A proper approach (longer, but more edifying) would look something like this:
If a and b have the same remainder when divided by 3, then |a - b| will be divisible by 3. We can see this algebraically. Suppose a = 3*something + 2 and b also = 3*(something else) + 2. Then
a - b =
(3*something + 2) - (3*something else + 2) =
3*something - 3*something else =
a multiple of 3
This will work if the remainders are both 1 as well (or both 0, obviously), so we'll always get a multiple of 3.
The question then reduces to "What is the probability that a and b have the SAME remainder when divided by 3?" From here, I'd just count the possibilities.
remainder 0: 3, 6, 9, ..., 18 (6 options)
remainder 1: 1, 4, ..., 19 (7 options)
remainder 2: 2, 5, ..., 20 (7 options)
So there are (6*5)/2 ways of choosing two numbers with remainder 0, (7*6)/2 ways of choosing two numbers with remainder 1, and (7*6)/2 ways of choosing two numbers with remainder 2.
That gives us 15 + 21 + 21 = 57 ways of choosing two numbers with the same remainder. There are (20*19)/2, or 190 ways of picking ANY pair of these numbers, so
Prob(divisible by 3) = 57/190
and
Prob(not divisible by 3) = 1 - 57/190 = 133/190
