Let $n, k, m \in \mathbb{N}$ with $k < n$, $m \leq n$ and $n$ mod $m$ = 0. $n$ is a number of balls where $k$ are red, the rest is blue. $m$ is a number of bins.
Now the balls will be randomly distributed so that each bin gets exactly $n/m$ balls.
What is $P(X:=$ "Each bin has at most one red ball".$)$
I came up with this solution:
Number all places and balls so we have $n!$ possible combinations to distribute the balls. For the legal assignements we have:
- Looking only at red balls: The first can choose $n$ places, the second $n – n/m$ since the places in the bin of the first ball are now illegal. The third $n – 2 \cdot n/m$ and so on untill the $k$-th ball can choose $n – (k-1) \cdot n/m$.
- Now the remaining blue balls can choose any of the remaining places. So this is $(n-k)!$
in total I get
$P(X) = \frac{n \cdot (n-1 \cdot n/m) \cdots (n – (k-1) \cdot m/n)\cdot(n-k)!}{n!}$
I tested my result and it also has the nice property to yield $P(X) = 0$ if $k>m$ and it seems to be right after computing the result for a few samples. But I still don't know if this is correct or if there is maybe a better solution, since I think the approach where I number all bins and balls is not necessary?
Best Answer
For convenience set $n=sm$ where $s$ is a positive integer.
Partition set $\{1,\dots,n\}$ in $m$ subsets of the form $\{(i-1)s+1,\dots is\}$ for $i=1,\dots, m$.
There are $\binom{n}{k}=\binom{sm}{k}$ ways to select $k$ integers.
Under the extra condition that the $k$ chosen integers are in distinct subsets that belong to the partition there are $\binom{m}{k}\times s^k$ ways to select $k$ integers.
That leads to a probability:$$\frac{\binom{m}{k}\times s^k}{\binom{n}{k}}$$