## Precalculus (10th Edition)

$\dfrac{1}{6}$
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 $6\cdot6=36$ possible outcomes. Out of the $36$ possible outcomes, the following have a sum of $7$: $[6,1],[5,2],[4,3],[3,4],[2,5][1,6]$ We know that $\text{probability}=\dfrac{\text{number of favorable outcomes}}{\text{number of all outcomes}}=\dfrac{n(F)}{n(S)}.$ Hence, $\text{probability of having a sum of$7$}=\dfrac{6}{36}=\dfrac{1}{6}$