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(D) The probability mass function of a binomial distribution is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.Given $n=4$ and $P(X=0) =

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$\frac{1}{81}$

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(D) The probability mass function of a binomial distribution is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.Given $n=4$ and $P(X=0) = \frac{16}{81}$.Substituting $k=0$ into the formula: $P(X=0) = \binom{4}{0} p^0 q^{4-0} = q^4$.So,$q^4 = \frac{16}{81} = (\frac{2}{3})^4$.This implies $q = \frac{2}{3}$.Since $p+q=1$,we have $p = 1 - \frac{2}{3} = \frac{1}{3}$.Now,we need to find $P(X=4)$:$P(X=4) = \binom{4}{4} p^4 q^{4-4} = 1 	imes (\frac{1}{3})^4 	imes 1 = \frac{1}{81}$.Thus,the correct option is $D$.

In a binomial distribution,if $n=4$ and $P(X=0)=\frac{16}{81}$,then $P(X=4)=$

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