Probability: Draw w/o replacement, 3 different kinds of balls, don't understand

[zeion]

Homework Statement

This was an example given in class but I didn't understand how the answer was derived.
Suppose we draw 3 balls without replacement from an urn with 2 red, 2 green, and 2 blue balls. Define the following random variables:

R = number of red balls drawn
G = number of green balls drawn
B = number of blue balls drawn

What is the probability mass function of R?

0 -> 24/120
1 -> 72/120
2 -> 24/120

Homework Equations

The Attempt at a Solution

I don't understand how to get those ratios.
For 0 red balls, there will be either 2G 1B or 2B 1G.

 

[LCKurtz]

Homework Statement

This was an example given in class but I didn't understand how the answer was derived.
Suppose we draw 3 balls without replacement from an urn with 2 red, 2 green, and 2 blue balls. Define the following random variables:

R = number of red balls drawn
G = number of green balls drawn
B = number of blue balls drawn

What is the probability mass function of R?

0 -> 24/120
1 -> 72/120
2 -> 24/120

Why not reduce those fractions?

I don't understand how to get those ratios.
For 0 red balls, there will be either 2G 1B or 2B 1G.

For example, to get 1 red you would have to choose 1 from the 2 red and 2 from the 4 non-red. That can be done in $\binom2 1 \binom 4 2$ ways. The total number of ways to choose 3 balls from the 6 is $\binom 6 3$ Dividing gives

$$P(2 \hbox{ red}) = \frac{\binom2 1 \binom 4 2}{\binom 6 3} = \frac 3 5$$
ways.

 

[HallsofIvy]

For the purposes of finding the probability of a certain number of red bals, the distinction between green and blue is irrelevant. So just think that there are 2 red and 4 "non-red" balls.

To get 0 red balls you have to get a non-red ball each time. There are 4 non-red out of 6 so the probability of a non-red ball on the first draw is 4/6= 2/3. Since we do not replace it there are 5 balls left, three of which are non-red. The probability of getting a non-red ball is now 3/5. Since we do not replace it, there are now 4 balls, two of which are non-red. The probability that the third ball drawn is also non-red is 2/4= 1/2.

For "one red ball", you could calculate the probability of "RNN" (red, non-red, non-red) in the same way, then use the fact that all three orders, "RNN", "NRN", and "NNR" are all the same.

For "two red balls", look at "RRN", "RNR", and "NRR".

 

[zeion]

For the purposes of finding the probability of a certain number of red bals, the distinction between green and blue is irrelevant. So just think that there are 2 red and 4 "non-red" balls.

To get 0 red balls you have to get a non-red ball each time. There are 4 non-red out of 6 so the probability of a non-red ball on the first draw is 4/6= 2/3. Since we do not replace it there are 5 balls left, three of which are non-red. The probability of getting a non-red ball is now 3/5. Since we do not replace it, there are now 4 balls, two of which are non-red. The probability that the third ball drawn is also non-red is 2/4= 1/2.

For "one red ball", you could calculate the probability of "RNN" (red, non-red, non-red) in the same way, then use the fact that all three orders, "RNN", "NRN", and "NNR" are all the same.

For "two red balls", look at "RRN", "RNR", and "NRR".

Thank you!
That made sense. I copied something like that in my notes but couldn't figure it out.