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.
Text Solution
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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)`
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