Question

A target is to be destroyed in a bombing exercise and there is a $75\%$ chance that a bomb will hit the target. Assuming that two direct hits are required to destroy the target completely, the minimum number of bombs to be chopped in order that the probability of destroying the target is not less than $99\%$, is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (C)

Let probability of hit the target $p=\frac{3}{4}$
So, probability to not hit the target $q=\frac{1}{4}$
Now, let ' $n$ number of bombs are dropped,
then probability of destroying the target $\ge \frac{99}{100}$ (given)
$\Rightarrow P(X\ge 2)\ge \frac{99}{100}$,
(here, $X$ represents the number of bombs dropped)
$\Rightarrow 1-[P(X=0)+P(X=1)]\ge \frac{99}{100}$
$\Rightarrow P(X=0)+P(X=1)\le 1-\frac{99}{100}$
$\Rightarrow {}^n{C}_0{p}^0{q}^n+{}^n{C}_1{p}^1{q}^{n-1}\le \frac{1}{100}$
$\Rightarrow \frac{1}{4^n}+n\frac{3}{4^n}\le \frac{1}{100}$
$\Rightarrow 4^n\ge 300n+100$
$\Rightarrow n\ge 6$