Imagine there are 'k' boxes. All boxes contain 'n' unique colored marbles. Assuming that we select 'm' marbles out of 'm' different boxes (one marble from each box), what is the probability that all of the selected marbles have the same color? (e.g. imagine we have five boxes (k = 5), in each box there is a red, blue, purple, and yellow marble (n = 4). Now we select 3 marbles from 3 different boxes (m = 3). What is the probability that all the 3 selected marbles have the same color)?
My solution:
The total possible ways of choosing 'm' marbles out of 'n' unique colored marbles from 'k' different boxes is:
A = ${n \choose m}\times{n \choose k}$
The desirable event can happen this many times:
B = $\left(\frac{1}{n^m}\right)$
So the answer will be: ${B\over A}$
I am not sure tho, so please correct me if I am wrong!
In fact, it is as good as choosing a marble each out of $m$ boxes regardless of $k$.
Hence, the probability you are looking for is simply $$\binom n1 \left(\frac1n\right)^m=n^{1-m}$$
Or to follow your thought process, your $A$ should have been
$$A=\binom km \cdot n^m $$
Then for $B$, you have to consider
the number of ways the $m$ boxes are chosen out of $k$,
the number of ways the first color may be chosen, and
the number of ways the subsequent $(m-1)$ balls are of the same color
So your $B$ is in fact $$B=\binom km \binom n1 (1)^{m-1}$$
So your probability similarly becomes $$\frac BA=n^{1-m}$$ Notice that the factor $\binom km$ vanishes. Hence, it may in fact be ignored from the start.