Finding a function of two random variables and calculating its pmf 
Suppose $X$ is a discrete random variable with countably infinitely
many outcomes, and $Y$ is a discrete random variable with a finite
number of outcomes. Give an example of $Z = f(X,Y)$ whose pmf is
calculable, and then calculate it. Your function $f$ must be a
function of both $X$ and $Y$, and both $X$ and $Y$ must be random.

I was thinking of letting $Y$ be a particular face that comes up on a roll of a die and letting $X$ be a geometric random variable. So $X$ is the number of trials until you get a certain number.
Is this a correct way of thinking about this? and if so, how would I calculate the pmf?
 A: I believe your way is correct. For specificity I would give the number $X$ is, such as it stops when the dice rolls a 1, and $X$ is the number of rolls previous to that 1, as well as that 1(so a geometric distribution, as you said). I would also say something like $Y$ is the number on the first roll. And then I would multiply them. For simplicity(from @user170231's comment), I would make the die for $X$ and $Y$ seperate so they are independent. It's your choice, though.
If you were to go with my specifications, then we could take some integer $n$(because the product of two integers is an integer). Then take all the factors of  $n$(I'll label them $g$) that are between 1 and 6 inclusive. For each of $g$, the probability $X(x = g)$ is $\frac{1}{6}$. Then $Y$ would be equal to $\frac{n}{g}$. The probability of this is the chances that $\frac{n}{g}$th roll is 1, and the others are not 1. So it would be $(\frac{5}{6})^{\frac{n}{g} - 1}* \frac{1}{6}$. So to get $Z(x = n)$ you would sum $\frac{1}{36}*(\frac{5}{6})^{\frac{n}{g} - 1}$ for all $g|n$ and $1 \leq g \leq 6$.
