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A target is to be destroyed in a bombing exercise. There is 75% chance that any one bomb will strike the target. Assume that two direct hits are required to destroy the target completely. The minimum number of bombs that must be dropped in order that the chance of destroying the target is greater than or equal to 99% is

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answer is 6.

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Detailed Solution

$\begin{array}{l} P (x \geq 2) \geq \frac{99}{100} \Rightarrow 1 - P (x = 0) - P (x = 1) \geq \frac{99}{100} \\ \Rightarrow \frac{1}{100} \geq \frac{3^{n} + 1}{4^{n}} \Rightarrow n = 6 \end{array}$

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