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%?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
$p=\frac{3}{4},\; q=1-p=\frac{1}{4}.$
$P(\text{at least one hit in } n \text{ shots})=1-q^n.$
$1-\left(\frac{1}{4}\right)^n>0.9.$
$\left(\frac{1}{4}\right)^n<0.1.$
$n\ln\frac{1}{4}<\ln 0.1.$
$n>\frac{\ln 0.1}{\ln \frac{1}{4}}.$
$n>\frac{-2.3026}{-1.3863}=1.66.$
$n=2.$
$\text{Minimum number of shots } = 2.$