# 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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