Twelve balls are distributed among three boxes, find the probability that the first box will contains three balls.

Options

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))`

Answer

B

Explanation

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))`