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Given a bag of six marbles with two colors, red and blue, what is the probability of drawing a set of three such that there is exactly one blue?

In the past, when given a problem such as this I would construct a tree diagram at each decision point and manually count the number satisfying the condition. But as examples become more complex, I find myself needing a more formal way of getting to the answer.

My thinking is that I have to solve the basic probability fraction of:

$\frac{\mathrm{num.\,of\,favorable\,outcomes}}{\mathrm{num.\,of\,total\,possible\,outcomes}}$

In our example, that would be:

$P(3\,marbles,\,1\,blue) = \frac{\mathrm{num.\,ways\,to\,draw\,3\,with\,one\,blue}}{\mathrm{num.\,ways\,to\,draw\,3\,marbles}}$

I think I can get the denominator with:

$_6{C}_3=20$

But for the numerator, absent of manually counting each possibility, I cannot think of a way.

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  • $\begingroup$ There are only $20$, not $120$, three-element subsets of six things, which is what $_6C_3$ counts. $\endgroup$ – whuber Oct 3 '16 at 0:41
  • $\begingroup$ any help on the numerator? $\endgroup$ – candles_and_oranges Oct 3 '16 at 1:53
  • $\begingroup$ Consider the full problem: how many 3-subsets are there with $k=0,1,2,3$ blue balls? Call these numbers $a(k)$. Because the numbers of blue and red balls are the same, and $k$ blue balls is equivalent to $3-k$ red balls, $a(0)=a(3)$ and $a(1)=a(2)$. Now $a(0)=1$ obviously. Also, all four values sum to $20$: $$20=a(0)+a(1)+a(2)+a(3)=1+a(1)+a(1)+1=2+2a(1),$$ whence $a(1)=9$ is the numerator you seek. $\endgroup$ – whuber Oct 3 '16 at 14:38
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The story behind the actual numerator could go like this: How many ways of selecting a $n = 3 \text{-member}$ committee (read: drawing $3$ balls) from a group of $N = 6$ people (read: total number of $6$ balls) with $k= 1$ chairperson (read: $1$ blue ball) voted among $K =3$ candidates (read: $3$ total blue balls) to chair the committee?

Well, given the $3$ candidates for chairmanship, there'll be ${K \choose k}={3 \choose 1}=3$ ways of selecting the chair. And for each of them, the $2$ ranking members among the $3$ non-chair-contenders can be chosen in ${N-K\choose n-k} = {3 \choose 2}$ ways. So the total number of ways is ${K \choose k}{N-K\choose n-k}=9.$


This experiment follows a hypergeometric distribution. From Wikipedia:

In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of $\displaystyle k$ successes in $\displaystyle n$ draws, without replacement, from a finite population of size $\displaystyle N$ that contains exactly $\displaystyle K$ successes, wherein each draw is either a success or a failure.

So the probability of drawing $1$ blue among a set of $3$ is the

$\small \Pr(1 \text{ success in } 3 \text{ draws without replacement}).$

In this case,

$N=\text{total}=\text{blue and red}=6;$

$K =\text{total no. successes}= \text{blue}=3;$

$n=\text{draws}=3;$ and

$k =\text{successes in n draws}= 1.$

The denominator is as you calculated, ${N \choose n} ={ 6 \choose 3}=20$.

The numerator is $\large {K\choose k }{N-K\choose n-k}={3\choose 1 }{6-3\choose 3-1}={3\choose 1 }{3\choose 2}=3\times 3=9$

And the probability $\Pr(1 \text{ blue in } 3 \text{ draws})=9/20.$

In fact, your question is exactly equivalent to the urn experiment proposed as an example in Wikipedia.

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