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(C) Let $p$ be the probability of success and $q$ be the probability of failure. Given that the experiment succeeds twice as often as it fails,we have

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$\frac{16}{27}$

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(C) Let $p$ be the probability of success and $q$ be the probability of failure. Given that the experiment succeeds twice as often as it fails,we have $p = 2q$. Since $p + q = 1$,we get $2q + q = 1$,which implies $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$. We need to find the probability of at least $3$ successes in $n = 4$ trials. Using the binomial distribution formula $P(X = k) = {^nC_k} p^k q^{n-k}$,the required probability is $P(X \ge 3) = P(X = 3) + P(X = 4)$. $P(X = 3) = {^4C_3} \left(\frac{2}{3}ight)^3 \left(\frac{1}{3}ight)^1 = 4 	imes \frac{8}{27} 	imes \frac{1}{3} = \frac{32}{81}$. $P(X = 4) = {^4C_4} \left(\frac{2}{3}ight)^4 \left(\frac{1}{3}ight)^0 = 1 	imes \frac{16}{81} 	imes 1 = \frac{16}{81}$. Therefore,$P(X \ge 3) = \frac{32}{81} + \frac{16}{81} = \frac{48}{81} = \frac{16}{27}$.

An experiment succeeds twice as often as it fails. Find the probability that in $4$ trials there will be at least three successes.

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