GATE 2025 DA Q.61: solution and answer

GATE 2025 · DAQ.61 · 2 marks · NAT

A bag contains 5 white balls and 10 black balls. In a random experiment, $n$ balls are drawn from the bag one at a time with replacement. Let $S_{n}$ denote the total number of black balls drawn in the experiment.

The expectation of $S_{100}$ denoted by $E [S_{100}] =$ ______

(Round off to one decimal place)

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Answer and solution

ANSWER

$E [S_{100}] = \frac{200}{3} \approx 66.7$

Official answer: 66.6 to 66.7Source: Official answer key, GATE 2025 DA (IIT Roorkee)

The bag holds $5 + 10 = 15$ balls, of which 10 are black.
Each draw is made with replacement, so the draws are independent and the bag is the same every time. On every draw $P (black) = \frac{10}{15} = \frac{2}{3}$ .
Let $X_{i}$ be the indicator that draw $i$ is black. Then $S_{100} = \sum_{i = 1}^{100} X_{i}$ , and $S_{100} \sim Binomial (100, \frac{2}{3})$ .
By linearity of expectation, $E [S_{100}] = \sum_{i = 1}^{100} E [X_{i}] = n p$ , because $E [X_{i}] = P (X_{i} = 1) = \frac{2}{3}$ .
$E [S_{100}] = 100 \times \frac{2}{3} = \frac{200}{3} \approx 66.667.$
Rounded to one decimal place, the answer is $66.7$ .

KEY IDEA

With replacement, every draw is an independent trial with the same success probability $p = 10/15 = 2/3$ . So $S_{n}$ is Binomial $(n, p)$ and $E [S_{n}] = n p$ . Linearity of expectation gives this directly, without needing independence.

WATCH OUT FOR

Using $p = 1/2$ , or counting black balls out of 10 instead of out of 15 (total balls). Another slip is treating the draws as without replacement, which does not change the expectation but is a wrong model for this question.

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 2 Discrete distributions questions, with notes