The probability of a shooter hitting a target is 3/4. How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 90%?

2

The correct answer is Option (2) → 2

$p=\frac{3}{4},\; q=1-p=\frac{1}{4}.$

$P(\text{at least one hit in } n \text{ shots})=1-q^n.$

$1-\left(\frac{1}{4}\right)^n>0.9.$

$\left(\frac{1}{4}\right)^n<0.1.$

$n\ln\frac{1}{4}<\ln 0.1.$

$n>\frac{\ln 0.1}{\ln \frac{1}{4}}.$

$n>\frac{-2.3026}{-1.3863}=1.66.$

$n=2.$

$\text{Minimum number of shots } = 2.$