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
Solution:
$\mathrm{P}(\mathrm{H})=\frac{1}{2}$
$\mathrm{P}(\overline{\mathrm{H}})=\frac{1}{2}$
Let total 'n' bomb are required to destroy the target
$1-{ }^{n} C_{n}\left(\frac{1}{2}\right)^{n}-{ }^{n} C_{1}\left(\frac{1}{2}\right)^{n} \geq \frac{99}{100}$
$1-\frac{1}{2^{n}}-\frac{n}{2^{n}} \geq \frac{99}{100}$
$\frac{1}{100} \geq \frac{n+1}{2^{n}}$
Now check for value of $\mathrm{n}$
$\mathrm{n}=11$