Probability and Statistics PYQs for GATE DA

GATE DA Probability and Statistics: 3 units and 11 chapters, weightage from 35 previous year questions across 3 papers, a study order by exam weight and 165 practice questions.

About Probability and Statistics Previous Year Questions (PYQs)

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

GATE DA Probability and Statistics Unit-wise Weightage from Past Papers

We counted every GATE DA Probability and Statistics previous year question in our bank (35 questions from 3 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Probability	4	11	31%	3.7
Random Variables	5	20	57%	6.7
Statistical Inference	2	4	11%	1.3

Suggested Probability and Statistics Study Order for GATE DA

Random Variables: 57% of past Probability and Statistics questions, about 6.7 per paper.
Probability: 31% of past Probability and Statistics questions, about 3.7 per paper.
Statistical Inference: 11% of past Probability and Statistics questions, about 1.3 per paper.

Start where the marks are. Units at the top of this list have appeared most often in past GATE DA papers.

Units in GATE DA Probability and Statistics

All Probability and Statistics chapters

One Solved Question from Each Probability and Statistics Chapter

Question 1 · Basic Probability, Events and Counting · 2026 MCQ

Suppose that a computer program provides a non-negative and integer-valued random solution to the equation

n_{1} + n_{2} + n_{3} + n_{4} = 20

.
Which of the following is the probability that all of

n_{1}, n_{2}, n_{3}, n_{4}

in the provided solution are positive?

A.
$(3 19) / (3 23)$
B.
$(4 20) / (4 24)$
C.
$(3 20) / (3 23)$
D.
$(4 19) / (4 24)$

Correct Answer:

Step-by-Step Solution

Insight: This is a classic Stars and Bars problem. We need the ratio of positive integer solutions to non-negative integer solutions for the same equation.

Exam route: Total non-negative solutions to $n_{1} + n_{2} + n_{3} + n_{4} = 20$ is $(4 - 1 20 + 4 - 1) = (3 23)$ . For positive solutions ( $n_{i} \geq 1$ ), give 1 to each variable, leaving a sum of 16. The number of non-negative solutions for the remaining 16 is $(4 - 1 16 + 4 - 1) = (3 19)$ . The probability is the ratio $(3 19) / (3 23)$ .

Learning route: The equation is $n_{1} + n_{2} + n_{3} + n_{4} = 20$ . For non-negative integers ( $n_{i} \geq 0$ ), we use the Stars and Bars formula: $(k - 1 n + k - 1)$ . Here $n = 20, k = 4$ , so total solutions = $(3 23)$ . For positive integers ( $n_{i} \geq 1$ ), we can substitute $m_{i} = n_{i} - 1 \geq 0$ . The equation becomes $m_{1} + m_{2} + m_{3} + m_{4} = 16$ . The number of solutions is $(4 - 1 16 + 4 - 1) = (3 19)$ . Since the program provides a random non-negative solution, each of the $(3 23)$ solutions is equally likely. The probability that all are positive is the ratio of favorable to total solutions: $(3 19) / (3 23)$ .

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

A clinic specializes in testing for a disease D. The result of the test can be either positive or negative.

A study revealed that if a person suffers from the disease D, the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease D, the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease D occurs in 30% of the individuals.

If a person tests positive for D in that clinic, the probability that he/she actually suffers from the disease D is __________ . (Rounded off to two decimal places)

Correct Answer:

0.77

Step-by-Step Solution

Insight: This is a textbook Bayes-reversal question — you know $P (+ ∣ D)$ and $P (D)$ , and you want $P (D ∣ +)$ . The denominator is the law of total probability over ${D, \overline{D}}$ .

Exam route:

Step 1 — Write priors and likelihoods:

$P (D) = 0.3, P (\overline{D}) = 0.7$

$P (+ ∣ D) = 0.8, P (+ ∣ \overline{D}) = 0.1$

Step 2 — Total probability of a positive test:

$P (+) = P (+ ∣ D) P (D) + P (+ ∣ \overline{D}) P (\overline{D}) = 0.8 (0.3) + 0.1 (0.7) = 0.24 + 0.07 = 0.31$

Step 3 — Bayes:

$P (D ∣ +) = \frac{P ( + ∣ D ) P ( D )}{P ( + )} = \frac{0.24}{0.31} = \frac{24}{31} \approx 0.7742$

Rounded to two decimal places: $0.77$ .

Learning route: The event "test positive" can happen in two mutually exclusive ways — the person has the disease and the test correctly flags it, or the person is healthy and the test falsely flags it. These two branches cover every way to see $+$ . Sum their joint probabilities to get $P (+)$ , then the diseased branch divided by the total gives the posterior.

Trap check: If you swap numerator and denominator and compute $P (+ ∣ D)$ instead of $P (D ∣ +)$ , you get $0.80$ — a classic reversal error.

Verification: $P (D ∣ +) + P (\overline{D} ∣ +) = 24/31 + 7/31 = 1$ . ✓

Question 3 · Independence and Expected Waiting Time · 2024 MCQ

A fair six-sided die (with faces numbered 1, 2, 3, 4, 5, 6) is repeatedly thrown
independently.
What is the expected number of times the die is thrown until two consecutive throws
of even numbers are seen?

A.
2
B.
4
C.
6
D.
8

Correct Answer:

Step-by-Step Solution

Insight: This is a "wait for a pattern" question — two consecutive evens. The trials are independent, so track progress toward the pattern using states, and write one equation per state.

Exam route:

Let $p = 1/2$ (probability of even on a fair die).

States: $S_{0}$ (no progress / last was odd), $S_{1}$ (last throw was even), $S_{2}$ (done — two consecutive evens).

Let $E_{i}$ = expected additional throws from state $S_{i}$ .

From $S_{0}$ : throw once. With prob $p$ get even $\to S_{1}$ ; with prob $1 - p$ get odd $\to S_{0}$ .

$E_{0} = 1 + p E_{1} + (1 - p) E_{0}$

From $S_{1}$ : throw once. With prob $p$ get even $\to S_{2}$ (done, 0 more); with prob $1 - p$ get odd $\to S_{0}$ .

$E_{1} = 1 + p \cdot 0 + (1 - p) E_{0}$

Substitute $p = 1/2$ :

$E_{0} = 1 + \frac{1}{2} E_{1} + \frac{1}{2} E_{0} \Rightarrow \frac{1}{2} E_{0} = 1 + \frac{1}{2} E_{1} \Rightarrow E_{0} = 2 + E_{1}$

$E_{1} = 1 + \frac{1}{2} E_{0}$

Substitute $E_{0} = 2 + E_{1}$ into the second:

$E_{1} = 1 + \frac{1}{2} (2 + E_{1}) = 2 + \frac{1}{2} E_{1} \Rightarrow \frac{1}{2} E_{1} = 2 \Rightarrow E_{1} = 4$

$E_{0} = 2 + 4 = 6$ .

Answer: 6 (option C).

Learning route:

A fair die has 3 even faces (2, 4, 6) and 3 odd faces (1, 3, 5), so $p = P (even) = 1/2$ .

We want the expected number of throws until we see two evens in a row.

Define states by how much of the target pattern we have completed:

$S_{0}$ : no useful progress (we are at the start, or the last throw was odd).
$S_{1}$ : the last throw was even (we are one step into the pattern).
$S_{2}$ : we have just seen two consecutive evens (absorbing state, we stop).

Let $E_{i}$ be the expected number of additional throws needed to reach $S_{2}$ starting from $S_{i}$ .

From $S_{0}$ , we spend 1 throw. With probability $1/2$ we get even and move to $S_{1}$ ; with probability $1/2$ we get odd and stay in $S_{0}$ . So $E_{0} = 1 + \frac{1}{2} E_{1} + \frac{1}{2} E_{0}$ .

From $S_{1}$ , we spend 1 throw. With probability $1/2$ we get even and reach $S_{2}$ (done, 0 more throws); with probability $1/2$ we get odd and fall back to $S_{0}$ (all progress lost). So $E_{1} = 1 + \frac{1}{2} (0) + \frac{1}{2} E_{0}$ .

Solving the first equation: $E_{0} - \frac{1}{2} E_{0} = 1 + \frac{1}{2} E_{1} \Rightarrow E_{0} = 2 + E_{1}$ .

Substituting into the second: $E_{1} = 1 + \frac{1}{2} (2 + E_{1}) = 2 + \frac{1}{2} E_{1} \Rightarrow E_{1} = 4$ .

Therefore $E_{0} = 2 + 4 = 6$ .

The expected number of throws is 6.

Wrong paths:

Option A (2): uses $E [T] = 1/ p = 2$ . This is the expected wait for a SINGLE even, not two consecutive.
Option B (4): uses $E [T] = 1/ p^{2} = 4$ . This treats two consecutive evens as two independent events, ignoring the sequential overlap of attempts.
Option D (8): uses $2/ p^{2} = 8$ . This incorrectly doubles the $1/ p^{2}$ formula, perhaps thinking "two evens means multiply by 2".

Question 4 · Continuous Probability and Geometric Probability · 2024 NAT

Let

X

be a random variable uniformly distributed in the interval [1, 3] and

Y

be a
random variable uniformly distributed in the interval [2, 4]. If X and Y are
independent of each other, the probability P(

X \geq Y

) is ______ (rounded off to
three decimal places).

Question 5 · Discrete Random Variables and Bernoulli-Binomial Models · 2026 NAT

Let

A_{5 \times 5}

be a matrix such that each of its elements follows

B er n o u l l i (p = 0.50)

distribution independently.

The probability that the row-sum of the second row and the column-sum of the third column are both equal to 3 is ________ . (Rounded off to two decimal places)

Question 6 · Continuous Random Variables, CDF and PDF · 2026 MSQ

Let

X

be a discrete valued random variable with cumulative distribution function

F (x)

.

Which of the following statements is/are correct?

A.
$F (x)$ is always a positive function.
B.
$F (x)$ is a non-decreasing function.
C.
$F (x)$ has jump discontinuity.
D.
$F (x)$ is a left continuous function.

Question 7 · Expectation, Variance, Covariance and Correlation · 2026 MCQ

Let

X

and

Y

be two independent random variables.

X

follows

B er n o u l l i (p = 0.3)

distribution and

Y

follows

N or ma l (μ = 0, σ^{2} = 100)

distribution.

Which of the following options is the variance of

(2 X - 1) Y

A.
100
B.
90
C.
49
D.
21

Question 8 · Standard Distributions: Normal, Exponential, Poisson and Chi-Square · 2026 NAT

Let $X$ be an exponentially distributed random variable with mean $λ (> 0)$ . If $P (X > 5) = 0.35$ , then the conditional probability $P (X > 10∣ X > 5)$ is ___________ . (Rounded off to two decimal places)

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Probability and Statistics PYQs for GATE DA

About Probability and Statistics Previous Year Questions (PYQs)

GATE DA Probability and Statistics Unit-wise Weightage from Past Papers

Suggested Probability and Statistics Study Order for GATE DA

Units in GATE DA Probability and Statistics

All Probability and Statistics chapters

One Solved Question from Each Probability and Statistics Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution