GATE 2025 DA Q.54: solution and answer

GATE 2025 · DAQ.54 · 2 marks · MSQ

Consider a coin-toss experiment where the probability of head showing up is $p$ . In the $i^{th}$ coin toss, let $X_{i} = 1$ if head appears, and $X_{i} = 0$ if tail appears. Consider $p = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ where $n$ is the total number of independent coin tosses. Which of the following statements is/are correct?

(A)
$E [p] = p$
(B)
$E [p] = \frac{p}{n}$
(C)
As $n$ increases, variance of $p$ decreases
(D)
Variance of $p$ does not depend on $n$

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Answer and solution

ANSWER

The correct options are (A) and (C): $E [p] = p$ and $Var (p) = \frac{p ( 1 - p )}{n}$ decreases as $n$ grows.

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

Each $X_{i}$ is a Bernoulli( $p$ ) variable, so $E [X_{i}] = p$ and $Var (X_{i}) = p (1 - p)$ .
Mean, by linearity of expectation: $E [p] = \frac{1}{n} \sum_{i = 1}^{n} E [X_{i}] = \frac{1}{n} \cdot n p = p$ . So $p$ is unbiased: (A) is true and (B) is false, because $\frac{p}{n}$ would only equal $p$ when $n = 1$ (or $p = 0$ ).
Variance: the tosses are independent, so variances add and $Var (c Y) = c^{2} Var (Y)$ . Hence $Var (p) = \frac{1}{n ^{2}} \sum_{i = 1}^{n} Var (X_{i}) = \frac{1}{n ^{2}} \cdot n p (1 - p) = \frac{p ( 1 - p )}{n}$ .
The variance has $n$ in the denominator, so for fixed $p$ with $0 < p < 1$ it decreases as $n$ increases: (C) is true, and (D) is false because the variance depends on $n$ .
Correct options: (A) and (C).

KEY IDEA

The sample proportion is the mean of $n$ independent Bernoulli( $p$ ) variables. Its mean is $p$ (unbiased), and its variance is $\frac{p ( 1 - p )}{n}$ , which shrinks as $n$ grows.

WATCH OUT FOR

Confusing the mean of the average with the mean of the sum. $E [\sum X_{i}] = n p$ , but dividing by $n$ gives $p$ , not $\frac{p}{n}$ . Another slip is forgetting that $\frac{1}{n}$ is squared in the variance, which gives $\frac{1}{n ^{2}} \cdot n = \frac{1}{n}$ instead of $\frac{1}{n} \cdot n = 1$ .

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GATE 2025 DA Q.54: solution and answer

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