Question

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

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

Answer 1

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

$12$
$11$
$13$

The probability of success in one strike is $p=\frac{1}{2}$.
$\Rightarrow$ Probability of failure $=q=\frac{1}{2}$
Now, probability of $r$ successes $P(x=r){=}^n{C}_r.{\left(\frac{1}{2}\right)}^r.{\left(\frac{1}{2}\right)}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^n$
According to the given condition, $P(x\ge 2)\ge 0.99$
$\Rightarrow 1-P(x<2)\ge 0.99$
$\Rightarrow 1-P(x=0)-P(x=1)\ge 0.99$
$\Rightarrow 1-0.99\ge \frac{1+n}{2^n}$
$\Rightarrow 2^n\ge 100+100n$
For $n=10$, $2^n<100+100n$
For $n=11,12,13$,
$2^n>100+100n$.