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(B) Let the probability of a person having a common cold be $p = \frac{10}{100} = \frac{1}{10}$.Then,the probability of a person not having a common c

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$0.9185$

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(B) Let the probability of a person having a common cold be $p = \frac{10}{100} = \frac{1}{10}$.Then,the probability of a person not having a common cold is $q = 1 - p = \frac{9}{10}$.We are selecting $n = 5$ persons. Let $X$ be the number of persons having a common cold. $X$ follows a binomial distribution $B(n, p) = B(5, 0.1)$.We need to find the probability that at most one person has a common cold,which is $P(X \le 1) = P(X = 0) + P(X = 1)$.Using the binomial probability formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:$P(X = 0) = {}^{5}C_{0} \left(\frac{1}{10}ight)^{0} \left(\frac{9}{10}ight)^{5} = 1 	imes 1 	imes \frac{59049}{100000} = 0.59049$.$P(X = 1) = {}^{5}C_{1} \left(\frac{1}{10}ight)^{1} \left(\frac{9}{10}ight)^{4} = 5 	imes \frac{1}{10} 	imes \frac{6561}{10000} = \frac{32805}{100000} = 0.32805$.Therefore,$P(X \le 1) = 0.59049 + 0.32805 = 0.91854$.Rounding to four decimal places,the probability is $0.9185$.

Out of $100$ people selected at random,$10$ have a common cold. If five persons are selected at random from the group,then the probability that at most one person will have a common cold is

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