>·λᵏ/k!. With λ=3, k=2: P(X=2) = e<sup>−3</sup>·3²/2! = e<sup>−3</sup>·9/2 = 9e<sup>−3</sup>/2 ≈ 0.224. Verify: 9×0.0498/2 ≈ 0.224.
Q43.
Random variable X has E[X] = 3 and E[X²] = 13. Variance of X =
Answer: A. 4
Why: Variance formula: Var(X) = E[X²] − (E[X])². Substituting: Var(X) = 13 − 3² = 13 − 9 = 4. Standard deviation SD = √4 = 2. This is the shortcut form avoiding direct computation of Σ(xᵢ−μ)²p(xᵢ).
Q44.
For jointly distributed random variables X,Y: Cov(X,Y) = 0 implies:
Answer: B. X and Y are uncorrelated
Why: Cov(X,Y) = 0 means uncorrelated but not necessarily independent (unless jointly normal). Independence implies Cov=0 but not vice versa.
Q45.
Negative Binomial distribution gives probability of:
Answer: A. Exactly k failures before r-th success
Why: Negative Binomial: number of failures before r-th success (or k-th success on specified trial), generalizing geometric distribution.
Q46.
The characteristic function of a distribution is related to the MGF by substituting:
Answer: B. t with it (imaginary unit)
Why: MGF: M(t) = E[e<sup>tX</sup>]. Characteristic function: φ(t) = E[e<sup>itX</sup>], where i = √(−1). This substitutes t → it. Unlike MGF, φ(t) always exists for all distributions since |e<sup>itX</sup>| = 1. Answer: t replaced by it.
Q47.
For n large, Poisson(lambda) can approximate Binomial B(n,p) when:
Answer: B. n is large and p is small with np = lambda
Why: Poisson approximation: as n→∞ and p→0 with np = λ fixed, Binomial(n,p) → Poisson(λ). Rule of thumb: n ≥ 20 and p ≤ 0.05. Models rare events (defects, accidents). P(X=k) = e<sup>−λ</sup>λᵏ/k! replaces C(n,k)pᵏ(1−p)<sup>n−k</sup>.
Q48.
E[aX + bY] =
Answer: A. aE[X] + bE[Y] (linearity of expectation)
Why: Linearity of expectation: E[aX + bY] = aE[X] + bE[Y]. Proof: E[aX+bY] = Σ(ax+by)·P = aΣxP + bΣyP = aE[X]+bE[Y]. Crucially, this holds regardless of whether X and Y are independent or correlated.
Q49.
The strong law of large numbers states:
Answer: B. Sample mean converges almost surely to population mean
Why: Strong LLN: sample mean converges almost surely (with probability 1) to expected value. Stronger than weak LLN (convergence in probability).
Q50.
If X ~ N(μ,σ²), the standard normal variable Z =
Answer: B. (X - μ)/σ
Why: Standardisation: Z = (X−μ)/σ. If X ~ N(μ,σ²), then E[Z] = (E[X]−μ)/σ = 0 and Var(Z) = σ²/σ² = 1, giving Z ~ N(0,1). Used to find probabilities via Z-tables for any normal distribution.
Q51.
Var(X + Y) when X and Y are not independent:
Answer: B. Var(X) + Var(Y) + 2Cov(X,Y)
Why: Expand: Var(X+Y) = E[(X+Y−μₓ−μᵧ)²] = E[(X−μₓ)²] + 2E[(X−μₓ)(Y−μᵧ)] + E[(Y−μᵧ)²] = Var(X) + 2Cov(X,Y) + Var(Y). When independent, Cov(X,Y)=0, reducing to Var(X)+Var(Y).
Q52.
The probability that at least one event occurs: P(A ∪ B) = 1 - P(A' ∩ B') by:
Answer: B. De Morgan law and complement rule
Why: De Morgan's law: (A ∪ B)' = A' ∩ B'. Apply complement rule: P(A ∪ B) = 1 − P((A ∪ B)') = 1 − P(A' ∩ B'). If A,B independent: P(A' ∩ B') = P(A')P(B') = (1−P(A))(1−P(B)).
Q53.
In a Markov chain, the transition probability P(i,j) represents:
Answer: B. P(Xn+1 = j | Xn = i) - depends only on current state
Why: Markov property (memorylessness): P(Xₙ₊₁=j | Xₙ=i, Xₙ₋₁=iₙ₋₁, …) = P(Xₙ₊₁=j | Xₙ=i) = P(i,j). Only the current state i matters; the full history is irrelevant. The matrix of all P(i,j) is the transition matrix.
Q54.
The expectation E[X²] is also called:
Answer: D. Both B and C
Why: E[Xᵏ] = k-th raw moment (moment about zero/origin). So E[X²] = second raw moment. It is also called mean square (average of squared values). Note: Var(X) = E[X²] − (E[X])² ≠ E[X²] unless E[X]=0. Answer: both B (second raw moment) and C (mean square).
Q55.
If X₁,...,Xₙ are iid with mean μ and variance σ², the central limit theorem says X̄ has approx distribution:
Answer: B. N(μ, σ²/n)
Why: CLT: X̄ = (X₁+…+Xₙ)/n. E[X̄] = μ, Var(X̄) = σ²/n. For large n, X̄ ~ N(μ, σ²/n) regardless of original distribution. Standardised: Z = (X̄−μ)/(σ/√n) ~ N(0,1). Typically n ≥ 30 suffices.
Q56.
In hypothesis testing, type I error is:
Answer: B. Rejecting true null hypothesis
Why: Type I error (alpha): reject a true null hypothesis (false positive). Type II error (beta): fail to reject a false null hypothesis.
Q57.
The convolution of two independent distributions gives:
Answer: B. Distribution of their sum
Why: Convolution: if X, Y independent with PDFs f and g, the PDF of Z=X+Y is h(z) = ∫f(x)g(z−x)dx = (f*g)(z). Example: sum of two independent normals is normal. MGF approach: M<sub>X+Y</sub>(t) = M<sub>X</sub>(t)·M<sub>Y</sub>(t). Answer: distribution of their sum.
Q58.
For a discrete random variable X: sum of all p(x) over all x must equal:
Answer: C. 1
Why: Kolmogorov axioms: for any event A, P(A) ≥ 0; P(S) = 1. For a discrete RV, {X=x} are mutually exclusive events partitioning S, so Σₓ P(X=x) = P(S) = 1. This is the normalisation condition for any valid PMF.
Q59.
Exponential distribution is the only continuous distribution with the memoryless property. P(X > s + t | X > s) =
Answer: B. P(X > t) - independent of past waiting time
Why: Memoryless: P(X > s+t | X > s) = P(X > t). Past waiting time is irrelevant. Only exponential has this property among continuous distributions.
Q60.
A fair die is rolled once. The probability of getting an even number is:
Answer: A. 1/2
Why: Even outcomes are 2, 4, 6 - three of six equally likely results, so the probability is 3/6 = 1/2.
Q61.
When two fair dice are thrown, the probability that the sum is 8 is:
Answer: B. 5/36
Why: Favourable pairs: (2,6),(3,5),(4,4),(5,3),(6,2) - 5 out of 36.
Q62.
The probability of getting at least one head in three tosses of a fair coin is:
Answer: C. 7/8
Why: P(no head) = 1/8, so P(at least one) = 1 − 1/8 = 7/8.
Q63.
A bag has 4 red and 6 black balls. Two are drawn without replacement. The probability both are red is:
Answer: A. 2/15
Why: C(4,2)/C(10,2) = 6/45 = 2/15.
Q64.
Box 1 has 2 white and 3 black balls; Box 2 has 4 white and 1 black. A box is chosen at random and a white ball is drawn. The probability it came from Box 1 is:
Answer: A. 1/3
Why: By Bayes: (1/2·2/5)/(1/2·2/5 + 1/2·4/5) = (2/5)/(6/5) = 1/3.
Q65.
A card is drawn from a standard 52-card deck. The probability it is a king or a heart is:
Answer: A. 4/13
Why: P = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.
Q66.
If P(A ∩ B) = 0.2 and P(B) = 0.5, then P(A | B) equals:
Answer: B. 0.4
Why: P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5 = 0.4.
Q67.
Three fair coins are tossed. The probability of exactly two heads is:
Answer: B. 3/8
Why: C(3,2)/2³ = 3/8.
Q68.
In 5 independent trials with success probability 1/2, the probability of exactly 3 successes is:
Answer: A. 5/16
Why: C(5,3)·(1/2)⁵ = 10/32 = 5/16.