The probability of hitting a target by three marksmen are 1/2, 1/3 and 1/4. Then find the probabi9lity that one and only one of them will hit the target when they fire simultaneously.

Explanation

We have `P(A)=1/2,P(B)=1/3,P(C)=1/4`
Clearly, the probabilities of A,B, C hitting the target the independent. P(exactly one of them hitting the target)
P(A is hatting and B, C missing the target)
+P(B is hatting and A,C missing the target)
+ P(C is hitting and A,B missing the target)
`=P(AnnB'nnC')+P(A'nnBnnC')+P(A'nnB'nnC')`
`=P(A)P(B')P(C')+P(A')P(B)P(C')+P(A')P(B')P(C')`
`=(1)/(2).(2)/(3).(3)/(4)+(1)/(2).(1)/(3).(3)/(4)+(1)/(2).(2)/(3).(1)/(4)=(11)/(24)`