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Question

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 C 613
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(A and B hit the target2 shots hit 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

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