GATE 2025 DA Q.21: solution and answer

GATE 2025 · DAQ.21 · 1 mark · MCQ

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

(A)
$ln 2$
(B)
$ln 4$
(C)
$ln 3$
(D)
$ln 0.25$

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Answer and solution

ANSWER

$λ = ln 2$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 DA (IIT Roorkee)

An exponential random variable with rate $λ$ (so $E [X] = 1/ λ$ ) has survival function $P (X \geq x) = e^{- λ x}$ for $x \geq 0$ . This follows from $P (X \geq x) = \int_{x}^{\infty} λ e^{- λ t} d t = e^{- λ x}$ .
Set $x = 2$ : $e^{- 2 λ} = 0.25 = \frac{1}{4}$ .
Take natural logs of both sides: $- 2 λ = ln \frac{1}{4} = - ln 4$ .
So $λ = \frac{ln 4}{2}$ .
Use $ln 4 = 2 ln 2$ to get $λ = \frac{2 ln 2}{2} = ln 2$ .
Check: $e^{- 2 l n 2} = 2^{- 2} = 0.25$ , as given. The answer is option (A).

KEY IDEA

For an exponential variable with rate $λ$ , the tail probability is $P (X \geq x) = e^{- λ x}$ . Set this equal to the given value and take natural logs to solve for $λ$ .

WATCH OUT FOR

Using the mean $1/ λ$ as if it were the rate, or writing $P (X \geq 2) = 1 - e^{- 2 λ}$ (the CDF) instead of $e^{- 2 λ}$ . That gives $e^{- 2 λ} = 0.75$ and $λ = \frac{1}{2} ln \frac{4}{3}$ , which is not an option. Choosing $ln 4$ by forgetting to divide by 2 is also common.

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GATE 2025 DA Q.21: solution and answer

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