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.