This is a problem I came up with to help me better understand other challenges I have been facing.
Suppose that $X \sim U(1,200)$ and $Y \sim U(1,50)$, both being discrete random variables. Now, suppose that we draw 5 realizations $x \in X$, not allowing repeated values: if a value is drawn more then once we simply repeat until getting a new value. We do exactly the same with $Y$, drawing 5 realizations $y \in Y$. What is the probability that we encounter exactly 3 identical values when comparing the 5 realizations of $X$ and the realizations of $Y$?
You are doing sampling without replacement. Suppose we pick the 5 numbers from $Y$ first; call this set $A$. There are $\binom{200}{5}$ ways to pick the 5 numbers from $X$, all of which we assume are equally likely. If these include exactly 3 numbers from $A$, then there are $\binom{5}{3}$ ways to pick those 3 numbers and $\binom{195}{2}$ ways to pick the remaining 2 numbers. So the probability that exactly 3 numbers from $X$ match numbers in $A$ is $$\frac{\binom{5}{3} \binom{195}{2}}{\binom{200}{5}} \approx 7.460 \times 10^{-5}$$