GATE 2020 EC Q.25: solution and answer

GATE 2020 · ECQ.25 · 1 mark · NAT

The two sides of a fair coin are labelled as 0 and 1. The coin is tossed two times independently. Let $M$ and $N$ denote the labels corresponding to the outcomes of those tosses. For a random variable $X$ , defined as $X = min (M, N)$ , the expected value $E (X)$ (rounded off to two decimal places) is ________.

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Answer and solution

ANSWER

$E (X) = \frac{1}{4} = 0.25$

Official answer: 0.25Source: Official answer key, GATE 2020 EC (IIT Delhi)

Sample space for two independent tosses of a fair coin: $(M, N) \in {(0, 0), (0, 1), (1, 0), (1, 1)}$ . The coin is fair and the tosses are independent, so each outcome has probability $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ .
Evaluate $X = min (M, N)$ on each outcome: $min (0, 0) = 0$ , $min (0, 1) = 0$ , $min (1, 0) = 0$ , $min (1, 1) = 1$ .
So $X$ takes only the values 0 and 1, and $X = 1$ only when both tosses show 1. By independence, $P (X = 1) = P (M = 1) P (N = 1) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ , and $P (X = 0) = 1 - \frac{1}{4} = \frac{3}{4}$ .
Definition of expectation for a discrete random variable, $E (X) = \sum x P (X = x)$ : $E (X) = 0 \cdot \frac{3}{4} + 1 \cdot \frac{1}{4} = \frac{1}{4} = 0.25$

KEY IDEA

For 0/1 variables, $min (M, N)$ equals the product $M N$ : it is 1 only when both are 1. So $X$ is Bernoulli with $p = P (M = 1) P (N = 1) = \frac{1}{4}$ , and the mean of a Bernoulli variable is $p$ . Here $E (X) = E (M N) = E (M) E (N) = \frac{1}{2} \cdot \frac{1}{2}$ , which uses the independence of the tosses.

WATCH OUT FOR

Answering 0.75 by finding $P (X = 0)$ when the question asks for $E (X)$ , or mixing up min with max: $max (M, N)$ is 1 in three of the four outcomes and gives $E = 0.75$ . A second mistake is writing $E (min (M, N)) = min (E (M), E (N)) = 0.5$ . Expectation does not commute with a nonlinear function like min.

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GATE 2020 EC Q.25: solution and answer

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