$12$ 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
Best Answer
Take the simplest model: One after the other of the balls is thrown at random into one of the boxes $a$, $b$, $c$. There are $3^{12}$ different possible histories for that, all of them equiprobable. The number of histories leading to a particular final content $(r,s,t)$ of the three boxes is the coefficient of the term $a^r b^s c^t$ in the expansion of $(a+b+c)^{12}$, i.e., is given by ${12!\over r!\>s!\>t!}$.
There is the question whether "one of the boxes" means "at least one of the boxes", or "exactly one of the boxes". Since one sentence later they talk about "exactly three balls" my working hypothesis is that "at least one of the boxes" is meant.
For the probability in question we have to consider the contents $$(3,9,0), (3,8,1), (3,7,2), (3,5,4)$$ each of them in six orders, and $(6,3,3)$ in three orders. The total number $N$ of "admissible" histories is therefore given by $$N=6{12!\over3!}\left({1\over9!}+{1\over8!}+{1\over 7!\>2!}+{1\over 5!\>4!}\right)+3{12!\over 6!\>3!\>3!}=282\,480\ ,$$ and the required probability $P$ is $$P={N\over 3^{12}}\doteq0.531536\ .$$