Question
There are three boxes ‘A’, ‘B’ and ‘C’. Box
‘A’ contains 6 black and 12 white cubes, box ‘B’ contains 6black, and 8 white cubes and box ‘C’ contains 5 black and 9 white cubes. 1 of the boxes is selected at random and 1 cube is withdrawn from it. If the cube withdrawn is black, then find the probability that it is from box ‘B’.Solution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
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