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(B) Let $n = 4$ be the number of scans and $p = 0.1$ be the probability of detecting the plane in a single scan. The probability of not detecting the

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$0.0523$

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(B) Let $n = 4$ be the number of scans and $p = 0.1$ be the probability of detecting the plane in a single scan. The probability of not detecting the plane is $q = 1 - p = 0.9$. Using the binomial distribution,the probability of detecting the plane $X$ times in $n$ scans is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$. We need to find the probability of detecting the plane at least twice,which is $P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]$. $P(X = 0) = \binom{4}{0} (0.1)^0 (0.9)^4 = 1 	imes 1 	imes 0.6561 = 0.6561$. $P(X = 1) = \binom{4}{1} (0.1)^1 (0.9)^3 = 4 	imes 0.1 	imes 0.729 = 0.2916$. Therefore,$P(X \ge 2) = 1 - (0.6561 + 0.2916) = 1 - 0.9477 = 0.0523$.

$A$ radar system can detect an enemy plane in one out of ten consecutive scans. The probability that it can detect an enemy plane at least twice in four consecutive scans is

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