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

Options

A. `(110)/(9)((2)/(3))^(10)`
B. `(9)/(110)((2)/(3))^(10)`
C. `(.^12C_(3))/(12^(3))xx2^(9)`
D. `(.^(12)C_(3))/(3^(12))`

Answer

A

Explanation

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)`