Determine the probability that at least $1$ three turns up when $3$ dice are rolled. The answer is $91/216$, I understand how $216$ is there because of $6\cdot 6\cdot 6$, $91$ though I don't what they got. 
I'm so confused on listing the outcomes, I feel like with $3$ die, there are far too many outcomes for me to write out. I believe it'd be something like
$(3,1,1)$
$(3,2,2)$
$(3,3,3)$
$(3,4,4)$
$(3,5,5)$
$(3,6,6)$  
 A: Well, consider probability that no three turns up.   That all three die show only one of $\{1,2,3,4,5\}$?  It is $$\mathsf P(N_3=0)~=~\frac{5^3}{6^3}$$
That's the complement event, so $$\mathsf P(N_3\geq 1) ~=~ 1 -\frac {5^3}{6^3} = \frac{91}{216}$$
That is all.
A: Well, using some notation, I will call $X$ the number of ones rolled in 3 rolls.
Then notice that we have $n = 3$ independent trials with probability $p = 1/6$ of success(rolling a one). Hence, $X$ follows a binomial distribution.
$X\sim\text{Binomial}(3, 1/6).$
Using the complement, we derive an easier solution;
$$P(X\geq 1) = 1-P(X = 0) = 1-\binom{3}{0}(1/6)^0(5/6)^3 = \frac{91}{216}.$$
Using a counting argument, we see that there are $6^3$ triplets possible, the number of triplets that do not contain a three  is $5^3$, and hence the probability of interest is
\begin{align*}
P(X\geq 1) &= 1-P(X = 0) \\
&= 1-\frac{\text{# of triplets with no threes}}{\text{# of all triplets}}\\
&= 1-\frac{5^3}{6^3} \\
&= \frac{6^3-5^3}{6^3} \\
&= \frac{91}{216}.
\end{align*}
A: If you're interested in counting the outcomes, consider the following.  Suppose only the first die has a 3.  The 2nd die can take any of the values 1, 2, 4, 5, or 6, and the third die can take any value from 1, 2, 4, 5, or 6.  Therefore, there are $5\times5$ = 25 possible outcomes.
Similarly, there are 25 possible outcomes in which the 2nd die is the only one with a 3, and 25 possible outcomes where where only the third die is a 3.  So, there are a total of 75 possible dice rolls where exactly one die has a 3.
Now, lets look at the number of possible outcomes where two of the die have 3's showing.  Suppose the first and second die have 3's showing.  The third die can be 1, 2 4, 5, or 6, so there are 5 possibilities.
The same can be said if the 1st and 3rd dice are 3's and if the 2nd and 3rd dice are 3's.  So, there are 15 possible outcomes where two of the dice show 3.
Finally, there is one possibility for all three dice showing a 3.
75 + 15 + 1 = 91.
Graham Kemp's answer is the easiest way to calculate this.  For example, imagine if you were trying to calculate this for 100 dice.  You could count the way I just did and it would probably take hours.  Or you can take Graham's approach and be done pretty quickly.  
You are basically asking how many outcomes are there where no 3's are rolled.  There are 5 possibilities for each die, so there are $5\times5\times5$ = 125 possible outcomes where no 3 is rolled.  This means that there are 216 - 125 = 91 outcomes where at least one 3 is rolled.
