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(A) For a binomial distribution,the mean is given by $\mu = np = 3$ and the variance is given by $\sigma^2 = npq = 2$.Dividing the variance by the mea

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$\left(\frac{2}{3}+\frac{1}{3}\right)^9$

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(A) For a binomial distribution,the mean is given by $\mu = np = 3$ and the variance is given by $\sigma^2 = npq = 2$.Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{3}$,which implies $q = \frac{2}{3}$.Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.Substituting $p = \frac{1}{3}$ into $np = 3$,we get $n 	imes \frac{1}{3} = 3$,which implies $n = 9$.The binomial distribution is given by $(q + p)^n = \left(\frac{2}{3} + \frac{1}{3}ight)^9$.

Find the binomial probability distribution whose mean is $3$ and variance is $2$.

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