## Algebra 2 (1st Edition)

$\frac{11}{18}$
We have $6\times6=36$ possible outcomes. Out of these, the possible ordered pairs are: $\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(5,1),(5,2),(6,1),(6,6)\}$. Thus $22$ satisfy the criteria, and the probability is: $\frac{22}{36}=\frac{11}{18}$