GATE 2024 DA Q.56: solution and answer

GATE 2024 · DAQ.56 · 2 marks · NAT

Let $X$ be a random variable uniformly distributed in the interval $[1, 3]$ and $Y$ be a random variable uniformly distributed in the interval $[2, 4]$ . If $X$ and $Y$ are independent of each other, the probability $P (X \geq Y)$ is ______ (rounded off to three decimal places).

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Answer and solution

ANSWER

$P (X \geq Y) = \frac{1}{8} = 0.125$ .

Official answer: 0.125Source: Official answer key, GATE 2024 DA (IISc Bengaluru)

Independence gives the joint density $f (x, y) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ on the rectangle $[1, 3] \times [2, 4]$ , which has area $2 \times 2 = 4$ .
So $P (X \geq Y) = \frac{area of the part of the rectangle where x \geq y}{4}$ .
Since $y \geq 2$ , the condition $x \geq y$ forces $x \geq 2$ . So $x \in [2, 3]$ and, for each such $x$ , $y \in [2, x]$ .
The area is $\int_{2}^{3} (x - 2) d x = [\frac{( x - 2 ) ^{2}}{2}]_{2}^{3} = \frac{1}{2}$ . This is a right triangle with legs of length $1$ .
Hence $P (X \geq Y) = \frac{1/2}{4} = \frac{1}{8} = 0.125$ .
Check by conditioning on $X$ : for $x \in [2, 3]$ , $P (Y \leq x) = \frac{x - 2}{2}$ , and for $x \in [1, 2)$ it is $0$ . So $P (X \geq Y) = \int_{2}^{3} \frac{1}{2} \cdot \frac{x - 2}{2} d x = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$ .

KEY IDEA

For independent uniform variables the point $(X, Y)$ is uniform on a rectangle, so the probability is the area of the region $x \geq y$ divided by the rectangle's area.

WATCH OUT FOR

Treating the two intervals as if they overlapped on $[2, 3]$ only and taking $P = \frac{1}{2}$ there, or integrating over $x \in [1, 3]$ and forgetting that $Y \geq 2$ makes $P (Y \leq x) = 0$ for $x < 2$ . The area must be divided by the full rectangle area $4$ , not by $1$ .

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GATE 2024 DA Q.56: solution and answer

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