The probability that a missile hits a target successfully is 0 . 75 . In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is not less than 0 .…

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The probability that a missile hits a target successfully is 0 . 75 . In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is not less than 0 . 95 , is ___________

Aicharya · Accepted Answer

Let the number of missiles be $n$, $P$ (of destroying the target) $\geq 0.95$ $C_{3}^{n} \left(\frac{3}{4}ight)^{3} \left(\frac{1}{4}ight)^{n-3} + C_{4}^{n} \left(\frac{3}{4}ight)^{4} \left(\frac{1}{4}ight)^{n-4} + \cdots + C_{n}^{n} \left(\frac{3}{4}ight)^{n} \geq 0.95$ $1 - \left\{C_{0}^{n} \left(\frac{3}{4}ight)^{0} \left(\frac{1}{4}ight)^{n} + C_{1}^{n} \left(\frac{3}{4}ight)^{1} \left(\frac{1}{4}ight)^{n-1} + C_{2}^{n} \left(\frac{3}{4}ight)^{2} \left(\frac{1}{4}ight)^{n-2}ight\} \geq 0.95$ $1 - \frac{95}{100} \geq \frac{1}{4^{n}} + \frac{3n}{4^{n}} + \frac{n(n-1)}{2} \cdot \frac{9}{4^{n}}$ $\frac{4^{n}}{20} \geq \frac{2 + 6n + 9n^{2} - 9n}{2}$ $\frac{2^{2n-1}}{5} \geq 9n^{2} - 3n + 2$ $2^{2n-1} \geq 5(9n^{2} - 3n + 2)$ $n = 3 \Rightarrow 32 \geq 5 	imes 74$ (not true) $n = 4 \Rightarrow 128 \geq 5 	imes 134$ (not true) $n = 5 \Rightarrow 512 \geq 5 	imes 212$ (not true) $n = 6 \Rightarrow 2048 \geq 5 	imes 308$ (true) So, $n = 6$.

The probability that a missile hits a target successfully is 0 . 75 . In order to destroy the target…

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