If I roll two fair dice, the probability that I would get at least one 6 would be.... $11$ out of $36$? I got this by writing down the number of possible outcomes ($36$) and then counting how many of the pairs had a $6$ in them: $(1,6)$, $(2,6)$, $(3,6)$, $(4,6)$, $(5,6)$, $(6,6)$, $(6,5)$, $(6,4)$, $(6,3)$, $(6,2)$, $(6,1)$. Is this correct?
That's right. The easier approach would be to calculate the chance of not rolling a $6$ - that's just $\frac56$ for the first die, and $\frac56$ for the second die, so by the product rule (as the events are independent), the probability is $\frac56 \cdot \frac56 = \frac{25}{36}$. 

Then the probability of rolling a $6$ is $1$ minus the probability of not rolling a $6$, which we just calculated: so it is $1-\frac{25}{36}=\frac{11}{36}$.

If calculating the probability of an event not occurring to calculate the probability of it occurring feels weird to you, you might want to read up on complementary events. The idea here is that the product rule can sometimes make probabilities smaller when that wouldn't make any sense --- in our example, if you had just multiplied $\frac16 \cdot \frac16 = \frac1{36}$, then that would've been obviously wrong. This is because when we work with probability, we're dealing with quantities in the $[0,1]$ interval, so multiplication usually makes things smaller, rather than larger.
The chance of getting a $6$ with the first dice is $1/6$ and the chance with the second dice is $1/6$ therfore the chance of getting a $6$ with either dice is $1/6 + 1/6 = 1/3$....not $11/36$!!!