Probability that red marbles are taken out before blue and green marbles? There is a bag with 10 red, 30 blue and 20 green marbles. You draw these marbles one by one out of the bag. What is the probability that when all of the red marbles are taken out, there is at least one green and at least one blue marble left in the bag?
What's wrong with the way I'm approaching this question:
I'm representing the order of drawing these out as a string. Since we want at least 1B and 1G left, lets say it looks like this:
XXX.... BG
The probability of this happening is A/B, where:
A = 58!/(10!)(29!)(19!) - this is because we have fixed the BG balls, so now we have the number of ways of arranging the remaining 58 balls, 29 of which are B, 19 of which are G, 10 of which are red
B = 60!/(20!)(30!)(10!) - all of the possible ways of arranging this
Now, we multiply by 2, because the last two can either be BG or GB; thus, it would be 2*(A/B). However, this answer is wrong. Where did I go wrong?
 A: The probability the last ball is blue is $\dfrac{30}{10+30+20} =\dfrac12$.
The probability the last ball is green is $\dfrac{20}{10+30+20} =\dfrac13$.
Ignoring the blues, the probability that the last ball from the red and greens is green is $\dfrac{20}{10+20} =\dfrac23$.
Ignoring the greens, the probability that the last ball from the red and blues is blue is $\dfrac{30}{10+30} =\dfrac34$.
So the probability that the last red is drawn before the last blue and the last green is $$\dfrac{1}{2}\times\dfrac{2}{3}+\dfrac{1}{3}\times\dfrac{3}{4} = \dfrac{7}{12}$$
A: There are two obvious approaches to this problem that I know of.  You can correct the approach you were taking by considering that the last red marble is followed by $b$ blue marbles and $g$ green marbles, where $1\leq b\leq30$ and $1\leq g\leq20$. The marbles that come after the last red one can be arranged in $$\frac{(b+g)!}{b!g!}$$ ways, and the marbles before the last red marbles can be arranged in $$\frac{(59-b-g)!}{(30-b)!(20-g)!9!}$$ ways.  This gives a probability of
$$\frac{30!20!10!}{60!}\sum_{b=1}^{30}\sum_{g=1}^{20}
\frac{(b+g)!(59-b-g)!}{b!g!(30-b)!(20-g)!9!}$$
The other approach is to find the probability of the complementary event and subtract from $1$.  The complementary event is that either the last marble is red, or that the last red marble is followed only by (one or more) blue marbles, or that the last red marble is followed only by (one or more) green marbles.
The probability that the last marble is red is obviously $\frac{10}{60}$, so in a manner similar to the first approach, we get $$\frac56-\frac{30!20!10!}{60!}\left(
\sum_{b=1}^{30}\frac{(59-b)!}{20!9!(30-b)!}+
\sum_{g=1}^{20}\frac{(59-g)!}{30!9!(20-b)!}\right)$$
Each of these ugly expressions works out to exactly $\frac7{12}$ which suggests that there's a neat way to do the problem.  With the benefit of hindsight, here it is.
Again we compute the probability of the complementary event, but we work from back to front throughout.  The probability that the last marble is red is $\frac16$ as before.  The probability that the last marble is blue is $\frac12$.  This case will give rise to a complementary event if and only if, as we proceed from back to front, the first non-blue marble we encounter is red.  The probability of this is $\frac13$ since there are $10$ red marbles and $20$ green.  The probability of this case is $\frac12\frac13=\frac16.$  Similarly the probability that the last marble is green is $\frac13$ and the probability that the last non-green marble is red is $\frac13$ giving a combined probability of $\frac13\frac14=\frac1{12}.$
The probability that the last red marble is followed by both a green and a blue marble -s
$$1-\frac16-\frac16-\frac1{12}=\frac7{12}$$
