Two archers, A and B, are shooting at the same target. A hits the target 75% of the time and B hits the target 25% of the time. Now suppose that both archers shoot one arrow at the target at the same time.
If exactly one arrow hits the target, what is the probability that it was shot by A?

P ( A wins / only 1 hits ) = 0.9

Step-by-step explanation:

Given:-

- Probability to hit target by Archer A, pa = 0.75

- Probability to hit target by Archer B, pb = 0.25

Find:-

f exactly one arrow hits the target, what is the probability that it was shot by A?

Solution:-

- We are required to compute the conditional probability such that Archer A hit the target given that only one arrow hits.

           P ( A wins / only 1 hits ) = P ( A win & 1 hit ) / P ( 1 hit )

- The probability if the target is only hit by Archer A is that A hits and B misses since the success of hitting the target is independent for both archers we can write:

          P ( A wins & 1 hit ) = pa*(1-pb)

                                        = (0.75)*(1 - 0.25) = 0.75^2

                                        = 0.5625

- The probability of only one of the archer hits the target will constitute of two cases. Case I: A hits and B misses ; Case II : A misses and B hits.

         P ( Only 1 hits ) = pa*(1-pb) + (1-pa)*pb

                                  =(0.75)*(1 - 0.25) + (1 - 0.75)*(0.25)

                                  = 0.75^2 + 0.25^2

                                  = 0.625

- Plug in the respective probabilities in the conditional probability expression:

        P ( A wins / only 1 hits ) = 0.5625 / 0.625

                                                = 0.9

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step-by-step explanation: