Basic Probability, Events and Counting PYQs for GATE DA

Solve 5+ Basic Probability, Events and Counting previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Probability, Events & Counting

Chapter Journey

1. Counting-Based Probability

The Foundation. Learn to count favorable vs total outcomes using permutations and combinations.

2. Coin Toss & Elementary Operations

Defining the Space. Sample spaces, unions, intersections, and complements.

3. Conditional Probability & Independence

Updating Beliefs. How new information changes probability. Bayes' Theorem.

4. Random Variables & Distributions

Quantifying Outcomes. Discrete/continuous variables, expectation, variance.

Why start here? In competitive exams, many problems look like probability questions but are actually counting puzzles in disguise. If you cannot count the sample space correctly, the probability formula yields incorrect results.

The Classical Definition: Probability as a Ratio

The Classical Approach

P (E) = \frac{n ( E )}{n ( S )}

Favorable Outcomes / Total Outcomes

Finite Sample Space: The total number of outcomes must be countable.
Equally Likely: No outcome is favored over another. (Fair coin vs Loaded coin).

The Real Challenge

Calculating $P (E)$ is rarely about plugging numbers. It is almost always a Counting Problem. We use Permutations and Combinations to find $n (S)$ and $n (E)$ .

Basic Probability, Events and Counting: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Probability and Statistics · 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 · Probability and Statistics · 2026 MCQ

Let

M

be a randomly chosen non-empty subset of

S = {1, 2, 3, \dots, 2026}

.
Which of the following is the probability that the product of all the elements of

M

is even?

A.
$2^{1013} (2^{1013} - 1) / 2^{2026}$
B.
$2^{1013} / 2^{2026}$
C.
$2^{1013} (2^{1013} - 1) / (2^{2026} - 1)$
D.
$1/ (2^{2026} - 1)$

Correct Answer:

Step-by-Step Solution

Insight: The product of a subset is even if at least one element is even. It is much easier to calculate the complement: the product is odd if and only if ALL elements in the subset are odd.

Exam route: Total non-empty subsets of $S$ is $2^{2026} - 1$ . There are 1013 odd numbers in $S$ . Non-empty subsets containing only odd numbers is $2^{1013} - 1$ . Probability of odd product is $\frac{2 ^{1013} - 1}{2 ^{2026} - 1}$ . Probability of even product is $1 - \frac{2 ^{1013} - 1}{2 ^{2026} - 1} = \frac{2 ^{2026} - 2 ^{1013}}{2 ^{2026} - 1} = \frac{2 ^{1013} ( 2 ^{1013} - 1 )}{2 ^{2026} - 1}$ .

Learning route: The set $S = {1, 2, \dots, 2026}$ contains 1013 odd numbers and 1013 even numbers. A subset product is odd if and only if every element in the subset is odd. The number of non-empty subsets formed entirely from the 1013 odd numbers is $2^{1013} - 1$ . The total number of non-empty subsets of $S$ is $2^{2026} - 1$ . The probability that a randomly chosen non-empty subset has an odd product is $\frac{2 ^{1013} - 1}{2 ^{2026} - 1}$ . Using the complement rule, the probability that the product is even is $1 - \frac{2 ^{1013} - 1}{2 ^{2026} - 1} = \frac{2 ^{2026} - 1 - ( 2 ^{1013} - 1 )}{2 ^{2026} - 1} = \frac{2 ^{2026} - 2 ^{1013}}{2 ^{2026} - 1}$ . Factoring out $2^{1013}$ gives $\frac{2 ^{1013} ( 2 ^{1013} - 1 )}{2 ^{2026} - 1}$ .

Question 3 · Probability and Statistics · 2026 MCQ

Let

L = n \to \infty lim k = 0 \sum n \frac{e ^{- n} n ^{k}}{k !}

Which of the following is the value of

L

A.
0.5
B.
1.0
C.
0
D.
$e^{- 1}$

Correct Answer:

Step-by-Step Solution

Insight: The sum is the cumulative distribution function (CDF) of a Poisson random variable with parameter n, evaluated at n. As n approaches infinity, the standardized Poisson converges to a standard normal distribution.

Exam route: The sum represents $P (Y_{n} \leq n)$ where $Y_{n} \sim Poisson (n)$ . Standardize the variable: $P (\frac{Y _{n} - n}{n} \leq 0)$ . By the Central Limit Theorem, this converges to $Φ (0) = 0.5$ .

Learning route: Recognize the summand $\frac{e ^{- n} n ^{k}}{k !}$ as the probability mass function of a Poisson distribution with parameter $λ = n$ . The sum from $k = 0$ to $n$ is exactly the probability $P (Y_{n} \leq n)$ . Since a Poisson(n) variable can be viewed as the sum of $n$ independent Poisson(1) variables, the Central Limit Theorem applies. As $n \to \infty$ , the standardized variable $Z_{n} = \frac{Y _{n} - n}{n}$ converges in distribution to $N (0, 1)$ . Therefore, the limit is $P (Z \leq 0) = Φ (0) = 0.5$ .

Question 4 · Probability and Statistics · 2024 MCQ

The probability of a boy or a girl being born is 1/2. For a family having only
three children, what is the probability of having two girls and one boy?

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

Correct Answer:

Step-by-Step Solution

Insight: This is a classic binomial probability problem where the probability of success (girl) is 0.5, and we want exactly 2 successes in 3 trials.

Exam route: 3 children, exactly 2 girls. Number of arrangements = 3C2 = 3. Total outcomes = 2^3 = 8. Probability = 3/8.

Learning route: The sample space for a family of 3 children has $2^{3} = 8$ equally likely outcomes (since prob of boy/girl is 1/2). We want exactly 2 girls and 1 boy. The favorable outcomes are GGB, GBG, and BGG. There are 3 such outcomes. The probability is the ratio of favorable to total outcomes: $3/8$ . Alternatively, using the binomial formula: $(2 3) (1/2)^{2} (1/2)^{1} = 3 \times 1/8 = 3/8$ .

Question 5 · Probability and Statistics · 2024 MCQ

Three fair coins are tossed independently.

T

is the event that two or more tosses
result in heads.

S

is the event that two or more tosses result in tails.
What is the probability of the event

T \cap S

A.
0
B.
0.5
C.
0.25
D.
1

Correct Answer:

Step-by-Step Solution

Insight: In 3 tosses, you cannot have 2 or more heads AND 2 or more tails simultaneously because that requires at least 4 tosses.

Exam route: T requires >= 2 heads, S requires >= 2 tails. Total coins = 3. Max heads + Max tails = 3. The intersection is empty. Probability = 0.

Learning route: The sample space for 3 fair coins has 8 equally likely outcomes. Event T (>= 2 heads) consists of {HHH, HHT, HTH, THH}. Event S (>= 2 tails) consists of {TTT, TTH, THT, HTT}. The intersection T ∩ S requires an outcome to have at least 2 heads and at least 2 tails, which means at least 4 coins. Since we only have 3 coins, T ∩ S = ∅. Therefore, P(T ∩ S) = 0.

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Basic Probability, Events and Counting PYQs for GATE DA

Chapter Roadmap: Probability, Events & Counting

1. Counting-Based Probability

2. Coin Toss & Elementary Operations

3. Conditional Probability & Independence

4. Random Variables & Distributions

The Classical Definition: Probability as a Ratio

The Real Challenge

Basic Probability, Events and Counting: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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