TITLE: $12$ Identical balls can be placed into $3$ identical boxes,
QUESTION [6 upvotes]: $12$ Identical balls can be placed into $3$ identical boxes, Then find probability that one of 
the boxes contain exactly  $3$ balls.
$\bf{My\; Try::}$ First we select $1$ bag out of $3$ and then put
$3$ balls into that bag and then put remaining balls into $2$ bag.
Which can be done by $\displaystyle \binom {3}{1}\times 1 \times \binom{1}{1}\times 1 = 3$
bcz here balls are identical . So we can arrange by only one ways.
But my answer is wrong. 
plz help me , How can I get correct answer.
Thanks

REPLY [5 votes]: Suppose the experiment consists of randomly assigning $12$ balls to $3$ boxes. There are $3^{12}$ ways to do so. We want to find the probability that (at least) one of the boxes contains exactly $3$ balls.
Let us number the boxes $1$, $2$ and $3$, and define $E_i$ as the event that box # $i$ contains exactly $3$ balls. We want to find the probability of the event $E_1 \cup E_2 \cup E_3$.
$\Pr(E_1 \cup E_2 \cup E_3) $
$= \Pr(E_1) + \Pr(E_2) + \Pr(E_3) - \Pr(E_1\cap E_2) - \Pr(E_1\cap E_3) - \Pr(E_2\cap E_3) + \Pr(E_1\cap E_2 \cap E_3)$
$= 3\Pr(E_1) - 3\Pr(E_1\cap E_2) + \Pr(E_1\cap E_2 \cap E_3) \ \ \ \ \color{blue}{\text{(By symmetry)}}$
We will now find $\Pr(E_1)$, $\Pr(E_1\cap E_2)$, and $\Pr(E_1\cap E_2 \cap E_3)$.
$\displaystyle \Pr(E_1)  = {12\choose 3}\left(\frac{2^{9}}{3^{12}}\right)$
$\displaystyle \Pr(E_1\cap E_2) = {12\choose 3} {9\choose 3}\left(\frac{1^{6}}{3^{12}}\right)$
$\displaystyle \Pr(E_1\cap E_2 \cap E_3) = 0$
Therefore,
$\Pr(E_1 \cup E_2 \cup E_3) $
$= 3\Pr(E_1) - 3\Pr(E_1\cap E_2) + \Pr(E_1\cap E_2 \cap E_3) $
$=  \displaystyle 3\times {12\choose 3}\left(\frac{2^{9}}{3^{12}}\right) - 3 \times {12\choose 3} {9\choose 3}\left(\frac{1^{6}}{3^{12}}\right)$
$\approx 0.5315$
Here is the code to find the above value. You will also find the computation of the required probability using simulation method in the code:
http://www.codeskulptor.org/#user42_g9uaFBlARwEdnL6_1.py