Roland Cubes tosses a fair die three times. What is the probability that the face with six dots comes up at least two times?
I tried to solve this by combinations but obviously I have done it incorrectly.
There are a total of $216$ combinations, but there are only $\binom{3}{2}$+$\binom{3}{3}$ ways to achieve AT LEAST $2$ die with $6$ dots.
This gives me a probability of $\frac{4}{216}$ or $\frac{1}{54}$
What is wrong with my thought process?
On
This is a Bernoulli trial, with probability of success for each trial $p = \frac16$
$$\begin{align*}P(\text{At least 2 of 3 rolls of 6}) &= P(\text{exactly 2 of 3 rolls of six}) + P(\text{exactly 3 of 3 rolls of six})\\&={3 \choose 2}\left(\frac{1}{6}\right)^2\left(\frac56\right)+{3 \choose 3}\left(\frac{1}{6}\right)^3\\&=\frac{2}{27}\end{align*}$$
There are $\binom{3}{2}\times 5$ ways to roll exactly two sixes and there is one way to roll three sixes. Therefore the probability is $\frac{16}{216}=\frac{2}{27}$