Four persons can hit a target correctly with probabilities respectively. If all hit at the target independently, then the probability that the target would be hit, is

Engineering

Mathematics

Addition Theorem on Probability

Basic Definitions Related to Probability

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

$\frac{7}{32}$

$\frac{1}{192}$

$\frac{25}{192}$

$\frac{25}{32}$

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Let persons be A, B, C, D

P (Hit) = 1 – P (none of them hits)

$= 1 - P (\bar{A} \cap \bar{B} \cap \bar{C} \cap \bar{D})$

$= 1 - P (\bar{A}) . P (\bar{B}) . P (\bar{C}) . P (\bar{D})$

$= 1 - \frac{1}{2} . \frac{2}{3} . \frac{3}{4} . \frac{7}{8}$

$= \frac{25}{32}$