So this is a question someone asked me about (there's a nasty rumor that I know something about math) and that is ostensibly a basic combinatorics question, though I'm getting conflicting answers:

Say we have a bag with $3$ green, $5$ blue and $4$ red balls. What is the probability of choosing three balls of the same color?

As near as I can tell, the best way to solve this is the following:

How many ways can we choose $3$ of the same color? Well, we can choose all 3 green balls (this is $1$ way) we could choose $3$ blue balls ($5 \choose 3$ ways to do this), or we could choose $3$ red balls (4 ways). Dividing by $12\choose 3$ total ways to do this, we get something like .0681... as our probability.

On the other hand, we could say the probability is P(drawing 3 green)+P(drawing 3 blue)+P(drawing 3 red) which to me looks something like $$\frac{3\times 2\times 1}{12^3}+\frac{5\times 4\times 3}{12^3}+\frac{4\times 3\times 2}{12^3}\approx.0521...$$

So what am I missing here?


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    $\begingroup$ Not $12^3$ in your last displayed equation, but $12\cdot11\cdot10$. You are not replacing the balls after selection. $\endgroup$ – David Mitra Dec 2 '11 at 21:23
  • $\begingroup$ Oho! Thankyou! What a mistake! If you'd like to post this as the answer, I will choose it. If not, I'll make it an answer, just so I can have some closure! $\endgroup$ – Jonathan Beardsley Dec 2 '11 at 21:26
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    $\begingroup$ @David, Oops, my apologies. I did not notice your comment till now. :) $\endgroup$ – Srivatsan Dec 2 '11 at 21:31

Your first answer is correct. Your second approach is almost right, but the denominator is off.

Let's take the probability that you draw 3 red balls. Yes, the first ball is red w.p. $4/12$. But once you draw a single red ball, the number of red balls, as well as the total number of balls goes down by 1. So the probability that second ball is also red (conditioned on the first ball being red) is $3/11$, not $3/12$. Similarly, for the third ball, this probability is $2/11$.

So you end up with $$ \frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}. $$ As a mental check, note that this is the same as $$ \frac{\binom{4}{3}}{\binom{12}{3}}. $$ Proceeding likewise, the final answer is $$ \frac{5 \cdot 4 \cdot 3 + 4 \cdot 3 \cdot 2 + 3 \cdot 2 \cdot 1}{12 \cdot 11 \cdot 10} = \frac{\binom53 + \binom43 + \binom33}{\binom{12}{3}}. $$


You can also use the density function for the hypergeometric distribution for this. http://en.wikipedia.org/wiki/Hypergeometric_distribution Briefly, for two colors (say, white and black) dhyper(a,b,c,d) is the probability of drawing a white balls when there are b white balls in the urn, c black balls, and you are making d draws without replacement.

Your second approach is correct (once you fix the denominator) and boils down to:


As you are only making three draws, you don't have to worry about overlap.


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