Question

In a precision bombing attack, there is $50\%$ chance that any one bomb will strike the target. Two precise 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

• A. $12$
• B. $11$
• C. $10$
• D. $13$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $12$
B $11$
D $13$

Let the bomb strike the target is $p=\frac{1}{2}$
Probability of failure, $q=\frac{1}{2}$
Let $n$ numbers of bomb is dropped to ensure $99\%$ chances or better destroying the target.
Then, the probability that at least $2$ bombs strike the target is greater than $0.99$.
Let $X$ denotes the number of bombs striking the target.

​​Then, $P(X=r)={}^n{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{n-r}={}^n{C}_r{\left(\frac{1}{2}\right)}^n,r=0,1,2,\cdots ,n$

According to the question,
$P(X\ge 2)\ge 0.99\Rightarrow 1-\left[P\right(X=0)+P(X=1\left)\right]\ge 0.99\Rightarrow 1-(n+1){\left(\frac{1}{2}\right)}^n\ge 0.99\Rightarrow \frac{1+n}{2^n}\le 0.01\Rightarrow 2^n\ge 100+100n\therefore n\ge 11$