Question

The probability of a bomb hitting a bridge is $1/2$. Two direct hits are needed to destroy it. The number of bombs required so that the probability of the bridge being destroyed is greater that $0.9$

Answer 1

Let n be the least number of bombs required and X the number of bombs that hit the bridge.

Then X follows a binomial distribution with parameters n and $p=\frac{1}{2}$.

$\Rightarrow q=1-p=\frac{1}{2}$

Now $P(X\ge 2)>0.9\Rightarrow 1-P(X<2)>0.9$

$\Rightarrow P(X=0)+P(X=1)<0.1$

${\Rightarrow }^n{C}_0{\left(\frac{1}{2}\right)}^n{+}^n{C}_1{\left(\frac{1}{2}\right)}^{n-1}\left(\frac{1}{2}\right)<\frac{1}{10}$

$\Rightarrow \frac{n+1}{2^n}<\frac{1}{10}\Rightarrow 10(n+1)<2^n$

By trial and error, we get $n\ge 7$.

Thus, the least value of $n$ is $7$.