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(D) This is a binomial distribution problem where $n = 5$ and the probability of success $p = 0.2$. The probability of failure is $q = 1 - p = 1 - 0.2

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$0.0064$

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(D) This is a binomial distribution problem where $n = 5$ and the probability of success $p = 0.2$. The probability of failure is $q = 1 - p = 1 - 0.2 = 0.8$. We need to find the probability that exactly $x = 4$ patients survive. The formula for binomial probability is $P(X = x) = {}^{n}C_{x} p^{x} q^{n-x}$. Substituting the values: $P(X = 4) = {}^{5}C_{4} (0.2)^{4} (0.8)^{5-4}$. $P(X = 4) = 5 	imes (0.0016) 	imes (0.8)$. $P(X = 4) = 5 	imes 0.00128 = 0.0064$.

The probability that a person who undergoes a certain operation will survive is $0.2$. If $5$ patients undergo similar operations,then the probability that exactly four will survive is

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