Question

The probability that $A$ hits a target is $\frac{3}{4}$. Then the minimum number of attempts $A$ should take so that the probability of hitting the target at least once is at least $0.99$, is

Answer 1

Let $p$ be the probability of hitting the target.
$p=\frac{3}{4}$
and $q=1-p=\frac{1}{4}$
$P(X=x)={}^n{C}_x{p}^x{q}^{n-x}$
We have to find the smallest $n$ such that
$P(X\ge 1)\ge 0.99\Rightarrow 1-P(X=0)\ge 0.99\Rightarrow 0.01\ge {}^n{C}_0{\left(\frac{3}{4}\right)}^0{\left(\frac{1}{4}\right)}^{n-0}\Rightarrow 4^n\ge 100$
Hence, the minimum value of $n$ is $4$.