There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order). We can solve for how many ways the $2$ numbers add up to an odd number and subtract the answer from $1$.  How to solve the problem: The probability of getting an odd number first is $\dfrac{4}{6}=\dfrac{2}{3}$. In order to make the sum odd, we must select an even number next. The probability of getting an even number is $\dfrac{2}{6}=\dfrac{1}{3}$. Now we multiply the two fractions: $\dfrac{2}{3}\times\dfrac{1}{3}=\dfrac{2}{9}$. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except switched, and by the Communitive Property of Multiplication, it does not matter if the equations are switched. Thus we do $\dfrac{2}{9}\times2=\dfrac{4}{9}$. This is the probability of getting an odd-number sum. In order to get the probability of getting an even number we do $1-\dfrac{4}{9}=\left (\text{C}\right )\dfrac{5}{9}$. 
The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? ． 
1. $\dfrac{4}{9}$ 
2. $\dfrac{1}{2}$ 
3. $\dfrac{5}{9}$ 
4. $\dfrac{3}{5}$ 
5. $\dfrac{2}{3}$ 