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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.

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If only one of the three dice has a 6 on in, the chance that two dice show a 6 is zero. – azimut Aug 9 '14 at 8:20
Bhoot - What do you mean? – StephanCasey Aug 9 '14 at 8:21
azimut - Sorry. I meant just one of the dice – StephanCasey Aug 9 '14 at 8:21
@user157220 After someone answers your question, you can accept it by clicking on the tick of the answer. This way, the website awards both you and the person who answered your question with reputation. It is the system on which this website is built on. – user105475 Aug 9 '14 at 8:32
Thanks. I just didn't know how to do that :) – StephanCasey Aug 9 '14 at 8:50
up vote 4 down vote accepted

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:

$$\frac{2}{3}·\frac{1}{6} = \frac{2}{18} = \frac{1}{9}$$

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Oh I see. I got the 2/3 but not the 1/6. Reason being was I thought that of the 2 dice there was a 1/12 chance of getting the 6 but the other dice isn't relevant? – StephanCasey Aug 9 '14 at 8:39
It's useful to ask yourself this two questions: How many outcomes are there? How many of those outcomes satisfy my requirements? – Darth Geek Aug 9 '14 at 8:41
Yeah. That's good advice. Also. I realize I wasn't thinking straight because the dice that we need to get a six on is not reliant on what the other dice shows. – StephanCasey Aug 9 '14 at 8:43
Exactly. Well done. – Darth Geek Aug 9 '14 at 8:44

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 $3/27=1/9$

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My instinct is telling me this is wrong. For instance, (Red,Green) is twice as likely as (Green,Green), but that doesn't factor into your calculation at all... – Niet the Dark Absol Aug 9 '14 at 11:09

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: $\frac{1}{3}\cdot \frac{1}{6}+ \frac{1}{3}\cdot \frac{1}{6}=\frac{2}{18}$

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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:

$P\left(E\right)=P\left(E\mid RG\right)P\left(RG\right)+P\left(E\mid GG\right)P\left(GG\right)=\frac{1}{6}\frac{2}{3}+0\frac{1}{3}=\frac{1}{9}$

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