There are three boxes ‘A’, ‘B’ and ‘C’. Box

Solution

Since, there are three bags, therefore, probability of selecting any one of the boxes = (1/3) i.e., P(A) = P(B) = P(C) = 1/3 Now, Total cubes in box ‘A’ = 6 + 12 = 18 Number of black cubes in box ‘A’ = 6 Let 1 box has been chosen and 1 black cube is chosen from it, the required probability = (1/3) × (6/18) = (1/9) Total cubes in box ‘B’ = 6 + 8 = 14 Number of black cubes in box ‘B’ = 6 Let 2 box has been chosen and 1 black cube is chosen from it, the required probability = (1/3) × (6/14) = (1/7) Total cubes in box ‘C’ = 5 + 9 = 14 Number of black cubes in box ‘C’ = 5 Let 3 box has been chosen and 1 black cube is chosen from it, the required probability = (1/3) × (5/14) = (5/42) Therefore, probability that box ‘B’ is chosen, and the cube chosen is black. = {(1/7)} ÷ {(1/9) + (1/7) + (5/42)} = (1/7) ÷ (41/126) = 18/41