jmc

We will use the inclusion-exclusion principle for this problem. The probability of getting a 5 on the first roll is obviously $\frac{1}{6}$, as it is on the second roll. So, the probability of getting a 5 on at least one of the rolls would appear to be $2\cdot \frac{1}{6}=\frac{1}{3}$. But this is not quite right. We have double counted the case of rolling a 5 twice. In that instance, we have included it in both the count for a 5 on the first roll and the second roll, when it only should have been included once overall. So, our answer is $\frac{1}{3}-\frac{1}{36}=\boxed{\frac{11}{36}}$.