I understand that 0+3 and 1+2 can get three. So why can't you do this

Prob of drawing a 0 Is 1/4 and prob of drawing a 3 is 1/3 so 1/3x1/4= 1/12

The same for the 1and 2 so 1/12 as well.

Adding 1/12+1/12=2/12=1/6

SM, I suppose you could.

In general, it's easier to list the permutations for a small universe problem like this. We have: (0,1), (0,2), (0,3), (1,0), (1,2), (1,3), (2,0), (2,1), (2,3), (3,0), (3,1), and (3,2). The pairs that have sum = 3 are (0,3), (1,2), (2,1), (3,0). That's 4 out of 12 possible combinations.

Your answer is wrong, because you need to add the combination of drawing 3 first, then 0. And add the combination of drawing 2 first then 1. then it's 1/12 + 1/12 + 1/12 + 1/12.

Remember, drawing 2 without replacement is the same as drawing 1 at a time until you draw 2, without replacement.