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Question

(C) Let $X$ be the number of components that survive the test. Here,$n = 4$ and $p = \frac{2}{3}$.Then $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.$X$

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$\frac{33}{3^4}$

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(C) Let $X$ be the number of components that survive the test. Here,$n = 4$ and $p = \frac{2}{3}$.Then $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.$X$ follows a binomial distribution $B(n, p)$,so $P(X = k) = \binom{n}{k} p^k q^{n-k}$.We need to find the probability that at most $2$ components survive,which is $P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$.$P(X = 0) = \binom{4}{0} (\frac{2}{3})^0 (\frac{1}{3})^4 = 1 	imes 1 	imes \frac{1}{81} = \frac{1}{81}$.$P(X = 1) = \binom{4}{1} (\frac{2}{3})^1 (\frac{1}{3})^3 = 4 	imes \frac{2}{3} 	imes \frac{1}{27} = \frac{8}{81}$.$P(X = 2) = \binom{4}{2} (\frac{2}{3})^2 (\frac{1}{3})^2 = 6 	imes \frac{4}{9} 	imes \frac{1}{9} = \frac{24}{81}$.Summing these probabilities: $P(X \le 2) = \frac{1}{81} + \frac{8}{81} + \frac{24}{81} = \frac{33}{81} = \frac{33}{3^4}$.

The probability that a certain kind of component will survive a given test is $\frac{2}{3}$. The probability that at most $2$ components out of $4$ tested will survive is

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