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Suppose that the contents of an urn are $w$ red balls, $x$ yellow balls, $y$ green balls, and $z$ blue balls collectively, where $w \geq 3$, $x\geq 1$, $y\geq 1$, and $z\geq 1$. We draw balls randomly from this urn without replacement.

What is the probability of our having drawn at least 1 yellow ball by (and including) the seventh draw, at least 1 green ball by (and including) the eight draw, and at least 3 red balls and 1 blue ball by (and including) the ninth draw?

Note that this is one single event and not four separate events.

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Honestly, more than I should likely have needed to. I thought initially that utilizing conditional probabilities and hypergeometric distributions would do the trick, but I've been unable so far to come up with anything. I've gone so far now as to head to the local university's library and check out some books on multivariate distributions as I suspect the multivariate inverse geometric distribution might work better for the purposes of solving this. – Roberto Jun 15 '12 at 6:03
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First of all, for clarity, I will use the following variables: $r$ for the number of red balls, $y$ for yellow, $g$ for green, and $b$ for blue. I will denote the total number of balls by $n$.

Out of the total number of permutations of 9 balls selected from the $n$ total, we only want to consider certain permutations. We want at least one of the first seven selections to be yellow, one of the remaining seven selections (one is taken by the yellow ball) to be green, three of the remaining seven selections to be red, and one of the 4 still remaining selections to be blue. The remaining four unspecified selections could be anything.

There are ${{y}\choose{1}}\cdot{{7}\choose{1}}$ ways of a yellow ball taking one of the first seven positions.

There are ${{g}\choose{1}}\cdot{{7}\choose{1}}$ ways of a green ball taking one of the first eight positions, after one has already been specified as yellow.

EDIT: There are ${{r}\choose{3}}\cdot{{b}\choose{1}}\cdot{{7}\choose{4}}\cdot{{4}\choose{1}}$ ways of three red balls and one blue taking four of the remaining seven positions. (And we choose one of the four positions to be taken by the blue ball.)

There are still three positions left, and we have already placed six balls, leaving us ${_{n-6}}P_3$ possible ways to fill the remaining positions.

Thus the total number of possible permutations that satisfy the requirements is ${{y}\choose{1}}\cdot{{7}\choose{1}}\cdot{{g}\choose{1}}\cdot{{7}\choose{1}}\cdot{{r}\choose{3}}\cdot{{b}\choose{1}}\cdot{{7}\choose{4}}\cdot{{4}\choose{1}}\cdot{_{n-6}}P_3$ out of a total of ${_n}P_9$ permutations.

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This is a much more elegant way of approaching this that what I had been attempting. I greatly appreciate this. – Roberto Jun 15 '12 at 6:07
It was late last night, and I made a small mistake. Please see the edited section above. – Andrew Parker Jun 15 '12 at 16:39
Apologies for the delay. Again, this was immensely helpful. – Roberto Jun 17 '12 at 2:31

Hint: "at least 1" = "not 0", "at least 3" = "not 0,1 or 2". Conditional probabilities may be useful.

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