A target is to be destroyed in a bombing exer | vedclass

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(C) Let the probability of a bomb hitting the target be $p = \frac{3}{4}$.Thus, the probability of a bomb missing the target is $q = 1 - p = \frac{1}{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let the probability of a bomb hitting the target be $p = \frac{3}{4}$.Thus, the probability of a bomb missing the target is $q = 1 - p = \frac{1}{4}$.Let $n$ be the number of bombs dropped. 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 $1 - [P(X = 0) + P(X = 1)] \geq 0.99$.$P(X = 0) + P(X = 1) \leq 0.01 = \frac{1}{100}$.Using the binomial distribution: $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.${}^{n}C_{0} (\frac{3}{4})^{0} (\frac{1}{4})^{n} + {}^{n}C_{1} (\frac{3}{4})^{1} (\frac{1}{4})^{n-1} \leq \frac{1}{100}$.$\frac{1}{4^{n}} + n \cdot \frac{3}{4} \cdot \frac{4}{4^{n}} \leq \frac{1}{100}$.$\frac{1 + 3n}{4^{n}} \leq \frac{1}{100} \Rightarrow 4^{n} \geq 100(3n + 1) = 300n + 100$.For $n = 5$: $4^{5} = 1024$ and $300(5) + 100 = 1600$. ($1024 For $n = 6$: $4^{6} = 4096$ and $300(6) + 100 = 1900$. ($4096 \geq 1900$, true).Thus, the minimum number of bombs is $6$.

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