# Drawing a ball question

Suppose we draw 3 balls without replacement from an urn with 2 red, 2 green, and 2 blue balls.

How can I find the probability of each 0 red, 1 red and 2 red? I know the number of sample space is 120 ( Since we draw 3 balls and 6 balls in total, it would be 6*5*4). But I'm not sure how to get 0 red probability and 1 red pro. and 2 red prob. Can someone please help me?

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If this is homework, please kindly add the [homework] tag. :) –  cardinal Oct 13 '11 at 0:55
I'm just reviewing for my test coming up now :) –  David Taeyang Lee Oct 13 '11 at 0:58
It says you're drawing without replacement, so if you stuck your hand in and grabbed three balls at once, what are the possible sets of colors you could get? –  cardinal Oct 13 '11 at 0:59
possible sets would be 120 cases.. isn't it? –  David Taeyang Lee Oct 13 '11 at 1:05
That is one way to go about it. It depends on whether you treat each individual ball as distinguishable or not. Let's say you decide to take this approach. Consider the probability of getting zero red balls. Now, we're going to draw three balls in sequence. There are six balls in the urn. How many choices do you have for the first draw such that you don't draw a red one? Now there are five in the urn. How many choices do you have? Now, you're ready for the last draw. How many choices do you have? So... –  cardinal Oct 13 '11 at 1:13

In this problem, all we care about is whether a ball is red or not. It is simpler to imagine we are green-blue colour blind, and that there are $2$ red balls and $4$ non-red balls.

The $120$ element sample space that you have chosen will work, but I think it is not the best choice. We first do the problem using a different sample space, and in a comment at the end we solve one part of the problem using your $120$ element sample space.

We are drawing (removing) $3$ balls from the urn. Imagine removing them all at once, or alternately, one at a time, but only looking at them at the end. Then there are $\binom{6}{3}$ possible outcomes, all equally likely.

How many of these outcomes result in $0$ red balls? We must have chosen $3$ balls from the $4$ non-red balls (and none from the red ones). There are $\binom{4}{3}$ ways of doing this. So the probability of $0$ red is $$\frac{\binom{4}{3}}{\binom{6}{3}}.$$

Next we calculate the probability of $1$ red. The red ball can be chosen in $\binom{2}{1}$ ways. For each choice, the needed $2$ non-reds can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{2}{1}\binom{4}{2}$. So the probability of $1$ red is $$\frac{\binom{2}{1}\binom{4}{2}}{\binom{6}{3}}.$$ A similar argument shows that the probability of $2$ red is $$\frac{\binom{2}{2}\binom{4}{1}}{\binom{6}{3}}.$$ The $\binom{2}{2}$ in the above expression is superfluous, but looks nice. In a similar way, for $0$ red, we could have written the numerator as $\binom{2}{0}\binom{4}{3}$.

Comment: It is possible to solve the problem by using your sample space of $120$. However, it makes things somewhat more complicated. For example, let's solve the $1$ red problem using the $120$ element sample space, where the order of selection matters. Then $1$ red can happen in three basic patterns: (i) RNN; (ii) NRN; and (iii) NNR. (Here R means red, N means non-red.)

How many ways can we get a type (i) pattern? The initial red can be chosen in $2$ ways, then the first non-red in $4$ ways, then the second in $3$ ways, for a total of $(2)(4)(3)=24$. The numbers for the other two patterns also turn out to be $24$, so the total number of choices, where order matters, is $(3)(24)$. Divide by $120$ to get the probability. We get $(3)(24)/120$, which simplifies to $3/5$.

Earlier, we had obtained the expression $\frac{\binom{2}{1}\binom{4}{2}}{\binom{6}{3}}$ for the probability of $1$ red. Calculate. We get $(2)(6)/20$, which simplifies to $3/5$.

You may think that the sample space of $120$ approach is simpler, and there is a good case for claiming that. However, in more complicated sampling problems, the binomial coefficient approach that we used turns out to be often more useful.

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And there is the identity $$\sum_k\binom{r}{k}\binom{n}{d-k}=\binom{r+n}{d}$$ that shows the sum of the probabilities of $0$, $1$, and $2$ red balls is $1$. –  robjohn Oct 13 '11 at 1:57