If the mean and the variance of a binomial distribution are 4 and 3 respectively, then the probability of six successes is

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$^{16} C_{6} {(\frac{1}{4})}^{6} {(\frac{3}{4})}^{10}$

$^{16} C_{6} {(\frac{1}{4})}^{10} {(\frac{3}{4})}^{6}$

$^{12} C_{6} {(\frac{1}{4})}^{6} {(\frac{3}{4})}^{6}$

$^{12} C_{6} {(\frac{1}{4})}^{6} {(\frac{3}{4})}^{10}$

answer is A.

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Detailed Solution

Given np = 4 and npq = 3

$\Rightarrow \frac{n p q}{n p} = \frac{3}{4} \Rightarrow q = \frac{3}{4} \Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4} .$

Also, $n \times \frac{1}{4} = 4 \Rightarrow n = 16$ .

Now, $P (X = 6) =^{16} C_{6} p^{6} q^{10}$

$=^{16} C_{6} {(\frac{1}{4})}^{6} {(\frac{3}{4})}^{10}$

$∴$ Option (a) is the correct answer.

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