## Precalculus (10th Edition)

$\frac{1}{36}$.
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. Hencem here we have a total of $6\cdot6=36$ outcomes. Out of the $36$ possible outcomes, the following have a sum of $12$: $[6,6]$. We know that $\text{probability}=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}=\frac{n(E)}{n(S)}.$ Hence, here probability of having a sum of $12$: $\text{probability}=\frac{1}{36}$.