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If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the product of one of two successes is

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$\frac{33}{2^{32}}$

$\frac{33}{2^{29}}$

$\frac{33}{2^{28}}$

$\frac{33}{2^{27}}$

answer is C.

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Detailed Solution

Mean + variance $= 24 \Rightarrow n p + n p q = 24$ …………. (i) np.npq =128 ……(ii)

$\Rightarrow p (x = 1) + p (x = 2) = 32_{C_{1}} p q^{31} + 32_{C_{2}} p^{2} q^{30} = \frac{33}{2^{28}}$

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