On average, one out of 10 persons is busy. If | vedclass

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(D) Let $n = 6$ be the number of trials and $p = 1/10$ be the probability of a person being busy. Then $q = 1 - p = 9/10$.Using the binomial distribut

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$55/(10)^6$

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(D) Let $n = 6$ be the number of trials and $p = 1/10$ be the probability of a person being busy. Then $q = 1 - p = 9/10$.Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$:For at least $5$ persons to be busy, we need $P(X \ge 5) = P(X = 5) + P(X = 6)$.$P(X = 5) = \binom{6}{5} (1/10)^5 (9/10)^1 = 6 	imes (1/10^5) 	imes (9/10) = 54/10^6$.$P(X = 6) = \binom{6}{6} (1/10)^6 (9/10)^0 = 1 	imes (1/10^6) 	imes 1 = 1/10^6$.$P(X \ge 5) = 54/10^6 + 1/10^6 = 55/10^6$.

On average, one out of $10$ persons is busy. If $6$ persons are selected at random, then the probability that at least $5$ of them will be busy is...

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