Let's say there are 2 red balls, 2 green balls, and 2 blue balls in an urn. We want to choose 3 balls from the urn without replacement. What is the probability that we select 1 ball of each color?
I tried to solve this problem as $\frac{2^3}{\binom63}$ and got it right, but I don't fully understand why the denominator is 6C3 if balls of the same color are identical to each other. For example, if the balls were numbered so that balls of the same color are distinguishable, {red 1, green 1, blue 1} and {red 2, green 1, blue 1} would be considered different. However, in this case, the balls are not numbered, but doesn't 6C3 still count {red 1, green 1, blue 1} and {red 2, green 1, blue 1} as different?
I'd appreciate any help!
With such a probability problem, you are allowed to regard balls of the same color as either distinguishable or indistinguishable, as long as you are consistent in computing the pertinent numerator and denominator.
Expressing the probability as $~\dfrac{N}{D},~$ you have to either :
assume that the balls of the same color are distinguishable, when calculating both $~N~$ and $~D.$
assume that the balls of the same color are not distinguishable, when calculating both $~N~$ and $~D.$
$\underline{\text{If the balls of the same color are distinguishable}}$
You have two (distinguishable) choices for the ball that will represent each color. So, $~N~ = 2^3.$
There are $~\displaystyle D = \binom{6}{3}~$ ways of selecting three distinguishable balls out of $~6.$
$\underline{\text{If the balls of the same color are not distinguishable}}$
Here, things get very tricky.
Instead of attacking the problem combinatorically, I will attack the problem as a series of events, and examine the probability that each event occurs.
Since the initial status of the colors of the balls is symmetric (i.e. there are two balls of each color), it does not matter which colored ball is selected first. Assume, without loss of generality that the first ball chosen is red.
In order for three colors to end up being chosen, two things now have to happen:
The second ball chosen must not be red.
Despite the fact that the green balls are indistinguishable from each other, and the blue balls are indistinguishable from each other, the fact is that there are now $5$ balls remaining, only one of which is red.
Therefore, the probability that the second ball chosen is not red is $~\dfrac{4}{5}.$
The third ball chosen must be different in color from the other two balls. Regardless of which two (different) colors are selected as the first and second balls, there will be four balls left, two of which are the third color.
Therefore, the probability that the third ball chosen is of the off-color is $~\dfrac{1}{2}.$
So, the overall probability is
$$\frac{4}{5} \times \frac{1}{2}.$$
Note
A case can be made that in the second case above, I didn't really regard balls of the same color as indistinguishable. This is a very tricky issue.
That is, once a red ball was chosen as the first ball, one might reason that there are three colors remaining, and so the probability that the second ball chosen is red is $~\dfrac{1}{3}~$ rather than $~\dfrac{1}{5}.$
It is true that there are only three colors remaining. However, despite the fact that the balls of the same color are to be regarded as indistinguishable, the three colors are not equally likely to be chosen.
Regardless of how you regard colors, when you have $1$ red ball, $2$ green balls, and $2$ blue balls, the probability of then selecting a red ball is only $~\dfrac{1}{5}.$