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, ...

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 

Answer Details

The probability that only one of the soldiers hits the target can be calculated by finding the sum of the probability that each soldier hits the target and the others miss.

Starting with soldier X, the probability that he hits the target and the others miss is: \(\frac{1}{3} \times \frac{2}{5} \times \frac{3}{4} = \frac{1}{15}\)

Similarly, the probability that soldier Y hits the target and the others miss is: \(\frac{2}{5} \times \frac{2}{3} \times \frac{3}{4} = \frac{4}{75}\)

And the probability that soldier Z hits the target and the others miss is: \(\frac{2}{3} \times \frac{4}{5} \times \frac{1}{4} = \frac{2}{45}\)

So the total probability that only one of the soldiers hits the target is: \(\frac{1}{15} + \frac{4}{75} + \frac{2}{45} = \frac{11}{45}\)

Therefore, the probability that only one of the soldiers hits the target is 0.24 or 24%.