# Probability of a random event

I am trying to calculate the probability of selecting one random dice colour from three dice rolls.

The first dice has 2 green faces and 4 red faces. The second dice has 1 green face and 5 red faces, The third dice has 3 green faces and 3 red faces. If I roll three of the dice once, and pick one at random, what is the probability of the dice being green?

So far I have created a sample space:

$S$ $=$ {$GGG, GRR, GGR, GRG, RRR, RGG, RGR, RRG$}

Which has all the possible orders of the dice rolls, where $R$ stands for red dice and $G$ stands for green dice.

How could I calculate the probability of picking a green die at random? Is it just best to calculate the probabilities of all the events with green dice and sum the probabilities?

Note: I am trying to use this sort of probability for a computer program that I am making, but I'm trying not sure how to do the probability. Any sort of help would be appreciated.

• It's more complicated than that. Generally for a sample space to be useful, the events in the sample space need to be equally likely and in your sample space GGG is less likely than RRR. You can fix it by making a much larger sample space taking in more detailed outcomes. An easier way to solve the problem would be to use a tree. The first level of the tree would be rolling the first die, etc. – Arby Mar 15 '17 at 12:28
• @Arby Thanks for your input. I have a very rusty background in probability, so I wasn't really sure if I was doing the right thing. – RoadRunner Mar 15 '17 at 12:32
• In addition, there's a fourth action you need to take into account, that you pick one of the three dice at the end. If you use a tree, it will actually have 4 levels. – Arby Mar 15 '17 at 12:34

It looks as if all $6+6+6=18$ faces are equally likely to be chosen in the end, and $2+1+3=6$ are green, making the answer $$\frac{6}{18}=\frac13$$

If you want a longer answer based on your sample space, try

$$\frac26 \times \frac16\times\frac36\times \frac33 + \frac26 \times \frac56\times\frac36\times \frac13 +\frac26 \times \frac16\times\frac36\times \frac23 + \frac26 \times \frac56\times\frac36\times \frac23 \\ + \frac46 \times \frac56\times\frac36\times \frac03 + \frac46 \times \frac16\times\frac36\times \frac23 + \frac46 \times \frac16\times\frac36\times \frac13 + \frac46 \times \frac56\times\frac36\times \frac13$$ but it is a little more complicated

• It may be obvious to you (and possibly to the OP, too), but it might help others to make clear that this intuition works because the dice have equal numbers of sides. If they had different numbers of sides (say, one of them was eight-sided), then those eight sides would each be less likely than any of the other twelve. – Brian Tung Mar 16 '17 at 0:52
• @BrianTung: yes, and I checked the numbers added up to $6$. My answer would also change if the dice were not equally likely to be chosen – Henry Mar 16 '17 at 0:54
• Thanks Henry. This answer is great :) – RoadRunner Mar 16 '17 at 1:48

The first dice has 2 green faces and 4 red faces. The second dice has 1 green face and 5 red faces, The third dice has 3 green faces and 3 red faces. If I roll three of the dice once, and pick one at random, what is the probability of the dice being green?

You know $\mathsf P(G\mid D=d)$ for each of the die, $d\in\{1,2,3\}$ and the probability for selecting any particular die are all $\mathsf P(D=d)=\tfrac 13$.   So use the Law of Total Probability.

$$\mathsf P(G) ~=~ \tfrac 13(\tfrac 26+\tfrac 16+\tfrac 36) ~=~ \tfrac {6}{18} ~=~ \tfrac 13$$

Which confirms Henry's intuitive approach, that there are 6 green among the 18 equally-likely-to-be-selected faces .

Your method neglected to include an indication of which die was chosen, and further the outcomes would not have equal probability weight.