There are $n+1$ containers, marked $C_0, C_1, C_2, ..., C_n$, each with $n$ balls inside. We know that container $C_i$ has exactly $i$ white balls, and $n-i$ black balls. The probability of picking a specific container is $\frac{1}{n+1}$. We randomly pick a container, and take out $r$ balls. Of which $k$ are white, and $r-k$ are black. What is the probability we took these balls out of container $C_m$.
The answer I received from peers is: $$ P(C_m) = \frac{\binom{m}{k} \binom{n-m}{r-k}}{\binom{n+1}{r+1}}$$
However, this solution fails for edge cases. For example:
Let $n=2$, $r=k=1$, and we want to check the probability for container $C_0$. Container $C_0$ has exactly 2 black balls and 0 white balls. We take out 1 ball, of which 1 is white. What's the probability the container is container $C_0$? Logically, we can conclude - $0$. Yet algebraically this doesn't make sense, because $\binom{m}{k}$ fails (You can't choose 1 out of 0).
So is the answer I received from my peers correct?
For natural numbers $m$ and $k$, $\binom mk$ is usually defined to be $0$ when $m\lt k$. See for instance Wikipedia and MathWorld.