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If the mean and variance of $a$ binomial distribution are $4$ and $3$ respectively, then $P (X = 1)$ is

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$\frac{1}{16}$

${(\frac{3}{4})}^{16}$

$4 {(\frac{3}{4})}^{15}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

$n p = 4, n p q = 3 \Rightarrow q = 3 / 4 \Rightarrow p = 1 / 4$

Thus, $n = 16$

$P (1) =^{16} C_{1} (\frac{1}{4}) {(\frac{3}{4})}^{15} = 4 {(\frac{3}{4})}^{15}$

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