The probability that a missile hits a target | vedclass

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Question

(A) Let $X$ be the number of successful hits,where $X \sim B(n, p)$ with $p = 0.75 = \frac{3}{4}$ and $q = 1 - p = 0.25 = \frac{1}{4}$.We require at l

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Let $X$ be the number of successful hits,where $X \sim B(n, p)$ with $p = 0.75 = \frac{3}{4}$ and $q = 1 - p = 0.25 = \frac{1}{4}$.We require at least $3$ successful hits to destroy the target,so $P(X \geq 3) \geq 0.95$.This is equivalent to $1 - P(X The probability mass function is $P(X=r) = {}^{n}C_{r} (\frac{3}{4})^r (\frac{1}{4})^{n-r}$.Substituting the values: ${}^{n}C_{0} (\frac{1}{4})^n + {}^{n}C_{1} (\frac{3}{4}) (\frac{1}{4})^{n-1} + {}^{n}C_{2} (\frac{3}{4})^2 (\frac{1}{4})^{n-2} \leq 0.05$.$\frac{1}{4^n} [1 + 3n + \frac{9n(n-1)}{2}] \leq 0.05$.$1 + 3n + 4.5n^2 - 4.5n \leq 0.05 	imes 4^n$.$4.5n^2 - 1.5n + 1 \leq 0.05 	imes 4^n$.For $n=5$: $4.5(25) - 1.5(5) + 1 = 112.5 - 7.5 + 1 = 106 \leq 0.05(1024) = 51.2$ (False).For $n=6$: $4.5(36) - 1.5(6) + 1 = 162 - 9 + 1 = 154 \leq 0.05(4096) = 204.8$ (True).Thus,the minimum number of missiles required is $6$.

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