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Problem: An urn holds 5 red balls and 3 white balls. They are drawn out one at a time (no replacement) until a total of 4 red balls have been taken out (and some unspecified number of white ones). Find the probability that exactly 6 balls have been taken out, showing the steps of your work.

How do you set up this problem? I'm a bit confused with the following statements in the problem: "They are drawn out one at a time (no replacement) until a total of 4 red balls have been taken out (and some unspecified number of white ones)" and "Find the probability that exactly 6 balls have been taken out". It kind of seems unclear. Is it suggesting me to find the probability of 6 balls that are specifically 4 red and 2 white?

I think your hindsight will help me a lot to understand the problem.

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    It's phrased a little strangely. Let me put it slightly differently. I'm going to pull balls out of the bag one at a time. I'm going to stop once I get 4 red balls. What's the probability that I will have taken exactly six balls out? – 2017-01-27
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    4 red, 2 white and the last drawn is red – 2017-01-27
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    @Callus ok now that I know better what the problem looks for. So it has to end with a red ball in order to actually 'stop'. so do I just add up the odds for each ball? – 2017-01-27

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Why don't you start with a simpler version of the problem that will tell you how to solve the more complex one.

Take $3$ red, $1$ white ... until $2$ red and ... probability of exactly $3$ balls have been taken out.

When you are drawing balls without replacement, the possible sequences are (each with equal probability):

$RRRW$, $RRWR$, $RWRR$, $WRRR$.

Call them $o_{1}$, $o_{2}$, $o_{3}$ and $o_{4}$ respectively. If you are drawing until $2$ red, you would stop at the second draw for $o_{1}$ and $o_{2}$ and at the third draw at $o_{3}$ and $o_{4}$. That means probability of $3$ balls have been taken out is $\frac{1}{2}$.

Enumerating the sequences for the original problem is tedious, but the number of ways to arrange $5$ red and $3$ white balls is multinomial coefficient $\binom{8}{3,5}$. Stopping at sixth ball means it is fourth red, meaning one red and one white remain. You can arrange one red and one white in a pair two times. The sixth ball is red. There are five balls prior out of $3$ red and $2$ white. You can do the rest and with this approach any similar question.

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    how do you get the combination model for 6 balls? Like I'm curious and I'm doing it right now but it seems cumbersome and it is hard to keep track what combinations are just 'mirrors' to the original version of the combination – 2017-01-27
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    I'd gotten a multinomial coefficient of 56. I understand the fourth red ball being the 'stop' of the sequence. However, how do I find the likelihood of the event happening exactly at 6 balls? – 2017-01-27
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    That is the other part of the comment. Fix sixth ball being fourth red. One red and one white remain and you can arrange them in 2 ways. Before the sixth ball are five other balls drawn random from three red and two white. Multinomial coefficient again. – 2017-01-27
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    So now it is a multinomial of (6! and 3!2!) which is 60. I use this and divide it by 56 (which is the original unarranged multinomial)? @Jan – 2017-01-27
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    Do you just divide it from arranged with unarranged components? – 2017-01-27
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    It is multinomial of $5!$, not $6!$. – 2017-01-27
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    My mistake. now everything makes much more sense. I got 10/56 when dividing the fixed coefficient versus the non-fixed multinomial coefficient. I think that must be it @Jan Thank you very much! – 2017-01-27
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The answer of $\frac{10}{56}$ you have mentioned in your comment is incorrect. Pl. study Jan's hints carefully.

Since only two colors are involved, for simplicity I shall use binomial coefficients instead of multinomial coefficients, and break into three parts:

  • Three red and two white in all orders

  • The fourth red (fixed )

  • the remaining red and white in all orders

Placing the reds using combinations, $Pr = \dfrac{\binom53\binom11\binom21}{\binom85}$