GATE DA Random Variables and Distributions Previous Year Questions

GATE DA 2025 | Question 19

MCQ · 1 mark · −0.33

Let \(X\) be a continuous random variable whose cumulative distribution function (CDF) \(F_{X}(x)\), for some \(t\), is given as follows:

\[F_{X}(x) = \begin{cases} 0 & x \leq t \\ \dfrac{x - t}{4 - t} & t \leq x \leq 4 \\ 1 & x \geq 4 \end{cases}\]

If the median of \(X\) is 3, then what is the value of \(t\)?

2 1 \(-1\) 0

Explanation

Answer: (A)

The median \(m\) is the point where the CDF reaches one half, \(F_{X}(m) = \frac{1}{2}\). Since 3 lies in the middle piece (\(t \leq 3 \leq 4\)),

\[\frac{3 - t}{4 - t} = \frac{1}{2} \;\Rightarrow\; 6 - 2t = 4 - t \;\Rightarrow\; t = 2.\]

Check: with \(t = 2\), \(X\) is uniform on \([2, 4]\), whose midpoint is 3. The other options fail; for example, \(t = 0\) gives \(F_{X}(3) = \frac{3}{4}\), so the median would be 2 instead.

Answer: A

It is given that \(P(X \geq 2) = 0.25\) for an exponentially distributed random variable \(X\) with \(E[X] = \frac{1}{\lambda}\), where \(E[X]\) denotes the expectation of \(X\). What is the value of \(\lambda\)?
(\(\ln\) denotes natural logarithm)

\(\ln 2\) \(\ln 4\) \(\ln 3\) \(\ln 0.25\)

Explanation

Answer: (A)

An exponential random variable with mean \(\frac{1}{\lambda}\) has rate \(\lambda\) and tail probability \(P(X \geq x) = e^{-\lambda x}\) for \(x \geq 0\). So

\[e^{-2\lambda} = 0.25 = \frac{1}{4} \;\Rightarrow\; 2\lambda = \ln 4 = 2\ln 2 \;\Rightarrow\; \lambda = \ln 2.\]

Check: \(e^{-2\ln 2} = 2^{-2} = 0.25\). Option B is the value of \(2\lambda\), not \(\lambda\), and option D is negative (\(\ln 0.25 = -\ln 4\)), which is impossible for a rate.

Answer: A

GATE DA 2025 | Question 39

MCQ · 2 marks · −0.66

Consider the cumulative distribution function (CDF) of a random variable \(X\):

\[F_X(x) = \begin{cases} 0 & x \le -1 \\ \frac{1}{4}(x+1)^{2} & -1 \le x \le 1 \\ 1 & x \ge 1 \end{cases}\]

The value of \(P(X^{2} \le 0.25)\) is

0.625 0.25 0.5 0.5625

Explanation

Answer: (C)

\(X^{2} \le 0.25\) means \(-0.5 \le X \le 0.5\). The CDF has no jumps (it rises continuously from 0 at \(x = -1\) to 1 at \(x = 1\)), so

\[P(-0.5 \le X \le 0.5) = F_X(0.5) - F_X(-0.5) = \frac{(1.5)^{2} - (0.5)^{2}}{4} = \frac{2}{4} = 0.5.\]

Two slips lead to the distractors: \(F_X(0.5) = 0.5625\) alone forgets the lower limit (D), and reading the event as \(|X| \le 0.25\) gives \(F_X(0.25) - F_X(-0.25) = 0.25\) (B).

Answer: C

GATE DA 2025 | Question 40

MCQ · 2 marks · −0.66

A random variable \(X\) is said to be distributed as \(Bernoulli(\theta)\), denoted by \(X \sim Bernoulli(\theta)\), if

\[P(X = 1) = \theta, \quad P(X = 0) = 1 - \theta\]

for \(0 \lt \theta \lt 1\). Let \(Y = \sum_{i=1}^{300} X_i\), where \(X_i \sim Bernoulli(\theta)\), \(i = 1, 2, \ldots, 300\) be independent and identically distributed random variables with \(\theta = 0.25\). The value of \(P(60 \le Y \le 90)\), after approximation through Central Limit Theorem, is given by
(Recall that \(\phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-\frac{t^{2}}{2}}\, dt\))

\(\phi(2) - \phi(-2)\) \(\phi(1) - \phi(-1)\) \(\phi(3) - \phi(-3)\) \(\phi(90) - \phi(60)\)

Explanation

Answer: (A)

\(Y\) is a sum of 300 independent Bernoulli variables with \(\theta = 0.25\), so

\(E[Y] = 300 \times 0.25 = 75\);
\(\operatorname{Var}(Y) = 300 \times 0.25 \times 0.75 = 56.25\), a standard deviation of \(7.5\).

By the Central Limit Theorem, \(\frac{Y - 75}{7.5}\) is approximately standard normal. The limits standardise to \(\frac{60 - 75}{7.5} = -2\) and \(\frac{90 - 75}{7.5} = 2\), so

\[P(60 \le Y \le 90) \approx \phi(2) - \phi(-2).\]

Option D puts the raw values of \(Y\) into \(\phi\) without standardising.

Answer: A

GATE DA 2025 | Question 41

MCQ · 2 marks · −0.66

For \(x \in \mathbb{R}\), the floor function is denoted by \(f(x) = \lfloor x \rfloor\) and defined as follows

\[\lfloor x \rfloor = k, \;\; k \le x \lt k + 1,\]

where \(k\) is an integer. Let \(Y = \lfloor X \rfloor\), where \(X\) is an exponentially distributed random variable with mean \(\frac{1}{\ln 10}\), where \(\ln\) denotes natural logarithm. For any positive integer \(\ell\), one can write the probability of the event \(Y = \ell\) as follows

\[P(Y = \ell) = q^{\ell}(1 - q)\]

The value of \(q\) is

0.1 0.01 0.5 0.434

Explanation

Answer: (A)

A mean of \(\frac{1}{\ln 10}\) means the rate is \(\lambda = \ln 10\), so \(P(X \ge t) = e^{-t \ln 10} = 10^{-t}\) for \(t \ge 0\).

\(Y = \ell\) happens exactly when \(\ell \le X \lt \ell + 1\), so

\[P(Y = \ell) = 10^{-\ell} - 10^{-(\ell + 1)} = (0.1)^{\ell}\,(1 - 0.1).\]

This has the form \(q^{\ell}(1 - q)\) with \(q = 0.1\), so the floor of an exponential variable is geometric. Option D, \(0.434 \approx \frac{1}{\ln 10}\), is the mean of \(X\), not \(q\).

Answer: A

GATE DA 2024 | Question 11

MCQ · 1 mark · −0.33

Consider the following statements:

(i) The mean and variance of a Poisson random variable are equal.
(ii) For a standard normal random variable, the mean is zero and the variance is one.

Which ONE of the following options is correct?

Both (i) and (ii) are true (i) is true and (ii) is false (ii) is true and (i) is false Both (i) and (ii) are false

Explanation

Answer: (A)

(i) A Poisson random variable with parameter \(\lambda\) has \(P(X = k) = \dfrac{e^{-\lambda}\lambda^{k}}{k!}\), and both its mean and its variance equal \(\lambda\). True.

(ii) The standard normal distribution \(N(0, 1)\) is, by definition, the normal distribution with mean 0 and variance 1; any normal variable is brought to it by \(Z = (X - \mu)/\sigma\). True.

Both statements hold.

Answer: A