<s>[INST] 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
$(3,1,1)$
$(3,2,2)$
$(3,3,3)$
$(3,4,4)$
$(3,5,5)$
$(3,6,6)$  
 [/INST] 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*}
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