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 or better of completely destroying the target can be

We have, the probability that the bomb strikes the target is p = 1/2. 
Let n be the number of bombs which should be dropped to ensure 99% chance or better of completely destroying the target. 
Then, the probability that out of n bombs at least two bombs strike the target is greater than 0.99. 
Let X denote the number of bombs striking the target. Then

$\mathrm{P}(\mathrm{X}=\mathrm{r}){=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{n}-\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{n}},\mathrm{r}=0,1,2,\dots ,\mathrm{n}$

We should have

$\begin{array}{l}\mathrm{P}(\mathrm{X}\ge 2)\ge 0.99\\ \Rightarrow \{1-\mathrm{P}(\mathrm{X}<2\left)\right\}\ge 0.99\\ \Rightarrow 1-\left\{\mathrm{P}\right(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1\left)\right\}\ge 0.99\\ \Rightarrow 1-\left\{(1+\mathrm{n})\frac{1}{2^{\mathrm{n}}}\right\}\ge 0.99\\ \Rightarrow 0.001\ge \frac{1+\mathrm{n}}{2^{\mathrm{n}}}\\ \Rightarrow 2^{\mathrm{n}}>100+100\mathrm{n}\Rightarrow \mathrm{n}\ge 11\end{array}$

Thus the minimum number of bombs is 11.