Removing balls from an Urn -- Probability

Question

Balls are randomly removed from an urn that initially contains 20 red and 10 blue balls.

(a) What is the probability that all of the red balls are removed before all of the blue ones have been removed?

This one I was able to figure out because it is essentially asking: "what is the probability that the last ball is blue?" Which is $\frac{1}{3}$

Now suppose that the urn initially contains 20 red, 10 blue, and 8 green balls.

(b) Now what is the probability that all of the red balls are removed before all of the blue ones have been removed?

I'm not too sure how to proceed with this one. I have seen other answers to this question on the site but I think they're answering it as "what is the probability that all red balls are removed before ANY blue balls are," which would be $\frac{10!20!}{30!}$. I don't think that is the correct answer since you could first remove all but one blue ball first, then all the red balls, and then remove the last blue ball. So would the answer to this question be the same as the above, i.e., $\frac{1}{3}$?

(c) What is the probability that the colors are depleted in the order blue, red, green?

I think this one would be $\frac{20!10!8!}{38!}$

(d) What is the probability that the group of blue balls is the first of the three groups to be removed?

And for the last one I was thinking of calculating the probability that the second to last is green and last is red + probability that second to last is red and last is green. But not too sure how to proceed from there.

Any help would be greatly appreciated!!! =)

Answer 1

b) indeed should be the same as the answer to a) (i.e. $\frac{1}{3}$) since the existence of the green balls has no effect on how fast the the red vs blue balls are being depleted ... but note that that is indeed not the same as $\frac{10!20!}{30!}$ ,... because the latter is the probability of first getting all the red ones out before getting any of the blue ones. Likewise your answer to c) is incorrect.

For c) ... I would think about it this way. First, by symmetry we can, instead of thinking of the colors being depleted in a certain order, think of the colors first appearing, but in the reverse order. That is, the probability of the colors being depleted in the order of blue, red, green, is the same probability of the colors making their first appearance in the order of green, blue, red. And I don;t know about you, but I would just find it easier to think about it that way. OK, so the first ball to appear needs to be green, and the probability of that is $\frac{8}{38}$. Now, after that, you can ignore the rest of the green balls, just like you did as in b), so after that it is simply a matter of getting a red one before a blue one, and the chances of that is $\frac{2}{3}$. So, the probability is:

$$\frac{8}{38} \cdot \frac{2}{3}$$

And yes, for d) follow your own suggestion, and now you know how to compute each of those probabilities.