Four persons can hit a target correctly
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:

  1. (1) $\frac{25}{192}$

  2. (2) $\frac{7}{32}$

  3. (3) $\frac{1}{192}$

  4. (4) $\frac{25}{32}$


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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