Question

Two marksmen shoot at a target simultaneously. Shooter A is known to have a 70% chance of hitting the target on any attempt. Person B has 40% accuracy. After the target is hit for the first time, it is revealed that A shot 5 shots while B shot 12. What is the probability that it was A who hit the target? What is the probability that person B hit the target? (Assume that accuracies of the shots remain the same and are independent of other shots by either person.)

Answer 1

A shot 5 shots while B shot 12.then

p(A)= 5/17 and p(B) = 12/17

then probability of hit the target will be

5/17*0.70+12/17*0.40

(3.5+4.8)/17

0.48823

now probability that A hits target = 5/17*0.70 /[0.48823]

0.20588/0.48823

0.4217

and probability that B hits target= 12/17*0.40/[0.48823]

0.28235/0.48823

0.5783

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