The probability of a man hitting a target is 1/2. How many 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) → 4
$\text{Let the probability of hitting the target in one shot be }p=\frac{1}{2}.$
$\text{Probability of missing in one shot: }q=1-p=\frac{1}{2}.$
$\text{Let the man fire }n\text{ times. Probability of missing all }n\text{ shots: }q^n=(\frac{1}{2})^n.$
$\text{Probability of hitting at least once: }1-q^n>0.9.$
$1-(\frac{1}{2})^n>0.9\ \Rightarrow\ (\frac{1}{2})^n<0.1.$
$\text{Take logarithm: }n\log\frac{1}{2}<\log 0.1\ \Rightarrow\ n>\frac{\log 0.1}{\log \frac{1}{2}}.$
$\log 0.1=-1\ (\text{base 10}),\ \log \frac{1}{2}=-0.3010.$
$n> \frac{-1}{-0.3010} \approx 3.32.$
$\text{Since }n\text{ must be an integer, }n=4.$
$\text{Answer: The man must fire at least 4 times.}$