GATE 2025 DA Q.41: solution and answer

GATE 2025 · DAQ.41 · 2 marks · MCQ

For $x \in R$ , the floor function is denoted by $f (x) = ⌊ x ⌋$ and defined as follows

⌊ x ⌋ = k, k \leq x < k + 1,

where $k$ is an integer. Let $Y = ⌊ X ⌋$ , where $X$ is an exponentially distributed random variable with mean $\frac{1}{l n 10}$ , where $ln$ denotes natural logarithm. For any positive integer $ℓ$ , one can write the probability of the event $Y = ℓ$ as follows

P (Y = ℓ) = q^{ℓ} (1 - q)

The value of $q$ is

(A)
0.1
(B)
0.01
(C)
0.5
(D)
0.434

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Answer and solution

ANSWER

The value of $q$ is $0.1$ , which is option (A).

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

An exponential random variable with mean $\frac{1}{λ}$ has rate $λ$ . Here the mean is $\frac{1}{l n 10}$ , so $λ = ln 10$ .
The event $Y = ℓ$ means $⌊ X ⌋ = ℓ$ , which holds exactly when $ℓ \leq X < ℓ + 1$ .
For an exponential variable, $P (X \geq t) = e^{- λ t}$ for $t \geq 0$ . Hence $P (Y = ℓ) = P (ℓ \leq X < ℓ + 1) = e^{- λ ℓ} - e^{- λ (ℓ + 1)} .$
Factor out $e^{- λ ℓ}$ : $P (Y = ℓ) = e^{- λ ℓ} (1 - e^{- λ}) = (e^{- λ})^{ℓ} (1 - e^{- λ}) .$
Comparing with $q^{ℓ} (1 - q)$ gives $q = e^{- λ}$ . So $Y$ is geometric with ratio $q$ .
Substitute $λ = ln 10$ : $q = e^{- l n 10} = \frac{1}{e ^{l n 10}} = \frac{1}{10} = 0.1.$ This is option (A).

KEY IDEA

The floor of an exponential variable is geometric. $P (Y = ℓ) = P (ℓ \leq X < ℓ + 1) = e^{- λ ℓ} (1 - e^{- λ})$ , so the ratio is $q = e^{- λ}$ , where $λ$ is the rate (the reciprocal of the mean).

WATCH OUT FOR

Taking the stated mean $\frac{1}{l n 10} \approx 0.434$ as the rate or as $q$ itself (option D). The mean is $\frac{1}{λ}$ , so $λ = ln 10$ and $q = e^{- λ} = 0.1$ .

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GATE 2025 DA Q.41: solution and answer

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