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?
 A: 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.
