📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!

    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and 8 white cubes and box ‘C’ contains 5 black and 9 white cubes. 1 of the boxes is selected at random and 1 cube is withdrawn from it. If the cube withdrawn is black, then find the probability that it is from box ‘B’.
    A 18/41 Correct Answer Incorrect Answer
    B 16/41 Correct Answer Incorrect Answer
    C 23/32 Correct Answer Incorrect Answer
    D 12/23 Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    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

    Practice Next
    More Probability Questions
    ask-question