Answer
$\frac{1}{18}.$
Work Step by Step
According to the Multiplication Principle, if we have $p$ selections for the first choice, $q$ selections for the second choice, then we have $p\cdot q$ different ways of selections.
Hence, here we have a total of $6\cdot6=36$ outcomes. Out of the $36$ possible outcomes, the following have a sum of $3$: $[2,1],[1,2]$.
We know that $\text{probability}=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}=\frac{n(E)}{n(S)}.$
Hence, the probability of having a sum of $3$: $\text{probability}=\frac{2}{36}=\frac{1}{18}.$