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GATE DA 2025 Q21: Exponential distribution: P(X ≥ 2) = 0.25, find λ

Question

It is given that P(X≥2)=0.25P(X \ge 2) = 0.25 for an exponentially distributed random variable XX with E[X]=1λE[X] = \dfrac{1}{\lambda}, where E[X]E[X] denotes the expectation of XX. What is the value of λ\lambda? (ln⁡\ln denotes natural logarithm)

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Answer: A (official GATE 2025 final answer key)

Concept used: Tail probability of an exponential distribution: P(X≥x)=e−λxP(X \ge x) = e^{-\lambda x}.

First step: Write P(X≥2)=e−2λ=0.25P(X \ge 2) = e^{-2\lambda} = 0.25 and take the natural logarithm of both sides.

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

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