<s>[INST] Rolling three dice and getting exactly two 4s - am I doing this right? 
Toss $3$ dice. What's the probability that a "4" will come up exactly
twice?

Since we need two of the dice to come up 4, and the third can be any number except 4 we can have the following options:
[4, 4, $\neq$ 4], [4, $\neq$ 4, 4], or [$\neq$ 4, 4, 4]. Right?
P of getting a 4 is $1/6$ and P of not getting a 4 is $5/6$ (assuming we have a $6$-sided die and each event is independent).
Let's take the first option: [4, 4, $\neq$ 4]
$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{5}{216}$  or $~0.02315$
And accounting for the other options is just multiplying by $3$ to get $0.06945$?
Did I do this correctly?
 [/INST] You are very close, but in order to solve it you need to use the binomial distribution: $$\binom{\textrm{No. of attempts}}{\textrm{number of successes}} \cdot  (\textrm{success chance})^{\textrm{No. of successes}} \cdot  (1 - \textrm{success chance})^{\textrm{No. of failures}}.$$ In your case, it is
$$\binom{3}{2} \cdot  \left(\frac{1}{6}\right)^{2} \cdot  \left(\frac{5}{6}\right)^{1} = 3 \cdot  \frac{1}{36} \cdot  \frac{5}{6} = \frac{15}{216} \approx 0.07=7\%$$
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