# Probability of Pulling Three Different Colors

You have a closed box with 10 red balls, 10 white balls and 10 blue balls. (Otherwise the balls are identical.)

Without looking in the box, you pull out exactly three balls. What is the probability that the three balls will all be a different color (i.e. 1 red, 1 white and 1 blue)? The order of color does not matter, only that all three balls are different.

The lesson here is the basic rules of figuring out the probability, but I wasn't sure I understand the correct math.

I think you multiply the probability of the outcomes. I thought it through this way:

1. Draw first ball
2. Probability of getting any different color on the next draw: 20/29
3. Assuming that happens, I think the probability of getting a third different color is now 10/28.
4. Then, do you multiply those outcomes? 20/29 * 10/28 = 200/812 = 0.2463

Is that the right answer?

• If this is homework you need to read stats.stackexchange.com/tags/self-study/info
– Carl
Feb 18, 2018 at 12:43
• The edit is good put please add the tag, too. Feb 19, 2018 at 6:13
• Juho, I already added the 'probability' tag. What other tags do you think fit the question? Feb 19, 2018 at 9:07
• If you're uncertain about the multiplication, apply your reasoning to smaller instances of the problem (try one ball of each color and then two balls of each color; and consider circumstances where there are just two colors instead of three) to see what might go wrong.
– whuber
Feb 19, 2018 at 15:13

## 1 Answer

Welcome to the site. This being a homework question, you should tag it as such. Want to share any progress you've made on your own?

Try making a list of all the possible orderings of draws that match the event for which you want to calculate the probability.

So, for example, there's going to be a probability of drawing the sequence RED, WHITE, BLUE. What other orderings can you find?

-------------EDIT----------------------------------------------------------------------------------------------------------

In all this explanation, I'm assuming that the draws are without replacement, that is once the balls are drawn they are not placed back in the box.

I think that what is confusing you is the last sentence of the question: "The order of color does not matter, only that all three balls are different".

There is a more general approach to this problem (it is formalized by the implementation of the multivariate hypergeometric distribution, for reference), but let's stick to this approach for simplicity.

Let's see what is wrong with your approach.

1. Draw first ball

Note that this also has a probability! But we must specify what the probability of drawing this first ball is (it's 10/30 for all colors). So we must consider this also in our equation (For the rest of the example, let's suppose it's red).

1. Probability of getting any different color on the next draw: 20/29

This deserves some comment. This is true, but why?

Because the event "drawing a blue ball" and "drawing a white ball" are disjoint events (there is no way you can draw a ball that is both white and blue at the same time). Since they are disjoint, the probability that you are naming is

$$P(B\cup W|R \: \text{at first draw}) =\\ P(B|R \: \text{at first draw}) + P(W|R \: \text{at first draw}) = \frac{10}{29} + \frac{10}{29} = \frac{20}{29}$$

which can be interpreted as the probability of drawing a blue ball or a white ball given than the first one was red. This is why now you cannot do what you state in point 3, since you have already considered the probability of getting a ball of blue or white color.

When the problem states that "the order of color does not matter, only that all three balls are different" what is meant is that you have to consider all the possible orderings of color. That is, we are not interested in the specific draw $\{R,B,W\}$ but in all of them, since all the possible orderings satisfy the event for which we want to calculate a probability, i.e the event "three balls will all be a different color". What configurations satisfy this event?

$\{R,W,B\} \\\{R,B,W\} \\\{B,W,R\} \\\{B,R,W\} \\ \dots$

Using the laws of probability, we can see that the probability of each of the above events can be written as (changing the appropriate order, but we notice they are all the same)

$$P(\{R,W,B\}) = P(R|W,B)P(W,B)=P(R|W,B)P(W|B)P(B)$$

where, in the above expression, you must add the text "at first draw", "at second draw", "at third draw" respectively after B, W and R.

Once you have these probabilities, then you can use the fact that the above events are disjoint to sum them all up, and there you have it.

I've told you what $P(B)$ is $\frac{10}{30}$. What are the other two components of the equation?

• I just wasn't sure I know the math on this. I think you multiply the probability of the outcomes. I thought it through this way: 1) Draw first ball 2) Probability of getting any different color on the next draw: 20/29 3) Assuming that happens, I think the probability of getting a third different color is now 10/28. 4) 20/29 * 10/28 = 200/812 = 0.2463 Is that the right answer? Feb 18, 2018 at 19:52
• I've added some more information that might help Feb 19, 2018 at 16:40