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
The probability of missing in one shot is 1/4. In n independent shots, “at least one hit” is the complement of missing every time:
\( \displaystyle P(\text{at least one hit}) = 1 - \left(\frac{1}{4}\right)^{n} \)
We need this to be more than 0.9, that is
\( \displaystyle \left(\frac{1}{4}\right)^{n} \lt 0.1 \)
\( \displaystyle n = 1:\ \frac{1}{4} = 0.25,\ \text{not less than } 0.1 \)
\( \displaystyle n = 2:\ \frac{1}{16} = 0.0625 \lt 0.1 \)
So he must fire at least 2 times (the probability is then 15/16, about 93.75%).