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(C) For a binomial distribution $X \sim B(n, p)$, the probability mass function is given by $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.Here, $n = 6$ an

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$X = 3$

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(C) For a binomial distribution $X \sim B(n, p)$, the probability mass function is given by $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.Here, $n = 6$ and $p = 1/2$.Since $p = 1/2$, the distribution is symmetric.For a symmetric binomial distribution, the maximum probability occurs at the mean, which is $E[X] = np = 6 	imes (1/2) = 3$.Alternatively, calculating for $k=3$: $P(X=3) = \binom{6}{3} (1/2)^3 (1/2)^3 = 20 	imes (1/64) = 20/64$.For $k=2$ or $k=4$: $P(X=2) = P(X=4) = \binom{6}{2} (1/2)^6 = 15 	imes (1/64) = 15/64$.Since $20/64 > 15/64$, the maximum probability occurs at $X = 3$.

Let $X \sim B(6, 1/2)$. Then the maximum probability occurs at

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