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(A) Let $X$ be the number of times the radar detects the plane in $n = 4$ scans. This follows a binomial distribution $B(n, p)$ where $n = 4$ and $p =

Vedclass · Accepted Answer

$0.9477$

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(A) Let $X$ be the number of times the radar detects the plane in $n = 4$ scans. This follows a binomial distribution $B(n, p)$ where $n = 4$ and $p = 0.1$ (probability of detection). The probability of not detecting the plane is $q = 1 - p = 0.9$. We want to find the probability that it cannot detect the plane at least two times,which is equivalent to $1 - P(	ext{detecting the plane } 2, 3, 	ext{ or } 4 	ext{ times})$. $P(X \ge 2) = P(X=2) + P(X=3) + P(X=4)$. $P(X=2) = {}^{4}C_{2} (0.1)^{2} (0.9)^{2} = 6 	imes 0.01 	imes 0.81 = 0.0486$. $P(X=3) = {}^{4}C_{3} (0.1)^{3} (0.9)^{1} = 4 	imes 0.001 	imes 0.9 = 0.0036$. $P(X=4) = {}^{4}C_{4} (0.1)^{4} (0.9)^{0} = 1 	imes 0.0001 	imes 1 = 0.0001$. $P(X \ge 2) = 0.0486 + 0.0036 + 0.0001 = 0.0523$. The required probability is $1 - P(X \ge 2) = 1 - 0.0523 = 0.9477$.

$A$ radar system can detect an enemy plane in one out of $10$ consecutive scans. The probability that it cannot detect an enemy plane at least two times in four consecutive scans is:

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