'a' can hit a target 4 times with 5 shots.'b' can hit 3 times with 5 shots and 'c' can hit 3 times with 5 shots as well. if two hits will damage the target (each person aims a target) then find probability that: i)target will be damaged. ii)two shots hit the target [ans with appropriate explanation will get the i got same ans for both question which is 57/125 and i am not sure about my ans]

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

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

where r the answers

step-by-step explanation: