Rolling at Least 2 2's on 3 fair dice Say we roll three 6-sided dice. 
What is the probability that at least two of the faces are a 2?
The sample space here is 216. 
Rolling a 2 on one dice is 1/6. 
Rolling two 2s is 1/6 * 1/6 = 1/36 
The third dice is at least two 2s so the probability it isn't a 2 is 5/6 and the probability it is a 2 is 1/6. 
So do we multiply 1/36 * 1/6 * 5/6 to get the answer?
5/1296 is the probability of rolling at least two 2s on 3 fair dice?
 A: The following rolls are all possible:
$$2,2,1 \\ 2,1,2 \\ 1,2,2 \\ \vdots \\ 2,2,6 \\ 2,6,2 \\ 6,2,2$$
The probability of getting exactly two $2$'s followed by a non-$2$: $$\dfrac{1}{6}\cdot \dfrac{1}{6}\cdot \dfrac{5}{6}$$
The probability of getting a $2$, a non-$2$, then a $2$:
$$\dfrac{1}{6}\cdot \dfrac{5}{6}\cdot \dfrac{1}{6}$$
The probability of getting a non-$2$, then two $2$'s:
$$\dfrac{5}{6}\cdot \dfrac{1}{6}\cdot \dfrac{1}{6}$$
The probability of getting three $2$'s:
$$\dfrac{1}{6}\cdot \dfrac{1}{6}\cdot \dfrac{1}{6}$$
Adding this up, the probability of at least two $2$'s is:
$$\dfrac{3\cdot 5}{6^3}+\dfrac{1}{6^3} = \dfrac{2}{27}$$
A: You can also split it into two separate independent parts - one is rolling exactly two 2's on 3 dice, and the other is rolling exactly three 2's on 3 dice.
For the first part, we use Bernoulli scheme, as we want precisely 2 successes in 3 trials, with success probability of 1/6:
$$\binom{3}{2} \cdot \left(\dfrac{1}{6}\right)^2 \cdot \frac{5}{6} = \dfrac{5}{72}$$
For the second part, we want all 3 trials to roll 2:
$$\left(\dfrac{1}{6}\right)^3 = \dfrac{1}{216}$$
As these two cases above are independent of each other, we can add their probabilities - so the sum is
$$\dfrac{5}{72} + \dfrac{1}{216} = \dfrac{2}{27}$$
as in the other answer.
