Question

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. Select the correct answer from above options

Answer 1

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)}`.