GATE 2025 DA Q.31: solution and answer

GATE 2025 · DAQ.31 · 1 mark · NAT

There are three boxes containing white balls and black balls.

Box-1 contains 2 black and 1 white balls.

Box-2 contains 1 black and 2 white balls.

Box-3 contains 3 black and 3 white balls.

In a random experiment, one of these boxes is selected, where the probability of choosing Box-1 is $\frac{1}{2}$ , Box-2 is $\frac{1}{6}$ , and Box-3 is $\frac{1}{3}$ . A ball is drawn at random from the selected box. Given that the ball drawn is white, the probability that it is drawn from Box-2 is ____________ (Round off to two decimal places)

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Answer and solution

ANSWER

Given that the ball drawn is white, the probability that it came from Box-2 is $P (B_{2} ∣ W) = \frac{1}{4} = 0.25$ .

Official answer: 0.25Source: Official answer key, GATE 2025 DA (IIT Roorkee)

Let $B_{i}$ be the event that Box- $i$ is chosen and $W$ the event that the ball drawn is white. The priors are $P (B_{1}) = \frac{1}{2}, P (B_{2}) = \frac{1}{6}, P (B_{3}) = \frac{1}{3}$ , and they sum to $1$ .
Read the likelihoods from the box contents (white balls out of total balls): $P (W ∣ B_{1}) = \frac{1}{3}, P (W ∣ B_{2}) = \frac{2}{3}, P (W ∣ B_{3}) = \frac{3}{6} = \frac{1}{2}$ .
By the law of total probability: $P (W) = \frac{1}{2} \cdot \frac{1}{3} + \frac{1}{6} \cdot \frac{2}{3} + \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} + \frac{1}{9} + \frac{1}{6} = \frac{3 + 2 + 3}{18} = \frac{4}{9} .$
By Bayes' theorem: $P (B_{2} ∣ W) = \frac{P ( W ∣ B _{2} ) P ( B _{2} )}{P ( W )} = \frac{\frac{2}{3} \cdot \frac{1}{6}}{\frac{4}{9}} = \frac{1/9}{4/9} = \frac{1}{4} .$
So the probability is $0.25$ to two decimal places.

KEY IDEA

The box is the hidden cause and the white ball is the evidence. Weight each box's chance of giving white by its prior, add these to get $P (W)$ , then take Box-2's share of that total (Bayes' theorem).

WATCH OUT FOR

Using the prior $\frac{1}{6}$ directly, or leaving out the prior and comparing only the likelihoods $\frac{1}{3}, \frac{2}{3}, \frac{1}{2}$ . That gives $\frac{2/3}{3/2} = \frac{4}{9} \approx 0.44$ . The unequal box probabilities must be included in both the numerator and the denominator.

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 3 Conditional & Bayes questions, with notes