The probability that a bomb will hit the targ | vedclass

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Question

(D) Let $p$ be the probability that a bomb hits the target, so $p = 0.8 = 4/5$.Let $q$ be the probability that a bomb misses the target, so $q = 1 - p

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$1 - (4/5)^6$

Vedclass · Answer

(D) Let $p$ be the probability that a bomb hits the target, so $p = 0.8 = 4/5$.Let $q$ be the probability that a bomb misses the target, so $q = 1 - p = 1 - 0.8 = 0.2 = 1/5$.We are dropping $n = 6$ bombs. Let $X$ be the number of bombs that miss the target.We want to find the probability that at least $1$ bomb misses the target, which is $P(X \ge 1)$.Using the complement rule, $P(X \ge 1) = 1 - P(X = 0)$.$P(X = 0)$ is the probability that no bombs miss the target, meaning all $6$ bombs hit the target.$P(X = 0) = p^6 = (4/5)^6$.Therefore, $P(X \ge 1) = 1 - (4/5)^6$.

The probability that a bomb will hit the target is $0.8$. Out of $6$ bombs dropped, what is the probability that at least $1$ will miss the target?

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