GATE 2025 DA Q.45: solution and answer

GATE 2025 · DAQ.45 · 2 marks · MCQ

A random experiment consists of throwing 100 fair dice, each die having six faces numbered 1 to 6. An event $A$ represents the set of all outcomes where at least one of the dice shows a 1. Then, $P (A) =$

(A)
$0$
(B)
$1$
(C)
$1 - (\frac{5}{6})^{100}$
(D)
$(\frac{5}{6})^{100}$

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Answer and solution

ANSWER

$P (A) = 1 - (\frac{5}{6})^{100}$ , which is option (C).

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

The event $A$ is "at least one of the 100 dice shows a 1". Its complement $A^{c}$ is "none of the 100 dice shows a 1", so $P (A) = 1 - P (A^{c})$ .
For one fair die, the faces 2, 3, 4, 5, 6 are not a 1, so $P (not 1) = \frac{5}{6}$ .
The 100 dice are thrown independently, so the probabilities multiply: $P (A^{c}) = (\frac{5}{6})^{100} .$ Equivalently, there are $5^{100}$ outcomes with no 1 out of $6^{100}$ equally likely outcomes.
Therefore $P (A) = 1 - (\frac{5}{6})^{100} .$
This value is strictly between 0 and 1, so options (A) and (B) are out. Option (D) is $P (A^{c})$ , not $P (A)$ . The answer is (C).

KEY IDEA

For an "at least one" event over independent trials, use the complement: P(at least one) = 1 − P(none), and P(none) is the product of the per-trial "none" probabilities.

WATCH OUT FOR

Choosing (D), which is the probability that no die shows a 1 (the complement), or trying to add probabilities such as 100 × 1/6, which overcounts outcomes where several dice show a 1.

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 5 Counting & axioms questions, with notes

GATE 2025 DA Q.45: solution and answer

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