How to find this conditional probability in sampling without replacement? Here is the complete question:
A club consists of $4n$ people of half men and half women. A woman is elected as the chair of a committee from this group. As all members are equally talented, she prefers to select
additional $n$ people at random from the remaining $4n-1$ people to work with her for a particular
project.
If $n$ people chosen happen to be of the same sex, what is the probability that all are 
(a) male 
(b) female?
Hi Everyone, I am taking 1st year statistics and this is an exam practice question that I am a little stuck on - would greatly appreciate some help!
Here is what I have so far:
For males, I have  thought about it this way:
Since a woman is elected as Chair, she would then have $2n$ men and $2n-1$ women to choose from.
The probability of selecting $n$ people that are all men for the project from the 1st three picks up to the $nth$ pick, I have:
$$\frac{2n}{4n-1} \cdot \frac{2n-1}{4n-2} \cdot \frac{2n-2}{4n-3} \cdot \space ...  \cdot \frac{2n-(n-1)}{4n-n}. $$
Similarly, for females I have the probability of selecting all women to be:
$$\frac{2n-1}{4n-1} \cdot \frac{2n-2}{4n-2} \cdot \frac{2n-3}{4n-3} \cdot \space ...  \cdot \frac{2n-n}{4n-n}. $$
Firstly, am I on the right track, and secondly how can I find a closed formula for this probability?
Many thanks in Advance!
 A: This problem is special, in the following way.  Label the $2n-1$ non-chair women with the numbers $1, 2, \ldots, 2n-1$ and label the $2n$ men with the numbers $1, 2, \ldots, 2n$.  Any all-female committee is determined by the $n$ labels of its members. Corresponding to such a committee are two distinct all-male committees: the one of men having those labels and the one of the men not having those labels.  Furthermore, exactly one of those all-male committees includes the man with label $2n$: call this the "second" committee and let the other choice of all-male committee be the "first" committee.
Notice that any two distinct "first" committees correspond to distinct all-female committees, that any two distinct "second" committees correspond to distinct all-female committees, and no "first" committee can also be a "second" committee or conversely.
As an illustration, consider the case $n=2$.  The $\binom{2n-1}{n}=3$ all-female committees and their $\binom{2n}{n}=6$ all-male counterparts are
$$\eqalign {
\text{Women} &  & \text{First men}; & \text{Second men}\\
\{1,2\} &\to &\{1,2\}; & \{3,4\}\\
\{1,3\} &\to &\{1,3\}; & \{2,4\}\\
\{2,3\} &\to &\{2,3\}; & \{1,4\}.
}$$
Therefore, there are exactly two possible all-male committees for each all-female committee.  If in fact all members of the committee are of the same sex, then it must be exactly twice as likely that they are male than that they are female.  The answer does not depend on $n$!
