Before drawing a tree, let’s think for a second
We just need to prove whether the following is true for some value $x$ and $y$: $ Pr(X=i \cap Y=j) = Pr(X=i) \cdot Pr(Y=j) $
Let’s do it for X=2 and Y=2
$ |S| = 36 $
$X=2$ for the pair $ (1,1) $
$ Pr(X=2) = \frac{1}{36} $
$Y=5$ for the pairs $ (6,1), (1,6) $
$ Pr(Y=5) = \frac{2}{36} $
Now, let’s find the case of $ X=2 \cap Y=5 $
When we look at overlapping pairs in $X=2$ and $Y=5$, we see that there is no intersection
This means that $ Pr(X=2 \cap Y=5) = 0 $
$ Pr(X=2 \cap Y=5) = Pr(X=2) \cdot Pr(Y=5) $
Since the equation is false, X and Y are not independent.