The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is

Solution :

Given that, for binomial distribution mean, np = 4 and variance, npq = 2.

\(\therefore\)  q = 1/2, but p + q = 1 \(\implies\) p = 1/2

and n \(\times\) \(1\over 2\) = 4 \(\implies\) n = 8

We know that,  P(X = r) = \(^nC_r p^r q^{n-r}\)

\(\therefore\)  P(X = 1) = \(^8C_1\)\(({1\over 2})^7\)\(({1\over 2})^1\)

= 1/32

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