A can hit a target 4 times out of 5 shots, B thrice in 4 shots and C twice in 3 shots, independent of each other. They fire a volley. Two shots hit the target. Then, the probability that it is C who has missed the target, is
A
15
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B
1330
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C
613
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D
25
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Solution
The correct option is C613 Given that
The chance of hitting by A,P(A)=45 and P(¯¯¯¯A)=15
The chance of hitting by B,P(B)=34 and P(¯¯¯¯B)=14
The chance of hitting by C,P(C)=23 and P(¯¯¯¯C)=13
P(AandBhit the target2shotshit the target) =P(A)⋅P(B)⋅P(¯¯¯¯C)P(A)⋅P(B)⋅P(¯¯¯¯C)+P(A)⋅P(¯¯¯¯B)⋅P(C)+P(¯¯¯¯A)⋅P(B)⋅P(C) =45×34×1345×34×13+45×14×23+15×34×23 =1212+8+6=613