GATE 2026 DA Q.34: solution and answer

GATE 2026 · DAQ.34 · 1 mark · NAT

Let $X$ be an exponentially distributed random variable with mean $λ (> 0)$ . If $P (X > 5) = 0.35$ , then the conditional probability $P (X > 10 ∣ X > 5)$ is ________. (Rounded off to two decimal places.)

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Answer and solution

ANSWER

$P (X > 10 ∣ X > 5) = P (X > 5) = 0.35$ .

Official answer: 0.34 to 0.36Source: Official answer key, GATE 2026 DA (IIT Guwahati)

For an exponential random variable with mean $λ$ , the survival function is $P (X > x) = e^{- x / λ}$ for $x \geq 0$ .
Apply the definition of conditional probability. Since ${X > 10} \subset {X > 5}$ , the intersection is ${X > 10}$ :
$P (X > 10 ∣ X > 5) = \frac{P ( X > 10 )}{P ( X > 5 )} = \frac{e ^{- 10/ λ}}{e ^{- 5/ λ}} = e^{- 5/ λ} .$
Note that $e^{- 5/ λ} = P (X > 5)$ . This is the memoryless property: $P (X > s + t ∣ X > s) = P (X > t)$ , here with $s = t = 5$ .
The problem gives $P (X > 5) = 0.35$ , so $P (X > 10 ∣ X > 5) = 0.35$ . The value of $λ$ is not needed (it would be $λ = 5/ ln (1/0.35) \approx 4.76$ ). To two decimal places the answer is $0.35$ .

KEY IDEA

The exponential distribution is memoryless, so $P (X > 10 ∣ X > 5) = P (X > 5)$ . The given $0.35$ is the answer directly, and $λ$ never has to be found.

WATCH OUT FOR

Solving for $λ$ first and then computing $P (X > 10)$ alone, which gives $0.3 5^{2} \approx 0.12$ . That forgets to divide by $P (X > 5)$ , so it is not a conditional probability.

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GATE 2026 DA Q.34: solution and answer

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