In a precision bombing attack, there is a 50% chance that any one bomb will strike the target two direct hits are required to destroy the target completely. The number of bombs which should be dropped to give a 99% chance to better of completely destroying the target can be

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Detailed Solution

Let x be the no. of bombs which strike the target of n.
Then x $~$ B(n, p) where p = probability that one
bomb fired strike the target $= \frac{50}{100} = \frac{1}{2}$
To find n such that $P (x \geq 2) > \frac{99}{100}$
$\begin{array}{l} \Rightarrow 1 - \frac{99}{100} > P (x = 0 or 1) \\ \Rightarrow \frac{1}{100} > [^{n} C_{0} +^{n} C_{1}] {(\frac{1}{2})}^{n} \\ \Rightarrow 2^{n} > (n + 1) \cdot 100 \Rightarrow n \geq 11 (by inspection) \end{array}$

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