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!

  • 2
    $\begingroup$ 1) Does the chair picks people sequentially or "all at once"? 2) You are searching for a conditional probability here, i.e. P(all men | all same sex) , P(all women | all same sex) $\endgroup$ Commented Sep 16, 2013 at 9:00
  • 1
    $\begingroup$ Also remember your Bayes; $P(M|SS) = \frac{P(SS|M)P(M)}{P(SS)}$; $P(SS|M) = 1$; and that $P(SS) = P(M) + P(F)$ (SS, same sex; M, all male; F, all female). $\endgroup$
    – January
    Commented Sep 16, 2013 at 9:19
  • 2
    $\begingroup$ One way to think about this problem is to note that "of the same sex" determines the collection of single-sex subsets of $4n-1$ people. There are $a=\binom{2n}{n}$ all-male subsets and $b=\binom{2n-1}{n}$ all-female subsets, all equally likely to be selected. Therefore the questions come down to this: you choose one item at random out of $a+b$ items. What is the probability that you choose one of the first $a$ items? What is the probability you choose one of the last $b$ items? Note how this circumvents having to consider sequential picking, conditional probabilities, or Bayes' Theorem. $\endgroup$
    – whuber
    Commented Sep 16, 2013 at 14:53
  • 1
    $\begingroup$ That's not a hypergeometric distribution, @BabakP: it's a Binomial distribution with parameters $n=1$ and $p=a/(a+b)$. That's what makes this approach so simple (and accessible even to novices in probability and statistics). $\endgroup$
    – whuber
    Commented Sep 16, 2013 at 15:24
  • $\begingroup$ You are right @Whuber, I misread the comment. $\endgroup$
    – user25658
    Commented Sep 16, 2013 at 15:25

1 Answer 1


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 $n$!

  • $\begingroup$ Hi Whuber, many thanks for your help. Just to clarify my understanding, am I on the same page as you when I say: the probability of all $n$ people being all men should be $\frac{\binom{2n}{n}}{\binom{2n}{n} + \binom{2n-1}{n}}$ (The total number of possible all men groups divided by the total number of possible all men groups + all women groups)? (however, I am left with a pretty unpleasant quotient - and can't seem to get a neat 2/3). Many thanks again! $\endgroup$ Commented Sep 17, 2013 at 12:30
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    $\begingroup$ I showed the answer must be $2/3$ in all cases: there is no calculation left to be done. You can indeed reduce the fraction you write down by expanding out all the factorials and canceling what you can: you will still get $2/3$ as the answer :-). This is a beautiful example of a "combinatorial proof." Other simple examples include showing $\binom{n}{k}=\binom{n}{n-k}$ by observing every $k$ subset of an $n$ set corresponds to an $n-k$ subset; and $\sum_{k=0}^n\binom{n}{k}=2^n$, which follows because both sides count all the subsets of an $n$ set. You don't have to do any arithmetic! $\endgroup$
    – whuber
    Commented Sep 17, 2013 at 14:47
  • $\begingroup$ @whuber: We don't have to count in the way you suggest correct? The point is that for each all female committee there are $2$ distinct all male committees? So for $n=2$, we could have done the counting as e.g.: $\{1,2 \} \to \{1,3 \}, \{2,4 \}$, $\{1,3 \} \to \{1,4 \}, \{2,3\}$, and $\{2,3\} \to \{2,4 \}, \{1,3 \}$? $\endgroup$ Commented Sep 17, 2013 at 18:57
  • 1
    $\begingroup$ @guest Of course not. Once you know that one collection of subsets has twice the cardinality of the other (say, $m$ and $2m$), then you can compute there are $(2m)!(2!)^{-m}m!$ distinct two-to-one mappings. E.g., with $n=2$ we have $m=3$ and find $540$ of them. The point is that you can easily discover that one collection of subsets has twice the cardinality of the other by explicitly constructing a single two-to-one mapping. If you have found another simple way of constructing such a mapping, that's fine: it would work just as well. $\endgroup$
    – whuber
    Commented Sep 17, 2013 at 19:45
  • 1
    $\begingroup$ Yes, you make a good point: not many people would have the insight under exam conditions to develop this combinatorial answer. (I confess I found it only after doing the arithmetic you describe and seeing the answer was always the same!) It was precisely the existence of such an interesting approach, leading to such a non-intuitive answer, that moved me to respond in the first place. If you want to make a final check on your logic, do the arithmetic to show that your quotient of binomial coefficients really does reduce to $2/3$ :-). $\endgroup$
    – whuber
    Commented Sep 17, 2013 at 21:13

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