12 balls are distributed among 3 boxes, then the probability that the first box will contain 3 balls is

Here, since each ball can be placed in any one of the $3$ boxes, therefore there are $3$ ways in which a ball can be placed in any one of 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.

Hence, the required probability is $\frac{{}^{12}{C}_3\times 2^9}{3^{12}}$.