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(C) For a binomial distribution,the mean is given by $np = \frac{4}{3}$  $(i)$ and the variance is given by $npq = \frac{8}{9}$  $(ii)$.Dividing equat

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$\frac{8}{27}$

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(C) For a binomial distribution,the mean is given by $np = \frac{4}{3}$  $(i)$ and the variance is given by $npq = \frac{8}{9}$  $(ii)$.Dividing equation $(ii)$ by equation $(i)$,we get:$\frac{npq}{np} = \frac{8/9}{4/3} \implies q = \frac{8}{9} 	imes \frac{3}{4} = \frac{2}{3}$.Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.Substituting $p = \frac{1}{3}$ into equation $(i)$:$n 	imes \frac{1}{3} = \frac{4}{3} \implies n = 4$.The probability mass function for a binomial distribution is $P(X=k) = {}^nC_k p^k q^{n-k}$.For $k=2$,$P(X=2) = {}^4C_2 	imes (\frac{1}{3})^2 	imes (\frac{2}{3})^{4-2}$.$P(X=2) = 6 	imes \frac{1}{9} 	imes \frac{4}{9} = \frac{24}{81} = \frac{8}{27}$.

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

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