Here are the probabilities.
$\displaystyle \begin{array}{*{20}c}
{P(2)} & = & {\frac{1}
{{36}}} \\
{P(4)} & = & {\frac{3}
{{36}}} \\
{P(6)} & = & {\frac{5}
{{36}}} \\
{P(8)} & = & {\frac{5}
{{36}}} \\
{P(10)} & = & {\frac{3}
{{36}}} \\
{P(12)} & = & {\frac{1}
{{36}}} \\
\end{array} $
The ways to get 10 are $\displaystyle \{(5,5),~(4,6),~(6,4)\}$ that is 3 out of 36.
Hello, AshleyT!
You've never rolled a pair of dice before, have you?
You have two dice, each has a choice of six numbers.
There are 36 possible outcomes:
. . $\displaystyle \begin{array}{cccccc}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}$
Now count the cases . . .
Sum is even: .$\displaystyle \begin{Bmatrix}(1,1) & (1,3) & (1,5) \\\
(2,2) & (2,4) & (2,6) \\ (3,1) & (3,3) & (3,5) \\ (4,2) & (4,4) & (4,6) \\ (5,1) & (5,3) & (5,5) \\ (6,2) & (6,4) & (6,6) \end{Bmatrix} \quad \text{ 18 cases}$
$\displaystyle \displaystyle{\text{Therefore: }\;P(\text{Sum is even}) \;=\;\frac{18}{36} \;=\;\frac{1}{2}$