Help Joe find a+b

There are only $2$ ways to get a sum that is a multiple of $3$ :

Case 1: Both numbers are divisible by $3$ . There are $4$ numbers in the set that are divisible by $3$ , thus there are $\binom{4}{2} = \frac{4\times 3}{2} = 6$ total possibilities for this case.

Case 2: One of the numbers leaves a remainder of $1$ and the other leaves a remainder of $2$ when divided by $3$ . There are $5$ numbers for each of the groups. Thus, by the rule of product, there are $5 \times 5 = 25$ total possibilities for this case.

In total there are $\binom{14}{2} = 91$ ways to pick $2$ distinct numbers from the set. Therefore the probability that the sum is divisible by $3$ is $\frac{6 + 25}{91} = \frac{31}{91}$ . Hence $a + b = 31 + 91 = 122$ .