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
Let him fire n times. By the complement rule, P(at least one hit) = 1 − P(no hit):
\( \displaystyle 1 - \left(\frac{1}{2}\right)^n \gt 0.9 \)
\( \displaystyle \left(\frac{1}{2}\right)^n \lt 0.1 \)
\(2^n \gt 10\)
Since \(2^3 = 8 \lt 10\) and \(2^4 = 16 \gt 10\), the least n is 4. With 4 shots, P(at least one hit) = \(1 - \frac{1}{16} = 0.9375\).