Combinatorial Probability and Independent Events PYQs for GATE CS

Solve 9+ Combinatorial Probability and Independent Events previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics

Binomial Models & Repeated Bernoulli Trials

Foundation of discrete distributions. Focus on fixed trials and binary outcomes.

Classical Counting Probability

Permutations, combinations, and sample spaces. Focus on equally likely outcomes.

Event Algebra and Independence

Union, intersection, conditional probability, and Bayes theorem.

Goal: Master the transition from simple counting to complex event dependencies.

The Bernoulli Trial: Binary Randomness

What is a Bernoulli Trial?

A Bernoulli trial is a random experiment with exactly two possible outcomes:

Success (

S

)

Probability

p

Failure (

F

)

Probability

q = 1 - p

Key Characteristics

Binary Outcome: Only two results are possible.
Fixed Probability: $p$ remains constant for every trial.
Independence: The outcome of one trial does not affect the next.

Intuition: Think of a light switch. It is either ON or OFF. That is a Bernoulli state.

Combinatorial Probability and Independent Events: Solved Questions with Step-by-Step Explanations (9 Problems)

Question 1 · Engineering Mathematics · 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 · Engineering Mathematics · 2026_Set2 MCQ

An unbiased six-faced dice whose faces are marked with numbers $1, 2, 3, 4, 5,$ and $6$ is rolled twice in succession and the number on the top face is recorded each time. The probability that the sum of the two recorded numbers is a prime number is ________

A.
$\frac{3}{36}$
B.
$\frac{13}{36}$
C.
$\frac{15}{36}$
D.
$\frac{19}{36}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a Classical Counting Probability problem. We need to count the number of outcomes where the sum of two dice is prime.

Step 1: Determine the Sample Space.

Two dice are rolled. Total outcomes $N = 6 \times 6 = 36$ .

Each outcome is an ordered pair $(d_{1}, d_{2})$ .

Step 2: Identify Possible Sums.

Minimum sum = $1 + 1 = 2$ .

Maximum sum = $6 + 6 = 12$ .

Prime numbers in range $[2, 12]$ are: $2, 3, 5, 7, 11$ .

Step 3: Count Favorable Outcomes for Each Prime Sum.

Sum = 2: $(1, 1)$ $\to$ 1 way.
Sum = 3: $(1, 2), (2, 1)$ $\to$ 2 ways.
Sum = 5: $(1, 4), (2, 3), (3, 2), (4, 1)$ $\to$ 4 ways.
Sum = 7: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ $\to$ 6 ways.
Sum = 11: $(5, 6), (6, 5)$ $\to$ 2 ways.

Step 4: Sum the Favorable Outcomes.

Total Favorable = $1 + 2 + 4 + 6 + 2 = 15$ .

Step 5: Calculate Probability.

$P (Prime Sum) = \frac{15}{36}$ .

Answer: $\frac{15}{36}$

Question 3 · Engineering Mathematics · 2025_Set2 NAT

A quadratic polynomial

(x - α) (x - β)

over complex numbers is said to be square invariant if

(x - α) (x - β) = (x - α^{2}) (x - β^{2})

. Suppose from the set of all square invariant quadratic polynomials we choose one at random.

The probability that the roots of the chosen polynomial are equal is __________. (rounded off to one decimal place)

Correct Answer:

0.5

Step-by-Step Solution

Key idea: This is a Classical Probability problem involving Algebraic Constraints. We must first determine the set of all "square invariant" polynomials (Sample Space) and then identify those with equal roots (Favorable Outcomes).

Step 1: Analyze the Square Invariant Condition.

The polynomial is $P (x) = (x - α) (x - β) = x^{2} - (α + β) x + α β$ .

The squared-root polynomial is $Q (x) = (x - α^{2}) (x - β^{2}) = x^{2} - (α^{2} + β^{2}) x + α^{2} β^{2}$ .

For $P (x) = Q (x)$ , the coefficients must match:

Sum of roots: $α + β = α^{2} + β^{2}$
Product of roots: $α β = α^{2} β^{2}$

Step 2: Solve the System of Equations.

From (2): $α β - (α β)^{2} = 0 ⟹ α β (1 - α β) = 0$ .

Case A: $α β = 0$ .

Case B: $α β = 1$ .

Step 3: Analyze Case A ( $α β = 0$ ).

Either $α = 0$ or $β = 0$ .

Substitute into (1): $α + β = α^{2} + β^{2}$ .

If $α = 0$ , then $β = β^{2} ⟹ β (β - 1) = 0 ⟹ β = 0$ or $β = 1$ .

Pairs $(α, β)$ : $(0, 0)$ and $(0, 1)$ .

By symmetry, if $β = 0$ , $α = 0$ or $α = 1$ . Pairs: $(0, 0)$ and $(1, 0)$ .

Distinct sets of roots ${α, β}$ : ${0, 0}$ and ${0, 1}$ .

Step 4: Analyze Case B ( $α β = 1 ⟹ β = 1/ α$ ).

Substitute into (1): $α + \frac{1}{α} = α^{2} + \frac{1}{α ^{2}}$ .

Let $s = α + \frac{1}{α}$ . Then $α^{2} + \frac{1}{α ^{2}} = s^{2} - 2$ .

Equation: $s = s^{2} - 2 ⟹ s^{2} - s - 2 = 0 ⟹ (s - 2) (s + 1) = 0$ .

So $s = 2$ or $s = - 1$ .

Subcase B1: $s = 2 ⟹ α + 1/ α = 2 ⟹ α^{2} - 2 α + 1 = 0 ⟹ (α - 1)^{2} = 0 ⟹ α = 1$ .

Then $β = 1$ . Pair: ${1, 1}$ .

Subcase B2: $s = - 1 ⟹ α + 1/ α = - 1 ⟹ α^{2} + α + 1 = 0$ .

Roots are complex cube roots of unity ( $ω, ω^{2}$ ).

Since $β = 1/ α$ , if $α = ω$ , $β = ω^{2}$ . If $α = ω^{2}$ , $β = ω$ .

The set of roots is ${ω, ω^{2}}$ . This is one unique polynomial.

Step 5: List All Valid Polynomials (Sample Space).

The polynomials are determined by their root sets:

Roots ${0, 0} ⟹ x^{2}$
Roots ${0, 1} ⟹ x (x - 1) = x^{2} - x$
Roots ${1, 1} ⟹ (x - 1)^{2} = x^{2} - 2 x + 1$
Roots ${ω, ω^{2}} ⟹ x^{2} + x + 1$

Total square invariant polynomials = 4.

Step 6: Identify Favorable Outcomes.

Condition: Roots are equal ( $α = β$ ).

${0, 0}$ : Equal. (Yes)
${0, 1}$ : Distinct. (No)
${1, 1}$ : Equal. (Yes)
${ω, ω^{2}}$ : Distinct. (No)

Favorable count = 2.

Step 7: Calculate Probability.

$P = \frac{2}{4} = 0.5$ .

Answer: 0.5

Question 4 · Engineering Mathematics · 2025_Set1 MCQ

A fair six-faced dice, with the faces labelled ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, and ‘6’, is rolled thrice. What is the probability of rolling ‘6’ exactly once?

A.
$\frac{75}{216}$
B.
$\frac{1}{6}$
C.
$\frac{1}{18}$
D.
$\frac{25}{216}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a Binomial Probability problem, recognizable because we have a fixed number of independent trials (rolling the die 3 times) and we are looking for a specific number of successes (rolling a '6').

Step 1: Identify the parameters of the Bernoulli trial.

Total trials ( $n$ ) = 3.
Success event: Rolling a '6'.
Probability of success ( $p$ ) = $\frac{1}{6}$ .
Probability of failure ( $q$ ) = $1 - p = \frac{5}{6}$ .
Desired number of successes ( $k$ ) = 1.

Step 2: Apply the Binomial Probability Formula.

The probability of getting exactly $k$ successes in $n$ trials is:

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

Step 3: Substitute the values and calculate.

$P (X = 1) = (1 3) (\frac{1}{6})^{1} (\frac{5}{6})^{3 - 1}$

$P (X = 1) = 3 \times \frac{1}{6} \times (\frac{5}{6})^{2}$

$P (X = 1) = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{75}{216}$

Answer: $\frac{75}{216}$

Question 5 · Engineering Mathematics · 2024_Set2 MCQ

When six unbiased dice are rolled simultaneously, the probability of getting all distinct numbers (i.e., 1, 2, 3, 4, 5, and 6) is

A.
$\frac{1}{324}$
B.
$\frac{5}{324}$
C.
$\frac{7}{324}$
D.
$\frac{11}{324}$

Correct Answer:

Step-by-Step Solution

Insight: Six dice, all distinct — this is exactly the permutation $6!$ over the sample space $6^{6}$ .

Exam route:

Total outcomes: $n (S) = 6^{6} = 46, 656$ .

All distinct means one of each face ${1, 2, 3, 4, 5, 6}$ , i.e. a permutation of 6 items: $n (A) = 6! = 720$ .

$P = \frac{720}{46 , 656}$ . Divide top and bottom by 144: $\frac{5}{324}$ .

Learning route:

This is a classical-counting question with distinct physical dice, so order matters.

Step 1 — Sample space. Six dice, each with 6 faces, all independent: $6^{6}$ equally likely outcomes.

Step 2 — Favourable outcomes. "All distinct" means each face $1, 2, 3, 4, 5, 6$ appears exactly once. Assigning faces to the 6 distinct dice is a permutation of 6 items: $6! = 720$ .

Step 3 — Reduce. $\frac{720}{46 , 656} = \frac{720 \div 144}{46 , 656 \div 144} = \frac{5}{324}$ .

Wrong path: If you treat the dice as indistinguishable and count only the 21 unordered sums/pairs, you get a completely wrong denominator.

Verification: $5 \times 46, 656 = 233, 280$ and $324 \times 720 = 233, 280$ , so the fraction is exact.

Question 6 · Engineering Mathematics · 2024_Set1 MSQ

Let $A$ and $B$ be two events in a probability space with $P (A) = 0.3$ , $P (B) = 0.5$ , and $P (A \cap B) = 0.1$ . Which of the following statements is/are TRUE?

A.
The two events $A$ and $B$ are independent
B.
$P (A \cup B) = 0.7$
C.
$P (A \cap B^{c}) = 0.2$ , where $B^{c}$ is the complement of the event $B$
D.
$P (A^{c} \cap B^{c}) = 0.4$ , where $A^{c}$ and $B^{c}$ are the complements of the events $A$ and $B$ , respectively

Correct Answer:

["B","C"]

Step-by-Step Solution

Insight: Compute $P (A \cup B)$ first with the addition theorem; everything else reduces to it.

Exam route:

Given: $P (A) = 0.3$ , $P (B) = 0.5$ , $P (A \cap B) = 0.1$ .

(A) Independent iff $P (A \cap B) = P (A) P (B) = 0.3 \times 0.5 = 0.15 \neq = 0.1$ . FALSE.

(B) $P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.3 + 0.5 - 0.1 = 0.7$ . TRUE.

(D) $P (A^{c} \cap B^{c}) = P ((A \cup B)^{c}) = 1 - P (A \cup B) = 1 - 0.7 = 0.3 \neq = 0.4$ . FALSE.

Learning route:

This is an event-algebra question: use the addition theorem and the identity $A \cap B^{c} = A ∖ (A \cap B)$ .

Step 1 — Union. $P (A \cup B) = 0.3 + 0.5 - 0.1 = 0.7$ .

Step 2 — Independence test. $P (A) P (B) = 0.15 \neq = 0.1$ , so not independent.

Step 3 — $A$ without $B$ . $P (A \cap B^{c}) = P (A) - P (A \cap B) = 0.2$ .

Step 4 — Neither. By De Morgan, $A^{c} \cap B^{c} = (A \cup B)^{c}$ , so $P = 1 - 0.7 = 0.3$ .

Wrong path: If you assume independence from the given numbers, you would get $P (A \cap B) = 0.15$ and break every other option.

Verification: $P (A \cap B) + P (A \cap B^{c}) + P (A^{c} \cap B) + P (A^{c} \cap B^{c}) = 0.1 + 0.2 + 0.4 + 0.3 = 1.0$ . The four atomic regions partition the sample space correctly.

Question 7 · Engineering Mathematics · 2024_Set1 MCQ

Consider a permutation sampled uniformly at random from the set of all permutations of ${1, 2, 3, \dots, n}$ for some $n \geq 4$ . Let $X$ be the event that 1 occurs before 2 in the permutation, and $Y$ the event that 3 occurs before 4. Which one of the following statements is TRUE?

A.
The events $X$ and $Y$ are mutually exclusive
B.
The events $X$ and $Y$ are independent
C.
Either event $X$ or $Y$ must occur
D.
Event $X$ is more likely than event $Y$

Correct Answer:

Step-by-Step Solution

Key idea: This problem tests the concept of Independence in the context of Random Permutations. We need to check if the occurrence of one event affects the probability of the other.

Step 1: Analyze Event X.

Event X: 1 occurs before 2 in a random permutation of ${1, \dots, n}$ .

By symmetry, in any random permutation, 1 is equally likely to be before 2 as it is to be after 2.

Therefore, $P (X) = \frac{1}{2}$ .

Step 2: Analyze Event Y.

Event Y: 3 occurs before 4 in the same permutation.

Similarly, by symmetry, $P (Y) = \frac{1}{2}$ .

Step 3: Check for Mutual Exclusivity and Exhaustiveness.

Can 1 be before 2 AND 3 be before 4 simultaneously? Yes (e.g., 1, 3, 2, 4). Thus, not mutually exclusive.

Must either X or Y occur? No (e.g., 2, 1, 4, 3 fails both). Thus, not exhaustive.

Step 4: Check for Independence.

Two events are independent if $P (X \cap Y) = P (X) P (Y)$ .

Consider the relative ordering of the four distinct elements ${1, 2, 3, 4}$ . There are $4! = 24$ equally likely relative orderings.

How many have 1 before 2 AND 3 before 4?

Choose 2 positions out of 4 for the pair (1,2): $(2 4) = 6$ ways. Once positions are chosen, 1 must be in the earlier spot and 2 in the later (1 way).
The remaining 2 positions are for 3 and 4. 3 must be in the earlier spot and 4 in the later (1 way).
So there are 6 favorable relative orderings.

$P (X \cap Y) = \frac{6}{24} = \frac{1}{4}$ .

Calculate product: $P (X) P (Y) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ .

Since $P (X \cap Y) = P (X) P (Y)$ , the events are independent.

Answer: The events X and Y are independent.

Question 8 · Engineering Mathematics · 2023 MSQ

Consider a random experiment where two fair coins are tossed. Let

A

be the event that denotes HEAD on both the throws,

B

be the event that denotes HEAD on the first throw, and

C

be the event that denotes HEAD on the second throw.
Which of the following statements is/are TRUE?

A.
$A$ and $B$ are independent.
B.
$A$ and $C$ are independent.
C.
$B$ and $C$ are independent.
D.
$Prob (B ∣ C) = Prob (B)$

Correct Answer:

["C","D"]

Step-by-Step Solution

Key idea: This problem tests the definition of Independence and Conditional Probability for simple coin toss events.

Step 1: Define Sample Space and Events.

Sample Space $S = {H H, H T, T H, T T}$ . Each has probability $1/4$ .

Event A (Head on both): ${H H}$ . $P (A) = 1/4$ .

Event B (Head on first): ${H H, H T}$ . $P (B) = 2/4 = 1/2$ .

Event C (Head on second): ${H H, T H}$ . $P (C) = 2/4 = 1/2$ .

Step 2: Check Option A: Are A and B independent?

Intersection $A \cap B = {H H} \cap {H H, H T} = {H H}$ .

$P (A \cap B) = 1/4$ .

Product $P (A) P (B) = (1/4) (1/2) = 1/8$ .

$1/4 \neq = 1/8$ . Not independent. Option A is FALSE.

(Intuition: If you know both are heads, it is certain the first is head. Dependence.)

Step 3: Check Option B: Are A and C independent?

Intersection $A \cap C = {H H} \cap {H H, T H} = {H H}$ .

$P (A \cap C) = 1/4$ .

Product $P (A) P (C) = (1/4) (1/2) = 1/8$ .

$1/4 \neq = 1/8$ . Not independent. Option B is FALSE.

Step 4: Check Option C: Are B and C independent?

Intersection $B \cap C = {H H, H T} \cap {H H, T H} = {H H}$ .

$P (B \cap C) = 1/4$ .

Product $P (B) P (C) = (1/2) (1/2) = 1/4$ .

$1/4 = 1/4$ . Match! So B and C are independent. Option C is TRUE.

Step 5: Check Option D: $P (B ∣ C) = P (B)$ .

This is the definition of independence between B and C.

Since we proved B and C are independent, this statement is TRUE.

Calculation: $P (B ∣ C) = \frac{P ( B \cap C )}{P ( C )} = \frac{1/4}{1/2} = 1/2$ .

$P (B) = 1/2$ .

$1/2 = 1/2$ . True.

Answer: Options C and D.

Question 9 · Engineering Mathematics · 2021_Set1 MCQ

There are five bags each containing identical sets of ten distinct chocolates. One chocolate is picked from each bag.

The probability that at least two chocolates are identical is ___________

A.
0.3024
B.
0.4235
C.
0.6976
D.
0.8125

Correct Answer:

Step-by-Step Solution

Insight: "at least two identical" is the complement of "all distinct", so count the distinct case and subtract from 1.

Exam route:

Total ways to pick one chocolate from each of 5 bags = $1 0^{5} = 100, 000$ .

Ways to get all 5 chocolates distinct (slot-filling) = $10 \times 9 \times 8 \times 7 \times 6 = 30, 240$ .

$P (all distinct) = \frac{30 , 240}{100 , 000} = 0.3024$ .

$P (at least two identical) = 1 - 0.3024 = 0.6976$ .

Learning route:

This is a complementary-counting question, signalled by the phrase "at least two".

Step 1 — Build the sample space. Each bag is independent and has 10 distinct chocolates, so picking one from each of 5 bags gives $1 0^{5}$ equally likely outcomes.

Step 2 — Count the complementary event. "All distinct" means no two of the five picked chocolates share a label. The first pick has 10 choices, the second must avoid the first (9 choices), the third avoids the first two (8 choices), and so on: $10 \times 9 \times 8 \times 7 \times 6 = 30, 240$ .

Step 3 — Subtract from 1. $P (at least two identical) = 1 - \frac{30 , 240}{100 , 000} = 1 - 0.3024 = 0.6976$ .

Wrong path: If you forget to complement, you land on 0.3024 (option A), which is the probability that all are distinct — the exact opposite of what is asked.

Verification: $0.6976 + 0.3024 = 1.0000$ , so the two complementary events partition the sample space correctly.

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Combinatorial Probability and Independent Events PYQs for GATE CS

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics

The Bernoulli Trial: Binary Randomness

What is a Bernoulli Trial?

Key Characteristics

Combinatorial Probability and Independent Events: Solved Questions with Step-by-Step Explanations (9 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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