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The probalities of A, B, C solving a problem are ⅓, ¼ and `.^(1)//_(6)` respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.

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Correct Answer - `31/72`
Given `P(A)=1/3, P(B)=1/4` and `P(C)=1/6`
`implies P(bar(A))=(1-1/3)=2/3, P(bar(B))=(1-1/4)=3/4` and `P(bar(C))=(1-1/6)=5/6`.
`:.` required probability
`=P` (exactly one of them solves the problem)
`=P[(A nn bar(B) nn bar(C)) os (bar(A) nn B nn bar(C)) or (bar(A) nn bar(B) nn C)]`
`= P(A nn bar(B) nn bar(C))+P (bar(A) nn B nnn bar(C))+P(bar(A) nn bar(B) nn C)`
`={P(A)xxP(bar(B))xxP(bar(C))}+{P(bar(A))xxP(B)xxP(bar(C))}+{P(bar(A))xxP(bar(B))xxP(C)}`.

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