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Continuous and Standard Probability Distributions PYQs for GATE CS

Solve 3+ Continuous and Standard Probability Distributions previous year questions for GATE CS with answers and detailed solutions. Free sample questions belo

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Question 1

2026 Slot Set2 PYQ

Level 3: Exam Standard

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 ?

Question 2

2025 Slot Set1 PYQ

Level 3: Exam Standard

Consider a probability distribution given by the density function P (x) . P (x) = {C x^{2}, 0, for 1 \leq x \leq 4 for x l t; 1 or x g t; 4 The probability that x lies between 2 and 3, i.e., P (2 \leq x \leq 3) is __________. (rounded off to three decimal places)

Question 3

2021 Slot Set1 PYQ

Level 3: Exam Standard

The lifetime of a component of a certain type is a random variable whose probability density function is exponentially distributed with parameter 2. For a randomly picked component of this type, the probability that its lifetime exceeds the expected lifetime (rounded to 2 decimal places) is __________.

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Continuous and Standard Probability Distributions PYQs for GATE CS

Solve 3+ Continuous and Standard Probability Distributions previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Continuous and Standard Probability Distributions

Chapter Journey

Continuous & Standard Probability Distributions

What you will master: Understanding the shape, parameters, and applications of key continuous distributions used in engineering and computer science.

1. Exponential Lifetime Distribution Modeling time between events, memoryless property. Key for reliability contexts.
2. Normal Distribution Identification Recognizing the Gaussian PDF, mean, and variance. High weightage.
3. Density Normalization & Interval Probabilities Finding constants for valid PDFs and integrating over intervals. Fundamental skill.

Topic Hero: Exponential Lifetime Distribution

Topic 1 of 3

Exponential Lifetime Distribution

The exponential distribution models the time elapsed until a specific event occurs. It is widely used in reliability engineering and queuing theory to represent "waiting times" or component lifespans.

Key Learning Points:

PDF and CDF formulas
Mean, Variance, and Memoryless Property
Solving classic exam problems

Continuous and Standard Probability Distributions: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Engineering Mathematics · 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 2 · Engineering Mathematics · 2025_Set1 NAT

Consider a probability distribution given by the density function

P (x)

P (x) = {C x^{2}, 0, for 1 \leq x \leq 4 for x l t; 1 or x g t; 4

The probability that

x

lies between 2 and 3, i.e.,

P (2 \leq x \leq 3)

is __________. (rounded off to three decimal places)

Correct Answer:

0.30

Step-by-Step Solution

Insight: First normalize the PDF to find $C$ , then integrate over the specific interval $[2, 3]$ .

Exam route: $\int_{1}^{4} C x^{2} d x = 1 ⟹ C (64/3 - 1/3) = 1 ⟹ C = 1/21$ . $P (2 \leq x \leq 3) = \int_{2}^{3} \frac{1}{21} x^{2} d x = \frac{1}{63} (27 - 8) = 19/63 \approx 0.30$ .

Learning route:

Normalize the PDF: The total area under the density function must be 1. $\int_{1}^{4} C x^{2} d x = 1$ .
Evaluate the integral: $C [x^{3} /3]_{1}^{4} = C (64/3 - 1/3) = 21 C = 1 ⟹ C = 1/21$ .
Set up the target probability: $P (2 \leq x \leq 3) = \int_{2}^{3} \frac{1}{21} x^{2} d x$ .
Evaluate the integral: $\frac{1}{21} [x^{3} /3]_{2}^{3} = \frac{1}{63} (3^{3} - 2^{3}) = \frac{1}{63} (27 - 8) = 19/63$ .
Calculate the decimal: $19/63 \approx 0.30158$ .
Round to 2 decimal places: 0.30.

Question 3 · Engineering Mathematics · 2021_Set1 NAT

The lifetime of a component of a certain type is a random variable whose probability density function is exponentially distributed with parameter 2. For a randomly picked component of this type, the probability that its lifetime exceeds the expected lifetime (rounded to 2 decimal places) is __________.

Correct Answer:

0.37

Step-by-Step Solution

Insight: For any exponential distribution, the probability of exceeding the mean is always $e^{- 1} \approx 0.37$ .

Exam route: $λ = 2 ⟹ E [X] = 0.5$ . $P (X > 0.5) = e^{- 2 (0.5)} = e^{- 1} \approx 0.37$ .

Learning route:

Identify the distribution: Exponential with rate parameter $λ = 2$ .
Find the expected lifetime: $E [X] = 1/ λ = 1/2 = 0.5$ .
Set up the probability: We need the probability that the lifetime exceeds the expected lifetime, so $P (X > 0.5)$ .
Use the survival function: For an exponential distribution, $P (X > t) = e^{- λ t}$ .
Calculate: $P (X > 0.5) = e^{- 2 \times 0.5} = e^{- 1} \approx 0.3678$ .
Round to 2 decimal places: 0.37.

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Questions calibrated to the real exam

Notes written for recall, not for volume

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Honest progress

Unlock the whole course

Continuous and Standard Probability Distributions PYQs for GATE CS

Chapter Roadmap: Continuous and Standard Probability Distributions

Chapter Journey

Topic Hero: Exponential Lifetime Distribution

Exponential Lifetime Distribution

Continuous and Standard Probability Distributions: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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