Give him a chance, he needs it

The probability of missing the target $n$ times in a row is $P = (1 - 3/4)^n$ . So, $(1 - P)$ is the probability of hitting the target at least once after $n$ shots.

We are asked when $(1 - P)$ is greater than $99/100$ . Solving for $n$ :

$1 - (\frac{1}{4})^n > \frac{99}{100}$

$\implies\frac{1}{100} > \frac{1}{4^n}$

$\implies 4^n > 100$ .

Because $4^3 = 64 < 100$ and $4^4 = 256 > 100$ , the minimum number of times Sanjeet must fire is $n = 4$ .