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!