InterviewSolution
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InterviewSolution
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Twelve balls are distributed among three boxes, find the probabilitythat the first box will contains three balls.A. `(2^(9))/(3^(12))`B. `(.^(12)C_(3)xx2^(9))/(3^(12))`C. `(.^(12)C_(3)xx2^(12))/(3^(12))`D. `(.^(12)C_(3))/(12^(3))` |
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Answer» Correct Answer - B Since each ball can be put into any one of the three boxes. So, the total number of ways in which 12 balls can be put into three boxes is `3^(12)`. Out of 12 balls, 3 balls can be chosen in `.^(12)C_(3)` ways. Now, remaining 9 balls can be put in the remaining 2 boxes in `2^(9)` ways. So, the total number of ways in which 3 balls are put in the first box and the remaining in other two boxes is `.^(12)C_(3)xx2^(9)`. Hence, required probability `=(.^(12)C_(3)xx2^(9))/(3^(12))` |
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