Probability of the same position!

Question

If we have $n$ different boxes in a room, each with $x$ slots (positions $1,2,\ldots,x$) each and we have $(x-1)$ red balls and $1$ white ball in each box on different slots. What is the probability that more than $y$ balls are at the same slot position in the whole room?

Answer 1

You have $n$ boxes, each with $x$ slots. We can now ask for the probability to find exactly $y$ white balls at a given slot. Note that it doesn't matter which slot we choose. We have a probability of $p=1/x$ to find the white ball for each box. The problem can be described via the Binomial distribution. The probability to find exactly $y$ white balls in $n$ boxes with a probability of $p$ for each single box is \begin{align} f(y,n,p) = \begin{pmatrix} n \\ y\end{pmatrix}p^y(1-p)^{n-y} \end{align} If we want to know with which probability we get more than $y$ white balls, we simply add up all corresponding probabilities. \begin{align} f(\text{more than }y,n,p) = \sum_{j=y+1}^n f(j,n,p) \end{align}