probabilty doubt

[jack]

12 balls are distributed at random among 3 boxes. What is the prob that the first box will contain 3 balls?

[Arushi :))]

Since there are 12 balls,
each ball has 3 boxes as options.
1 st ball can be put in 3 ways .
similarly all the balls can be distributed in 3 ways.
So total number of ways = (3)^12
If we want first box to contain 3 balls, then  
it can be done in 12C3 ways.
now corresponding to each of these ways,
remaining 9 balls can be distributed in 2 ways. ( we cant put them in box 1)
So favorable ways are 12C3 x (2)^9
P =12C3 x (2)^9/ (3)^12= 0.0235

[jack]

thank you so much