Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose
[tex]1\leq x\leq 3,\quad 1\leq y \leq 5[/tex]
We have [tex]P(X=x)=\frac{1}{3}[/tex] and [tex]P(Y=y)=\frac{1}{5}[/tex], because the dice are fair.
Now we use the assumption of independence to claim that
[tex] P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}[/tex]
Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:
This implies that the probabilities of the outcomes of [tex]W=X+Y[/tex] are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5
