GATE 2025 IN Q.52: solution and answer

GATE 2025 · INQ.52 · 2 marks · NAT

The probability of a student missing a class is 0.1. In a total number of 10 classes, the probability that the student will not miss more than one class is ___ (rounded off to two decimal places).

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Answer and solution

ANSWER

The probability that the student misses at most one of the 10 classes is $1.9 (0.9)^{9} \approx 0.74$ .

Official answer: 0.72 to 0.76Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Let $X$ be the number of classes missed out of 10. Treat each class as an independent trial with miss probability $p = 0.1$ . Then $X \sim Binomial (n = 10, p = 0.1)$ and $q = 1 - p = 0.9$ .
"Will not miss more than one class" means $X \leq 1$ , so we need $P (X = 0) + P (X = 1)$ .
The binomial pmf is $P (X = k) = (k n) p^{k} q^{n - k}$ . This gives $P (X = 0) = (0 10) (0.1)^{0} (0.9)^{10} = (0.9)^{10}$ and $P (X = 1) = (1 10) (0.1) (0.9)^{9} = 10 \times 0.1 \times (0.9)^{9} = (0.9)^{9}$ .
Take out the common factor $(0.9)^{9}$ : $P (X \leq 1) = (0.9)^{10} + (0.9)^{9} = (0.9)^{9} (0.9 + 1) = 1.9 (0.9)^{9}$
With $(0.9)^{9} = 0.387420489$ : $P (X \leq 1) = 1.9 \times 0.387420489 = 0.736099 \approx 0.74$ (to two decimal places).

KEY IDEA

Treat the 10 classes as independent trials with a fixed miss probability of 0.1. The number missed is then $Binomial (10, 0.1)$ . "Not more than one" means $X \leq 1$ , so add $P (X = 0)$ and $P (X = 1)$ . Here $(1 10) (0.1) = 1$ , so $P (X = 1) = (0.9)^{9}$ and the whole sum factors to $1.9 (0.9)^{9}$ .

WATCH OUT FOR

Reading "not more than one" as "exactly one" gives only $P (X = 1) = (0.9)^{9} \approx 0.39$ . Reading it as "none" gives only $(0.9)^{10} \approx 0.35$ . A third slip is swapping $p$ and $q$ , using 0.9 as the miss probability, which gives a tiny wrong value. Another is using the Poisson approximation with $λ = n p = 1$ : it gives $2 e^{- 1} \approx 0.7358$ . That happens to round to 0.74 here, but it is an approximation, and the exact binomial is what the question asks for.

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