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

Explanation

Let us consider the following events:
`E_(1):A hits B,P(E_(1))=2//3`
`{:(E_(1):A hits B,P(E_(1))=2//3,),(E_(2):B hits A,P(E_(2))=1//2,),(E_(3):C hits A,P(E_(3))=1//3,):}`
E: A is his
`thereforeP(E)=P(E_(2)uuE_(3))`
`=1-P(barE_(2)nnbarE_(3))`
`=1-P(barE_(2).PbarE_(3))`
`=1-(1)/(2).(2)/(3)=2/3`
Now, `P((E_(2)nnbarE_(3))//E)=(P(E_(3)nnbarE_(3)))/(P(E))`
`[because P(E_(2)nnbarEnnE)=P(E_(2)nnbarE_(3)),` i.e., B hits A and A is hit = B hits A]
`=(P(E_(2))P(barE_(3)))/(P(E))`
`=(1/2xx2/3)/(2/3)=1/2`