Probability and Statistics PYQs for GATE CS

GATE CS Probability and Statistics: 5 chapters, 25 previous year questions (24% of Engineering Mathematics), 408 practice questions and one solved question from each chapter.

About Probability and Statistics Previous Year Questions (PYQs)

25 previous year questions from Probability and Statistics in GATE CS, grouped by chapter with the exam year, answer key and step-by-step solution for each.

Probability and Statistics Weightage in GATE CS

Probability and Statistics accounts for 25 of 105 Engineering Mathematics previous year questions in our bank (24%), about 2.5 per paper across 10 papers.

Probability and Statistics Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Combinatorial Probability and Independent Events	Binomial Models and Repeated Bernoulli Trials, Classical Counting Probability, Event Algebra and Independence	9	36%	141
Conditional Probability and Bayes Theorem	Bayes Theorem and Posterior Inference, Conditional Probability in Sequential Sampling, Conditional Events in Repeated Coin Tosses	5	20%	82
Random Variables, Expectation, Variance and Covariance	Expectation of Continuous Geometric Quantities, Expectation Bounds for Products of Random Variables, Variance, Covariance and Standard Deviation, Discrete Expectation and Decision Problems	6	24%	104
Continuous and Standard Probability Distributions	Exponential Lifetime Distribution, Normal Distribution Identification, Density Normalization and Interval Probabilities	3	12%	48
Urn Models and Reinforcement Processes	Polya Urn Reinforcement Processes	2	8%	33

One Solved Question from Each Probability and Statistics Chapter

Question 1 · Combinatorial Probability and Independent Events · 2026_Set2 MCQ

A day can only be cloudy or sunny. The probability of a day being cloudy is $0.5$ , independent of the condition on other days. What is the probability that in any given four days, there will be three cloudy days and one sunny day?

A.
$\frac{1}{4}$
B.
$\frac{3}{4}$
C.
$\frac{2}{3}$
D.
$\frac{3}{8}$

Correct Answer:

Step-by-Step Solution

Insight: Independent days with two outcomes = binomial with $n = 4, k = 3, p = 0.5$ .

Exam route:

$n = 4$ days, $k = 3$ cloudy, $p = 0.5$ , $q = 0.5$ .

$P (X = 3) = (3 4) (0.5)^{3} (0.5)^{1} = 4 \times (0.5)^{4} = 4 \times \frac{1}{16} = \frac{1}{4}$ .

Learning route:

This is a binomial question: fixed $n$ , binary outcome, independent trials, constant $p$ .

Step 1 — Identify parameters. $n = 4$ , $k = 3$ , $p = 0.5$ .

Step 2 — Apply the binomial PMF.

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$

Step 3 — Substitute. $(3 4) = 4$ . $(0.5)^{3} \times (0.5)^{1} = (0.5)^{4} = \frac{1}{16}$ .

Step 4 — Multiply. $4 \times \frac{1}{16} = \frac{1}{4}$ .

Wrong path: If you forget the binomial coefficient $(3 4)$ , you get $(0.5)^{4} = \frac{1}{16}$ , which is not among the options — a signal you missed the arrangements.

Verification: The four arrangements are CCCS, CCSC, CSCC, SCCC. Each has probability $(0.5)^{4} = \frac{1}{16}$ . $4 \times \frac{1}{16} = \frac{1}{4}$ .

Question 2 · Conditional Probability and Bayes Theorem · 2026_Set2 NAT

Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all previous coin tosses. Let

E_{1}

be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let

E_{2}

be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails.

The conditional probability

P (E_{1} ∣ E_{2})

is equal to ____________. (rounded off to one decimal place)

Correct Answer:

0.5

Step-by-Step Solution

Insight: This is a conditional probability problem with overlapping events in repeated trials. The shared variable (Toss 2) prevents independence.

Exam route: Use the reduced sample space method. Count outcomes for E2 (denominator), then count outcomes where both E1 and E2 hold by splitting on the shared Toss 2 (numerator).

Learning route:

Step 1: Understand the events.

E1: Tosses {2, 4, 6} have >= 2 Heads.
E2: Tosses {1, 2, 3, 5} have exactly 2 Heads and 2 Tails.

Step 2: Calculate the denominator N(E2).

Tosses {1, 2, 3, 5} must have 2H, 2T. Number of ways = C(4, 2) = 6.
Tosses {4, 6} are unconstrained by E2. Number of ways = 2^2 = 4.
Total N(E2) = 6 * 4 = 24.

Step 3: Calculate the numerator N(E1 ∩ E2) by conditioning on the shared Toss 2.

Case A: Toss 2 is Head (H).

For E2 to hold, the remaining tosses in its set {1, 3, 5} must have exactly 1H, 2T. Ways = C(3, 1) = 3.
For E1 to hold, the remaining tosses in its set {4, 6} must have >= 1H (since Toss 2 is already H). Ways = Total - 0H = 4 - 1 = 3.
Ways for Case A = 3 * 3 = 9.

Case B: Toss 2 is Tail (T).

For E2 to hold, {1, 3, 5} must have exactly 2H, 1T. Ways = C(3, 2) = 3.
For E1 to hold, {4, 6} must have >= 2H (since Toss 2 is T). Ways = 1 (both must be H).
Ways for Case B = 3 * 1 = 3.
Total N(E1 ∩ E2) = 9 + 3 = 12.

Step 4: Compute the conditional probability.

P(E1 | E2) = N(E1 ∩ E2) / N(E2) = 12 / 24 = 0.5.

Verification: Plug back into the definition. P(E1 ∩ E2) = 12/64, P(E2) = 24/64. Ratio is 12/24 = 0.5. Matches perfectly.

Question 3 · Random Variables, Expectation, Variance and Covariance · 2025_Set2 NAT

The unit interval

(0, 1)

is divided at a point chosen uniformly distributed over

(0, 1)

R

into two disjoint subintervals.

The expected length of the subinterval that contains 0.4 is ___________. (rounded off to two decimal places)

Correct Answer:

0.74

Step-by-Step Solution

Insight: The length of the subinterval containing a fixed point $c$ depends on whether the random cut $X$ is to the left or right of $c$ . This requires a piecewise function and splitting the integral.

Exam route: Use the derived formula for the expected length containing $c$ in a unit interval: $E [L] = c - c^{2} + 0.5$ . For $c = 0.4$ , $E [L] = 0.4 - 0.16 + 0.5 = 0.74$ .

Learning route:

Define the random variable: Let $X \sim U (0, 1)$ be the cut point. The PDF is $f_{X} (x) = 1$ for $0 < x < 1$ .
Define the target variable $L$ (length containing $c = 0.4$ ):

If $X < 0.4$ , the interval containing $0.4$ is $(X, 1)$ , so $L = 1 - X$ .
If $X > 0.4$ , the interval containing $0.4$ is $(0, X)$ , so $L = X$ .

Set up the expectation integral:

$E [L] = \int_{0}^{0.4} (1 - x) (1) d x + \int_{0.4}^{1} (x) (1) d x$ .

Evaluate the integrals:

First part: $[x - x^{2} /2]_{0}^{0.4} = 0.4 - 0.08 = 0.32$ .
Second part: $[x^{2} /2]_{0.4}^{1} = 0.5 - 0.08 = 0.42$ .

Sum the results: $0.32 + 0.42 = 0.74$ .

Question 4 · Continuous and Standard Probability Distributions · 2026_Set2 MCQ

The probability density function

f (x)

of a random variable

X

which takes real values is

f (x) = \frac{1}{3 2 π} exp (- \frac{x ^{2}}{18}), x \in (- \infty, + \infty)

Which one of the following statements is correct about the random variable

X

A.
$X$ is an exponential random variable
B.
$X$ is a normal random variable
C.
$X$ is a Poisson random variable
D.
$X$ is a uniform random variable

Correct Answer:

Step-by-Step Solution

Insight: The PDF has the form $C exp (- k x^{2})$ , which is the algebraic signature of a Normal distribution.

Exam route: Match $exp (- x^{2} /18)$ to $exp (- (x - μ)^{2} /2 σ^{2})$ . $μ = 0$ , $2 σ^{2} = 18 ⟹ σ = 3$ . Coefficient is $1/ (3 2 π)$ , matching $1/ (σ 2 π)$ . Thus, Normal.

Learning route:

Identify the form: The PDF is defined over $(- \infty, \infty)$ and contains the term $exp (- x^{2} /18)$ .
Recall standard forms: The Normal PDF is $\frac{1}{σ 2 π} exp (- \frac{( x - μ ) ^{2}}{2 σ ^{2}})$ .
Match the exponent: $- \frac{x ^{2}}{18} = - \frac{( x - 0 ) ^{2}}{2 σ ^{2}}$ . This gives $μ = 0$ and $2 σ^{2} = 18 ⟹ σ^{2} = 9 ⟹ σ = 3$ .
Verify the coefficient: The standard coefficient is $\frac{1}{σ 2 π} = \frac{1}{3 2 π}$ , which perfectly matches the given PDF.
Conclusion: $X$ is a normal random variable with mean 0 and variance 9.

Question 5 · Urn Models and Reinforcement Processes · 2026_Set1 MCQ

An urn contains one red ball and one blue ball. At each step, a ball is picked uniformly at random from the urn, and this ball together with another ball of the same color is put back in the urn. The probability that there are equal number of red and blue balls after two steps is

A.
$\frac{1}{4}$
B.
$\frac{1}{3}$
C.
$\frac{1}{2}$
D.
$\frac{2}{3}$

Correct Answer:

Step-by-Step Solution

Insight: Equal red and blue after 2 steps means exactly 1 red and 1 blue were drawn — use exchangeability or direct enumeration.

Exam route: Start: 1R, 1B. After 2 steps, total = 4. Equal counts need 2R and 2B, so exactly $k = 1$ red drawn. By exchangeability: $P = (1 2) \cdot 1^{(1)} \cdot 1^{(1)} / 2^{(2)} = 2 \cdot 1 \cdot 1/6 = 1/3$ . Select B.

Learning route:

This is a Polya Urn state question, identifiable by the reinforcement mechanism (draw, replace, add one of the same colour) and the question asking about the urn's composition after a fixed number of steps.

Setup: Initial state: 1 red, 1 blue, total 2. After 2 draws, total = $2 + 2 = 4$ balls. For equal red and blue counts, we need 2 red and 2 blue, which means exactly 1 red and 1 blue were drawn in the 2 draws.

Method 1 — Direct enumeration:

List all possible sequences of 2 draws:

RR: $P = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$ → final: 3R, 1B (not equal)
RB: $P = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ → final: 2R, 2B (equal ✓)
BR: $P = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ → final: 2R, 2B (equal ✓)
BB: $P = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$ → final: 1R, 3B (not equal)

$P (equal) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

Method 2 — Exchangeability formula:

$P (k reds in n draws) = \frac{( k n ) \cdot r ^{(k)} \cdot b ^{(n - k)}}{( r + b ) ^{(n)}}$

With $r = 1, b = 1, n = 2, k = 1$ :

$P = \frac{( 1 2 ) \cdot 1 ^{(1)} \cdot 1 ^{(1)}}{2 ^{(2)}} = \frac{2 \cdot 1 \cdot 1}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3}$

Wrong path walkthrough: A student who treats draws as independent coin flips computes $P (exactly 1 head in 2 flips) = (1 2) (1/2)^{2} = 1/2$ , matching option C. The mistake is at the independence assumption: in a Polya Urn, the second draw's probability depends on the first draw's outcome. The reinforcement mechanism makes draws dependent, reducing the probability of mixed outcomes.

Generalisation: In a Polya Urn, reinforcement amplifies early outcomes, making "all same colour" more likely and "mixed" less likely compared to independent draws.

Verification: The four sequence probabilities sum to $\frac{1}{3} + \frac{1}{6} + \frac{1}{6} + \frac{1}{3} = 1$ . ✓

Answer: B) $\frac{1}{3}$

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Probability and Statistics PYQs for GATE CS

About Probability and Statistics Previous Year Questions (PYQs)

Probability and Statistics Weightage in GATE CS

Probability and Statistics Chapter Matrix

More from Engineering Mathematics

One Solved Question from Each Probability and Statistics Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution