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(D) Let $p(F) = q$ and $p(S) = p$. Given $p = 2q$. Since $p + q = 1$,we have $2q + q = 1$,which gives $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{

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$\frac{256}{729}$

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(D) Let $p(F) = q$ and $p(S) = p$. Given $p = 2q$. Since $p + q = 1$,we have $2q + q = 1$,which gives $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$. This is a binomial distribution with $n = 6$,$p = \frac{2}{3}$,and $q = \frac{1}{3}$. The probability of at least $5$ successes is $P(X \geq 5) = P(X = 5) + P(X = 6)$. Using the formula $P(X = k) = {^nC_k} p^k q^{n-k}$: $P(X = 5) = {^6C_5} \left(\frac{2}{3}ight)^5 \left(\frac{1}{3}ight)^1 = 6 	imes \frac{32}{243} 	imes \frac{1}{3} = \frac{192}{729}$. $P(X = 6) = {^6C_6} \left(\frac{2}{3}ight)^6 \left(\frac{1}{3}ight)^0 = 1 	imes \frac{64}{729} 	imes 1 = \frac{64}{729}$. Therefore,$P(X \geq 5) = \frac{192}{729} + \frac{64}{729} = \frac{256}{729}$.

An experiment succeeds twice as often as it fails. The probability of at least $5$ successes in the $6$ trials of this experiment is

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