Revisiting old problem a bit more formally (hopefully) Removing balls from an Urn -- Probability
I asked the above a year ago, and I think I never fully understood the answers to my questions, partially because I wasn't thinking about it too formally. I was wondering if someone could help me redo these questions, stating clearly the probability space (what $\Omega$, $F$, and $P$) to use and corresponding events.
For example, a) asked:

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?

Intuitively, it makes sense for it be $1/3$. But I want to state clearly the probability space I am using. Let $\Omega$ correspond to the collection of 20 red balls and 10 blue balls. Let $F=2^{\Omega}$, and let $P$ be the real valued function defined $\forall$ $E$ $\in F$ with $P(E)=\frac{|E|}{|\Omega|}=\frac{|E|}{30}$. Let A be the event that all of the red balls are removed before all of the blue ones have been removed. Let B be the event that the last ball is blue. Then $|A|=|B|=10 \implies P(A)=1/3.$

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

Here I am a bit confused. I was thinking of letting A be the event that all non green balls are depleted before green balls. And B be the event that all blue balls are depleted before all red balls. The answer is $\frac{2}{3}\cdot\frac{8}{38}.$ But what probability space would you use? In particular what $\Omega$ and what P (P is a function defined $\forall$ $E$ $\in F$ for which $P(E)=\frac{|E|}{?}$).
** Just a thought: for c) can we condition on an event and so, the probability measure might change? Especially since the person who answered mentioned the word "ignore."
 A: I can't make any sense of your probability space for part a).  I would say the probability space is the set of all sequences of $20$ red balls and $10$ blue balls, each of which has a probability of ${20!10!\over30!}$  Then the probability that a a blue ball is $${29!\over20!9!}\cdot{20!10!\over30!}=\frac13$$
For the remaining parts the probability space is the set of all sequences of $20$ red balls, $10$ blue balls, and $8$ green balls.
I think you really should try to understand the more intuitive explanations given in the linked answer, though.  You'll find it a lot easier to solve problems with that kind of reasoning, I believe.
EDIT
(In response to OP's comment.)  Okay, but as I said, I don't think it's the best way to go about these problems.  We want to compute the probability that blue is depleted first, red is depleted second and green depleted last.  In any such sequence, the last red ball must come after the last blue ball, so if we just consider the subsequences of red and blue balls, there are $${29!\over10!19!}$$ possibilities as above.  
Now we have to place $8$ green balls into the sequence, and there are $31$ possible places to put them.  This sounds like stars and bars but we require that at least one green ball go in the last place.  So, we put a green ball in the last place, and then distribute the remaining $7$ green balls in $31$ places in $${31+7-1\choose7}={37\choose7}$$ ways.  This gives a probability of  $${29!\over10!19!}\cdot{37!\over30!7!}\cdot{10!8!20!\over38!}={29!\over30!}\cdot{20!\over19!}\cdot{10!\over10!}\cdot{8!\over7!}\cdot{37!\over38!}=\frac23\cdot{8\over38}$$
Don't you agree that the original solution is more insightful?
