There are $6\cdot6 = 36$ possible outcomes for the pair of tosses. A 3 can be obtained in two ways: either one die shows 1 and the other shows 2. There are 6 possible doubles, one for each number from 1 to 6. However, double 1's and double 2's have been counted twice (in the 3-total category and in the doubles category), so we must correct for this. So, there are $2+6-2 = 6$ ways to roll a 3 or doubles, meaning that the probability is $\dfrac{6}{36} = \boxed{\dfrac{1}{6}}$.
