Probability question involving colored balls Probability question.  I’ve been trying to figure this out for a while and need a little help.  This is not a homework problem just something I made up but can’t figure it out on my own.
I have a total of 30 balls(10 red, 10 green, 10 blue) I dump all 30 balls into a large bin.
I have 4 smaller containers.  I randomly take 2 balls out and put them in each of the 4 smaller containers (I remove 8 balls total).  What are my chances of having 2 bins or more of just red balls?  Can you please show the math behind the answer.
 A: Let $E$ be the event of interest ( at least 2 bins have 2 red balls). Let $R$ be the amount of red balls extracted (out of 8). Then
$$ P(E) = \sum_{R=0}^8 P(E \mid R) P(R)$$
Now,
$$P(R)= \frac{\binom{10}{R} \binom{20}{8-R}}{\binom{30}{8}}$$
And $P(E \mid R)=0$ for $R\le 3$ and $P(E \mid R)=1$ for $R\ge 6$.
You need only to compute $P(E \mid R=4)$ and $(E \mid R=5)$... can you go on from here?
A: You could use an inclusion-exclusion strategy:
Call the bins A,B,C,D, and let AB represent the event that both A and B are all red (with no requirements on the other bins); ABC the event that A,B,C are all red (with no requirements on D); etc.
First, let's find P(AB). The order in which we fill the bins doesn't matter - we might as well have put the first four removed balls into bins A and B, and the probability these are all red is $\frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}$. The same goes for any other pair of bins, so P(AB)+P(AC)+P(AD)+P(BC)+P(BD)+P(CD) $= 6P(AB) =6\cdot\frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}$.
This sum does some over-counting. Any outcome where exactly two bins are all red falls into just one of the six cases, so we've counted it correctly, but any outcome in which three bins are all red appears in three of the two-bin events (e.g. ABC is counted in each of AB,AC,BC), so by adding the six probabilities, we've over-counted these outcomes by a factor of 3. Let's correct that by un-counting them twice, by subtracting 2(P(ABC)+P(ABD)+P(ACD)+P(BCD)) $=2\cdot 4\cdot P(ABC)=8 \cdot \frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}\frac{6}{26}\frac{5}{25}$ (because P(ABC) equals the probability that the first six balls we look at are all red).
Now our adjusted sum correctly accounts for the exactly-two-bin and exactly-three-bin outcomes, and we just need to fix it for ABCD, which was "accidentally" included in all the events considered above. We added its probability 6 times and subtracted it 8 times, so we should add it back 3 times to get its count back to 1.
So the probability that at least two bins are all red is
$$6\cdot\frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}-8\cdot \frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}\frac{6}{26}\frac{5}{25}+3\cdot \frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}\frac{6}{26}\frac{5}{25}\frac{4}{24}\frac{3}{23}.$$
