Probability that $m$ balls will fall into first box

Question

Suppose we have $n$ balls that are randomly distributed into $N$ boxes. Find the probability that $m$ balls will fall into the first box. Assume that all $N^m$ arrangements are equally likely.

Attempt:

First, we notice that for the first box, we have $n$ choices, and for the second box we also have $n$ choices, and so on. Thus, we have $n^N$ ways to place the balls into the boxes. Pick $m$ balls out of the total $n$ balls, that gives ${n \choose m}$. In how many ways can these $m$ balls go into the first box? Well, in just ${1 \choose 1 }= 1 $ ways. Thus,

$$ P = \frac{ {n \choose m } }{n^N } $$

Is this correct?

Answer 1

The number of ways to choose the $m$ items in the first box is $\binom{n}{m}$ and the number of ways to put the remaining $n-m$ items into the remaining $N-1$ boxes is $(N-1)^{n-m}$; therefore, the number of ways to put $m$ items into the first box is $$ \binom{n}{m}(N-1)^{n-m} $$ So the probability is $$ \frac{\binom{n}{m}(N-1)^{n-m}}{N^n}=\frac{\binom{n}{m}}{N^m}\left(1-\frac1N\right)^{n-m} $$ Note that the Binomial Theorem guarantees that $$ \begin{align} \sum_{m=0}^n\binom{n}{m}(N-1)^{n-m} &=((N-1)+1)^n\\ &=N^n \end{align} $$

Answer 2

If $X$ is the number of balls that fall into the first box, then $P(X=m)$ follows the binomial distribution, where $p=1/N$.

Answer 3

If we're just concerned with the first box, and all other boxes are equally likely, it's just a binomial with $n$ trials, for which $p=1/N$.

$$\Pr(m=k) \,=\, {n \choose k}\,p^{k}(1-p)^{n-k} \,=\, {n \choose k}\,N^{-k}\left(1-\frac1N\right)^{n-k}.$$

Answer 4

I think that the problem is different if the balls are indistinguishable. If we have 2 balls and 3 boxes we have 6 ways.

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In this case we have $ \frac{(2+(3-1)!)}{2! (3-1)!}=6 $