Answer:
i) 93/125
ii) 57/125
Step-by-step explanation:
If we assume each shooter gets one shot, that shot hits the target with probability equal to the average hit rate, and all shots are independent, then we seem to have ...
p(A) = probability A hits the target = 4/5
p(B) = probability B hits the target = 3/5
p(C) = probability C hits the target = 3/5
ii) Then the probability exactly two shots will hit the target is ...
... p(2) = p(A)·p(B)·p(C') +p(A)·p(B')·p(C) +p(A')·p(B)·p(C)
... = 4·3·2/125 +4·2·3/125 +1·3·3/125 = (24 +24 +9)/125
... p(2) = 57/125
i) The probabiilty the target will be damaged is the probability that 2 or 3 shots will hit the target. The probability that 3 shots will hit is ...
... p(3) = p(A)·p(B)·p(C) = 4·3·3/125 = 36/125
So ...
... p(damage) = p(2) +p(3) = (57+36)/125
... p(damage) = 93/125
