Question

The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. Find the least number of bombs required so that the probability of the bridge being destroyed is greater than 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=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.