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Official PYQ · 2006Q21Probability: Random Variables and DistributionsMCQ1 mark

GATE CS 2006 Q21 — Probability: Random Variables and Distributions

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For each element in a set of size 2n2n, an unbiased coin is tossed. The 2n2n coin tosses are independent. An element is chosen if the corresponding coin toss was a head. The probability that exactly nn elements are chosen is

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Answer: A

Let XX be the random variable representing the number of elements chosen from the set of size 2n2n.

Since each element is chosen independently based on the toss of an unbiased coin, the probability of choosing any single element is p=12p = \frac{1}{2}, and the probability of not choosing it is q=1−p=12q = 1 - p = \frac{1}{2}.

Thus, XX follows a Binomial distribution with parameters N=2nN = 2n and p=12p = \frac{1}{2}:

X∼Binomial(2n,12)X \sim \text{Binomial}\left(2n, \frac{1}{2}\right)

The probability that exactly nn elements are chosen is given by the probability mass function of the Binomial distribution:

P(X=n)=(2nn)pnq2n−nP(X = n) = \binom{2n}{n} p^n q^{2n - n}

Substituting p=12p = \frac{1}{2} and q=12q = \frac{1}{2}:

P(X=n)=(2nn)(12)n(12)nP(X = n) = \binom{2n}{n} \left(\frac{1}{2}\right)^n \left(\frac{1}{2}\right)^n P(X=n)=(2nn)(12)2nP(X = n) = \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} P(X=n)=(2nn)22n=2nCn4nP(X = n) = \frac{\binom{2n}{n}}{2^{2n}} = \frac{^{2n}\mathrm{C}_n}{4^n}

Therefore, the correct option is A.

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