Two fair six-sided dice are rolled. What is the probability that the sum of the two dice is at least 10?

There are $6\cdot6=36$ equally likely outcomes. The sums that are at least $10$ are: sum 10 with $(4,6),(5,5),(6,4)$; sum 11 with $(5,6),(6,5)$; and sum 12 with $(6,6)$. That is $3+2+1=6$ outcomes, so the probability is $\frac{6}{36}=\frac{1}{6}$, choice A. Choice B ($\frac{1}{12}=\frac{3}{36}$) counts only sums of 11 and 12. Choice C ($\frac{1}{9}=\frac{4}{36}$) leaves out two favorable outcomes. Choice D ($\frac{5}{36}$) misses one outcome, usually $(5,5)$ or one of the ordered pairs.