Twelve balls are distribute among three boxes. The probability that the first box contains three balls is a.`(110)/9(2/3)^(10)` b. `(9)/110(2/3)^(10)` c. `((12)C_3)/(12^3)xx2^9` d. `((12)C_3)/(3^(12))`

Answer

`.^(12)C_(3)xx (2^(9))/(3^(12))`

Explanation

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 for the first box. 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 balls in other two boxes is `.^(12)C_(3) xx 2^(9)`. Hence, required probability is `.^(12)C_(3) xx 2^(9)//3^(12)`.