Question 1162983
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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.)
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We are given that EITHER A hits the target first time  at his 5-th shot,

                  OR     B hits the target first time at his 12-th shot,


but we do not know exactly who did it first, A or B.



        Also notice that this "OR" is <U>EXCLUSIVE</U>: either A or B, but not both.



The probability that A hits the target first time with his 5-th shoot  <U>AND</U>  B do not hit target with his 12 shoots is  


    P(A, 5+;  B, 12-) = 0.7*(1-0.7)^4 * (1-0.4)^12 = 0.7*0.3^4 * 0.6^12 = 0.00001234.



The probability that B hits the target first time with his  12-th shoot <U>AND</U>  A do not hit target with his 5 shoots  is


    P(B, 12+;  A, 5-) = 0.4*(1-0.4)^11 * (1-0.7)^5 = 0.4*0.6^11 * 0.3^5 = 0.00000353.

 


The probability that it is A who hits the target first is  


    P(A, 5+;  B, 12-) / ( P(A, 5+;  B, 12-) + P(B, 12+;  A,5-) ) = {{{0.00001234/(0.00001234 + 0.00000353)}}} = 0.7778.



The probability that it is B who hits the target first is  


    P(B, 12+;  A, 5-) / ( P(A, 5+;  B, 12-) + P(B, 12+;  A,5-) ) = {{{0.00001234/(0.00001234 + 0.00000353)}}} = 0.2222.



<U>ANSWER</U>.  The probability that A hits the target first is  0.7778.

         The probability that B hits the target first is  0.2222,
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Solved.