Confused in a probability question 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.
 A: 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.
A: 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, $Pr = \dfrac{\binom53\binom11\binom21}{\binom85}$
