A fair 6 sided dice is rolled 4 times. What is the probability that at least 3 of the numbers will be either 1 or 6? I'd really love a sanity check here as I walk through what I believe is the solution.
Total possible outcomes = $6^4 = 1296$
Possible combinations of 3 rolls being either 1 or 6 = $({}_4C_3)\cdot2 = (4)\cdot2 = 8$
Also take into account all 1's and all 6's = $1 + 1 = 2$
Answer = $\frac{8+2}{ 1296} = \frac{10}{1296} = \mathbf{\frac{5}{648}} $
Really appreciate the help! :)

I would use a binomial probability:
$$ \begin{align*}
P(\text{at least 3 are 1 or 6}) &= P(\text{exactly 3 are 1 or 6}) + P(\text{exactly 4 are 1 or 6}) \\
&= {}_4C_3 \left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^1 + {}_4C_4 \left(\dfrac{2}{6}\right)^4\left(\dfrac{4}{6}\right)^0 \\
&= 4 \left(\dfrac{1}{3}\right)^3\left(\dfrac{2}{3}\right) + \left(\dfrac{1}{3}\right)^4 \\
&= \dfrac{8+1}{81} \\
&= \dfrac{1}{9} \\
\end{align*} $$
