# Probability of Picking sets of Colored Balls

This is an example from Carol Ash's, "Probability Tutoring Book" which I don't understand.

You have 40 white, 50 red and 60 black balls. Pick 20 without replacement. Find the probability of getting 10 white, 4 red and 6 black.

She gives the answer $\frac{\binom{40}{10}\binom{50}{4}\binom{60}{6}}{\binom{150}{20}}$

She makes the point that it doesn't matter if the balls are distinguishable or not.

Why? Could someone explain this question from a combinatorics point of view?

Bonus: Could generating functions be used to solve part of this?

EDIT

I did a quick review of my discrete math notes from last year and this is how I would think to approach the problem.

Total number of possibilities would be the solution to $b+r+w=20$, namely $D(3,20)=\binom{22}{2}$, the number of ways to put 20 balls into slots labeled white, red, black.

Then I would take the number of permutations of 20 distinct balls divided by the desired number of white, red and black balls.

So I would get $\frac{\frac{!20}{!10!4!6}}{\binom{22}{2}}$. Does this seem correct? It's very different on the face from the given answer.

EDIT 2

What you’re missing is that even if there is nothing to distinguish one white ball from another, there are still 99 different white balls in the bag.

That gets to the heart of what I'm missing here. If I were asked, without connection to probability, how many sets of 20 balls I can get from 99 indistinguishable white balls and 1 red ball I would answer 2, like you said. But that would be incorrect because, in reality, there are 99 different white balls in the bag, as you pointed out. I was implicitly thinking of making a list of 20 from the options white or red with at most one appearance of red in the list, but that's a fundamentally different problem.

If I restated the problem as pick a committee of 20 from 99 men and one woman I wouldn't have made that mistake, because it's obvious that even though they're all "men", each man is distinguishable from each other man.

So in summary, it's important to not get confused between

1. How many unordered lists of length $k$ you can make from a set of $n$ traits / types ( such as white, red, black, or man, woman, child ) which is $D(n,k)$
2. Picking sets / committees of size $k$ from a set of $n$ items where some items might have a trait that seems to make them indistinguishable, such as white, red and black balls.
• Do you want the answer to the question explained, or do you want to know why it doesn't matter if the balls are distinguishable or not, or is it both? Dec 24, 2012 at 10:45
• @MichaelAlbanese Both. Dec 24, 2012 at 11:02

There are $150$ balls altogether, and you’re choosing $20$ of them; a set with $150$ members has $\binom{150}{20}$ $20$-element subsets, so there are $\binom{150}{20}$ sets of $20$ balls that you could draw, all of them equally likely to be drawn. That accounts for the denominator of the fraction: it’s the number of equally likely possible outcomes.

Similarly, there are $\binom{40}{10}$ different sets of $10$ white balls, $\binom{50}4$ different sets of $4$ red balls and $\binom{60}6$ different sets of $60$ black balls. There are $$\binom{40}{10}\binom{50}4\binom{60}6$$ ways to combine one of the $\binom{40}{10}$ possible sets of $10$ white balls, one of the $\binom{50}4$ possible sets of $4$ red balls, and one of the $\binom{60}6$ possible sets of $6$ black balls, so there are $$\binom{40}{10}\binom{50}4\binom{60}6$$ successful outcomes, where successful means having $10$ white balls, $4$ red balls, and $6$ black balls. As usual, the probability of a successful outcome is the ratio of successful outcomes to equally likely possible outcomes, or

$$\frac{\binom{40}{10}\binom{50}4\binom{60}6}{\binom{150}{20}}\;.$$

Nothing in the analysis changes in any way if we paint numbers from $1$ through $150$ on the balls to give them unique identities: we still have to count the number of sets of $10$ white balls, the number of sets of $4$ red balls, the number of sets of $6$ black balls, and the number of sets of $20$ balls of whatever colors, and we still have to perform the same calculations with these numbers.

Added: I’m afraid that the calculation in the edit makes little sense. To see this more clearly, let’s look at a simpler example. Suppose that there are $99$ white balls and $1$ red ball, and you draw $20$ balls at random without replacement. By your reasoning there are $\binom{21}1=21$ possible outcomes, ranging from $20$ white and no red balls to no white and $20$ red balls. This, however, is clearly not the case, since there is only $1$ red ball in the bag. Okay, suppose that you take this limitation into account: then by your approach there are exactly two possible outcomes: $20$ white balls, and $19$ white balls and $1$ red balls, so the denominator of your fraction will be $2$.

For the outcome of drawing $20$ white balls the numerator will be $\frac{20!}{20!}=1$, so you would conclude that the probability of getting $20$ white balls is $\frac12$. Is that reasonable? Notice that you’d get the same result if there were $999$ white balls and $1$ red ball, or $999999$ white balls and $1$ red ball. I think that it’s pretty clear that this cannot be right.

Worse, your numerator for the outcome of drawing $19$ white balls and the red ball will be $\frac{20!}{1!19!}=20$, and the ‘probability’ of this outcome will be $\frac{20}2=10$. That certainly can’t be right!

What you’re missing is that even if there is nothing to distinguish one white ball from another, there are still $99$ different white balls in the bag, and a $99$-element set has $\binom{99}{20}$ different $20$-element subsets. Every one of these $\binom{99}{20}$ sets is a different outcome, even if you have no way to tell which of them you actually got. Thus, in the simplified problem the probability of getting $20$ white marbles is actually

$$\frac{\binom{99}{20}}{\binom{100}{20}}=\frac{99!}{20!79!}\cdot\frac{20!80!}{100!}=\frac{80}{100}=\frac45\;,$$

while the probability of getting $19$ white balls and the red ball is simply the probability of getting the red ball, $$\frac{20}{100}=\frac15\;.$$

• "Nothing in the analysis changes in any way if we paint numbers from 1 through 150 on the balls to give them unique identities". This is the part I don't get in this problem. Maybe seeing my edit to the OP will help with understanding what point of view I'm coming from. For me the above analysis kind of makes sense if all the balls are distinct, but not the other way around. Dec 25, 2012 at 15:25
• @Robert: I’ve made a rather extensive addition to my answer to address confusion; see if it helps. Dec 25, 2012 at 18:01
• OK, I'm pretty sure I understand now. Could you please just take a quick look at my new edit to the OP to make sure. Dec 26, 2012 at 11:25
• @Robert: Yes, that looks better. Another way to put it: when you’re counting (say) the number of ways to fill an order for $20$ doughnuts when at least $20$ of each of $3$ types are available, taking $6$ plain doughnuts can be thought of as taking the first $6$ in the queue: you’re not actually choosing which $6$ plain doughnuts you take. In this problem, though, you really are choosing $6$ specific black balls from the $60$ in the bag, and there are $\binom{60}6$ possible choices. The balls are indistinguishable, but they still have separate identities. Dec 26, 2012 at 11:49
• The balls are indistinguishable, but they still have separate identities. That's what caused me all the confusion. Dec 26, 2012 at 15:21