InterviewSolution
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InterviewSolution
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Twelve balls are placed in three boxes. The probability that the first box contains three balls isA. `(110)/(9)((2)/(3))^(10)`B. `(9)/(110)((2)/(3))^(10)`C. `(.^12C_(3))/(12^(3))xx2^(9)`D. `(.^(12)C_(3))/(3^(12))` |
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Answer» Correct Answer - A Since each ball can be placed in any one of the 3 boxes, therefore there are 3 ways in which a ball can be palced in any one of the three boxes. Thus, there are `3^(12)` ways in which 12 balls can be placed in 3 boxes. The number of ways in which 3 balls out of 12 can be put in the box is `.^(12)C_(3)`. The remaining 9 balls can be placed in 2 boxes in `2^(9)` ways. So, required probability is `(.^(12)C_(3))/(3^(12)) 2^(9) = (110)/(9) ((2)/(3))^(10)` |
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