If the mean and variance of a binomial variate $X$ are $\frac{4}{3}, \frac{8}{9}$ respectively, then $P(X=2)=$

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Aicharya · Accepted Answer

For a binomial variate x,

From Eqs. $(i)$ and $(i i)$, we have $q=\frac{2}{3}, p=\frac{1}{3}$ and $n=4$ Now, $P(X=2)={ }^4 C_2 p^2 q^2$ $ =6\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^2=\frac{24}{81}=\frac{8}{27} $

If the mean and variance of a binomial variate $X$ are $\frac{4}{3}, \frac{8}{9}$ respectively, then $P(X=2)=$

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