Three fair dice are rolled one time. What is the probability of at least one $6$? I think the answer is $10/216$, but I am not sure. I did it by brute force though, and would like to know the background, as well as knowing if $10/216$ is correct.
Edit: now i think via brute force, the answer is 93/216. if you roll a 1 on any dice there are 3x6 = 18 ways of getting a 6 using any number of dice. similarly for the other numbers.. giving 5 x 18 = 90. now add in the three 6's which adds three more ways... giving 93 total. is the answer 93/216 correct? and if so is there a more elegant was of doing it than manually writing down all the combinations? :) 

Roll the three dice. Record the result as $(a,b,c)$ where $a$ is the number on the blue die, $b$ the number on the white, and $c$ the number on the red. All $6^3$ such sequences are equally likely.
There are $5^3$ sequences consisting of non-$6$, and therefore $216-125$ sequences with at least one $6$. Thus the probability of at least one $6$ is $\frac{216-125}{216}$.
Note that $216-125$ is $91$, not your $93$. You counted everything correctly, except there is only one sequence $(6,6,6)$, not three.
For this problem, I prefer to work directly with probabilities. The probability of a non-$6$ is $5/6$, so the probability of three of them in  a row is $(5/6)^3$, so the probability of at least one $6$ is $1-(5/6)^3$.
