Question

The probability of a bomb hitting a bridge is $\frac{1}{2}$ and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge beeing destroyed is greater then 0.9, is

• A. 8
• B. 7
• C. 6
• D. 9

Answer 1

The correct option is A 8

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 parameter n and $p=\frac{1}{2}$
Now, $P(X\ge 2)>0.9\Rightarrow 1-P(X<2)>0.9$
$\Rightarrow P(X=0)+P(X=1)<0.1$
$\Rightarrow {}^n{C}_0{\left(\frac{1}{2}\right)}^n+{}^n{C}_1{\left(\frac{1}{2}\right)}^{n-1}\left(\frac{1}{2}\right)<0.1\Rightarrow 10(n+1)<2^n$
This gives $n\ge 8$