Three soldies, X, Y and Z have probabilities \(\frac{1}{3}, \frac{1}{5}\) and \(\frac{1}{4}\) respectively of hitting a target. If each of them fires once, find, correct to two decimal places, the probability that only one of them hits the target
Explanation
Note that P(X not hitting the target) = \(\frac{2}{3}\) P(Y not hittiing the target) = \(\frac{4}{5}\) and P(Z not hitting the target) = \(\frac{3}{4}\) So that, P(only one) = (\(\frac{1}{3} \times \frac{4}{5} \times \frac{3}{4}\)) + (\(\frac{2}{3} \times \frac{1}{5} \times \frac{3}{4}\)) + (\(\frac{2}{3} \times \frac{4}{5} \times \frac{1}{4}\)) = \(\frac{12}{60} + \frac{6}{60} = \frac{13}{30}\) = 0.43, correct to two decimal places.
