Q:

Choosing a ball randomly from each urn

I am stucked at this combinatorics/probability problem:

There are 10 urns, each contains 8 balls numbered 1,2,...,8
If we randomly choose 1 ball from each urn, what is the probability that the total number of 1's will be exactly 3 and the total number of 4's will be exactly 2 ?

I have tried several times to solve this problem but I got incorrect probabilities.
Thanks for any hint/help.

A:

Let $X$ denote the number of $1$'s, let $Y$ denote the number of $4$'s and let let $Z$ denote the number of other numbers. Then:$$P(X=3\wedge Y=2\wedge Z=5)=\frac{10!}{3!2!5!}\left(\frac18\right)^3\left(\frac18\right)^2\left(\frac68\right)^5$$
First a selection concerning: from wich urn are the $1$'s, $4$'s and others taken? There are $\frac{10!}{3!2!5!}$ possibilities for that. Then the question: by a fixed selection what is the probability that this indeed happens? It is $\left(\frac18\right)^3\left(\frac18\right)^2\left(\frac68\right)^5$.

