Question

In a bombing attack, there is $50\%$ chance that a bomb will hit the target.

At least two independent hits are required to destroy the target completely.

Then the minimum number of bombs, that must be dropped to ensure that there is at least $99\%$ chance of completely destroying the target, is

Answer 1

Find the required number of bombs that must be dropped.

It is given that there is $50\%$ chance that a bomb will hit the target.

So the probability of success is $p=\frac{1}{2}$ and the probability of failure is also $q=\frac{1}{2}$.

At least two independent bombs must hit the target so that it is destroyed completely.

Let us assume that $n$ number of bombs are dropped to ensure that there is at least $99\%$ chance of completely destroying the target.

Let $x$ denotes number of bombs striking the target in $n$ chances. Then

$$\begin{gathered} P(X=x)={}^{n}C_x{q}^{n-x}{p}^n \\ ={}^{n}C_x{\left(\frac{1}{2}\right)}^{n-x}{\left(\frac{1}{2}\right)}^n \end{gathered}$$

Now, according to question the inequality is $P(x\ge 2)\ge 0.99$.

Now, $P\left(x\ge 2\right)$ can be found as:

$\begin{array}{rcl}P\left(x\ge 2\right)& =& 1-P(x=0)-P(x=1)\\ & =& 1-{}^n{C}_0{\left(\frac{1}{2}\right)}^{n-0}{\left(\frac{1}{2}\right)}^0-{}^n{C}_1{\left(\frac{1}{2}\right)}^{n-1}{\left(\frac{1}{2}\right)}^1\\ & =& 1-1\times {\left(\frac{1}{2}\right)}^n\times 1-n{\left(\frac{1}{2}\right)}^{n-1+1}\\ & =& 1-{\left(\frac{1}{2}\right)}^n-n{\left(\frac{1}{2}\right)}^n\\ & =& 1-\left(1+n\right){\left(\frac{1}{2}\right)}^n\end{array}$

So the inequality can be solved as:

$$\begin{gathered} P(x\ge 2)\ge 0.99 \\ \Rightarrow \begin{array}{l}1\end{array}-\left(1+n\right){\left(\frac{1}{2}\right)}^n\begin{array}{rcl}& \ge & \end{array}0.99 \\ \begin{array}{rcl}& \Rightarrow & \end{array}-\begin{array}{l}\frac{n+1}{2^n}\end{array}\begin{array}{rcl}& \ge & \end{array}-0.01 \\ \Rightarrow \begin{array}{l}\frac{n+1}{2^n}\end{array}\begin{array}{rcl}& \le & \end{array}0.01 \\ \Rightarrow \begin{array}{l}\frac{n+1}{0.01}\end{array}\begin{array}{rcl}& \le & \end{array}\begin{array}{l}{2}^n\end{array}\begin{array}{l}\end{array} \\ \Rightarrow {\begin{array}{l}2\end{array}}^n\begin{array}{rcl}& \ge & \end{array}100n+100 \\ \Rightarrow \begin{array}{l}n\end{array}\begin{array}{rcl}& \ge & \end{array}\begin{array}{l}11\end{array} \end{gathered}$$

So the minimum number of bombs, that must be dropped to ensure that there is at least $99\%$ chance of completely destroying the target, is $11$.

Hence, the answer is $11$.