GATE 2025 EE Q.15: solution and answer

GATE 2025 · EEQ.15 · 1 mark · MCQ

Consider discrete random variables $X$ and $Y$ with probabilities as follows:

P (X = 0 and Y = 0) = \frac{1}{4}

P (X = 1 and Y = 0) = \frac{1}{8}

P (X = 0 and Y = 1) = \frac{1}{2}

P (X = 1 and Y = 1) = \frac{1}{8}

Given $X = 1$ , the expected value of $Y$ is

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

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Answer and solution

ANSWER

$E [Y ∣ X = 1] = \frac{1}{2}$ , option (B).

Official answer: BSource: Official answer key, GATE 2025 EE (IIT Roorkee)

Check that the joint PMF is valid: $\frac{1}{4} + \frac{1}{8} + \frac{1}{2} + \frac{1}{8} = 1$ .
Find the probability of the conditioning event by adding the two $X = 1$ entries over all values of $Y$ : $P (X = 1) = P (X = 1, Y = 0) + P (X = 1, Y = 1) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$ .
Conditional PMF $P (Y = y ∣ X = 1) = \frac{P ( X = 1 , Y = y )}{P ( X = 1 )}$ : $P (Y = 0 ∣ X = 1) = \frac{1/8}{1/4} = \frac{1}{2}$ and $P (Y = 1 ∣ X = 1) = \frac{1/8}{1/4} = \frac{1}{2}$ . These add to 1, as they should.
Conditional expectation: $E [Y ∣ X = 1] = \sum_{y} y P (Y = y ∣ X = 1) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$ .
So the expected value of $Y$ given $X = 1$ is $\frac{1}{2}$ , which is option (B).

KEY IDEA

Conditioning on $X = 1$ keeps only the $X = 1$ entries of the joint PMF, rescaled by $P (X = 1)$ so they add to 1. Then $E [Y ∣ X = 1] = \sum_{y} y P (Y = y ∣ X = 1)$ . Because $Y$ takes only the values 0 and 1, this equals $P (Y = 1 ∣ X = 1)$ .

WATCH OUT FOR

Using the joint probability $P (X = 1, Y = 1) = \frac{1}{8}$ as the answer without dividing by $P (X = 1)$ gives option (C). Taking the unconditional mean $E [Y] = P (Y = 1) = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$ ignores the condition altogether, and $\frac{5}{8}$ is not among the options.

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 33 Probability & statistics questions, with notes

GATE 2025 EE Q.15: solution and answer

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