Question:
Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is:
Correct Option: , 4
Solution:
$P$ (at least one hit the target) $=1-P$ (none of them hit the target)
$=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)$
$=1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{7}{8}=1-\frac{7}{32}=\frac{25}{32}$
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