Probability that the sum of two elements in a set is divisible by 3. Select 2 integers from the set $\{1,2,3,4,5,6,7,8\}$. What is the probability that their sum is divisible by 3? I am assuming this is without replacement although this is not explicitly stated. 
My answer: There are $\binom{8}{2}$ ways of selecting two elements. Only 6 of these pairs $(1,2), (4,5), (4,2), (7,8),(3,6), (2,7) $ are divisible by 3. Thus $P=6/\binom{8}{2}$. Is this correct?

We can make an explicit list and count. Or else note that there are $2$ numbers that have remainder $0$ on division by $3$ (list A) plus three numbers that have remainder $1$ (list B), plus three numbers that have remainder $2$ (list C). We must select $2$ numbers from list A ($1$ way) or $1$ each from lists B and C  ($3\times 3$ ways). The total is $10$. Divide by $\binom{8}{2}$.
