Answer:
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
