2 of 3 dice are selected randomly and thrown. What is the probability that one of the dice shows 6 1 red die with faces labelled 1, 2, 3, 4, 5, 6.
2 green dice labelled 0, 0, 1, 1, 2, 2.
Answer: 1/9
Please can you show me how to get the answer. I'm confused about joining the events of choosing 2 of 3 dice vs. getting the probability that one of the dice chosen will get a 6 when rolled.
Note: There is an equi-probable chance of getting any of the six sides on a given die.
 A: Out of the three possible choosings, two contain the die with a $6$. (Chance $2/3$)
If the die is selected, then there is a chance in six to get a six. (Chance $1/6$)
The total chance is:
$$\frac{2}{3}·\frac{1}{6} = \frac{2}{18} = \frac{1}{9}$$
A: Two ways to select a die: (Red,Green) or (Green,Green), out of which only (Red,Green) can show a 6.
Now total ways are: 
$$\begin{array}{|c|c|}\hline(Red,Green)&6\times3=18\\\hline(Green,Green)&3\times3=9\\\hline\end{array}$$ Total=27, Now for 6, ways are (Red,Green)$1\times3=3$;so probability is $3/27=1/9$
A: You can select the dices r,g1 and g2 in following combinations:
(r,g1);(g1,g2);(r,g2) Each combination has the probability of 1/3.
There are two combinations, which can show a 6:(r,g1);(r,g2).
Each of this two combination has a probability of 1/6 showing a 6.
All together it is: $\frac{1}{3}\cdot \frac{1}{6}+ \frac{1}{3}\cdot \frac{1}{6}=\frac{2}{18}$
A: Where it concerns the selection of the dice there are two possibilities: 
$RG$ with probability $\frac{2}{3}$
and $GG$ with probability $\frac{1}{3}$. 
(Actually if the green dice are indexed then there are $3$ possibilities with equal probability: $RG_1$, $RG_2$ and $G_1G_2$. That makes clear why the probability of $RG$ is twice the probability of $GG$.)
If $E$ denotes the event
that one of the dice shows $6$ then:
$P\left(E\right)=P\left(E\mid RG\right)P\left(RG\right)+P\left(E\mid GG\right)P\left(GG\right)=\frac{1}{6}\frac{2}{3}+0\frac{1}{3}=\frac{1}{9}$
