The probability of a shooter of hitting the target is \(\displaystyle\frac{1}{4}\). The minimum number of fire needed so that the probability of hitting the target at least once is greater than \(\displaystyle\frac{7}{16}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
The probability of missing in one shot is \(\frac{3}{4}\). In n independent shots, use the complement of “no hit”:
\( \displaystyle P(\text{at least one hit}) = 1 - \left(\frac{3}{4}\right)^n \)
We need
\( \displaystyle 1 - \left(\frac{3}{4}\right)^n \gt \frac{7}{16} \)
\( \displaystyle \left(\frac{3}{4}\right)^n \lt \frac{9}{16} = \left(\frac{3}{4}\right)^2 \)
Since \(\frac{3}{4} \lt 1\), this needs n > 2, so the smallest n is 3. For n = 2 the probability is exactly \(1 - \frac{9}{16} = \frac{7}{16}\), which is not greater than \(\frac{7}{16}\). For n = 3 it is \(1 - \frac{27}{64} = \frac{37}{64}\), which is more than \(\frac{7}{16} = \frac{28}{64}\).