Consider the cumulative distribution function (CDF) of a random variable X: F_X(x) = cases 0 & x ≤ -1 \\ 14(x + 1)^2 & -1 ≤ x ≤ 1 \\ 1 & x ≥ 1 cases The value of P(X^2 ≤ 0.25) is

GATE DA 2025 Q39: Using a CDF to find P(X² ≤ 0.25)

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

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

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

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Answer: C (official GATE 2025 final answer key)

Concept used: Converting an event on $X^2$ into an interval on $X$ and using $P(a \le X \le b) = F_X(b) - F_X(a)$ .

First step: $X^2 \le 0.25$ means $-0.5 \le X \le 0.5$ , so compute $F_X(0.5) - F_X(-0.5)$ .

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Source: official GATE 2025 DA question paper and final answer key, organised by IIT Roorkee. Question paper · Answer key