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(A) Let $n$ be the number of bombs dropped. The probability of a bomb hitting the target is $p = \frac{1}{2}$,and the probability of missing is $q = 1

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) Let $n$ be the number of bombs dropped. The probability of a bomb hitting the target is $p = \frac{1}{2}$,and the probability of missing is $q = 1 - p = \frac{1}{2}$.The target is destroyed if there are at least $2$ hits. Let $X$ be the number of hits. We want $P(X \geq 2) \geq 0.99$.This is equivalent to $1 - P(X Using the binomial distribution,$P(X=k) = {}^{n}C_{k} p^k q^{n-k}$.$1 - [{}^{n}C_{0} (\frac{1}{2})^n + {}^{n}C_{1} (\frac{1}{2})^n] \geq 0.99$$1 - \frac{1 + n}{2^n} \geq 0.99$$\frac{1 + n}{2^n} \leq 0.01 = \frac{1}{100}$$2^n \geq 100(n + 1)$.Testing values for $n$:For $n=10$: $2^{10} = 1024$,$100(11) = 1100$. $1024 \geq 1100$ is false.For $n=11$: $2^{11} = 2048$,$100(12) = 1200$. $2048 \geq 1200$ is true.Thus,the minimum number of bombs required is $11$.

In a bombing attack,there is a $50 \%$ chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs that must be dropped to ensure that there is at least a $99 \%$ chance of completely destroying the target is:

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