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A committee consists of five Mexicans, two Asians, three African Americans, and two Caucasians. A subcommittee of 4* is chosen at random. What is the probability that all the ethnic groups are represented on the subcommittee?

Note: original posting said 5 but I meant 4. Also, I know, the race thing is weird, but it was in my book.

It seems to me like the answer should be $$\frac{\frac{5*2*3*2}{4!}}{\binom{12}{4}}$$ because we want both the numerator and the denominator to be unordered selection.

Alternatively, we would say that it is $$\frac{5*2*3*2}{12*11*10*9}$$ However, something in my reasoning is incorrect as the solution is just $$\frac{5*2*3*2}{\binom{12}{4}}$$ I am dreadfully confused.

source: http://www.math.illinois.edu/~psdey/stat20SU07/Solutions2_2007.pdf

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  • $\begingroup$ One good way to get unconfused is to look closely at a simpler version of the question. You might be able to make quick work of writing down all the possible ways of picking one Mexican and one Asian in a committee of size $2$ from a group of two Mexicans and one Asian, for instance. You should easily discover why your formula gives a value that's too small. $\endgroup$ – whuber♦ Apr 27 '17 at 22:01
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If we think of seating the people in 4 seats and insist that the first seat be occupied by a Mexican, the second by an Asian, the third by an African American and the fourth by a Caucasian, then the chance of this is: 5/12 x 2/11 x 3/10 x 2/9. In the problem you have set we don't care which person sits in which seat so long as one of each is present. There are 4! ways of re-arranging the people in the seats, so the answer to your question is 4! times the product I just gave, which once you re-arrange it is the same as the answer given in the book.

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  • $\begingroup$ this is a much clearer way to think about it $\endgroup$ – user2879934 Apr 28 '17 at 1:08
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There are $\binom{12}{5}$ possible committees.

Committees of $5$ that include at least one member from each group necessarily have just one group having two members on the committee, all other groups have just one. So, the number of such committees is $$\binom{5}{2}\cdot 2\cdot 3\cdot 2 + 5\cdot \binom{2}{2}\cdot 3\cdot 2 + 5\cdot 2 \cdot \binom{3}{2}\cdot 2 + 5\cdot 2 \cdot 3 \cdot \binom{2}{2}$$

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  • $\begingroup$ Sorry that is not what I am asking, please read the edit $\endgroup$ – user2879934 Apr 27 '17 at 15:45
  • $\begingroup$ @user2879934 I posted it before reading the edited version of your question (you made the edit while I was writing my answer.) $\endgroup$ – Dilip Sarwate Apr 27 '17 at 15:50
  • $\begingroup$ no problem, it was my bad $\endgroup$ – user2879934 Apr 27 '17 at 15:51
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The numerator is the number of ways to choose $4$ people from $12$ people such that exactly one person is chosen from each ethnic group. The numerator is thus ${5 \choose 1} {2 \choose 1}{3 \choose 2}{2 \choose 2}$. The denominator is clearly ${12 \choose 4}$.

Alternatively, you could assume that the subcommittee has a name for each of the positions - say president, vice president, secretary and treasurer. Then, the numerator would be ${5 \choose 1} {2 \choose 1}{3 \choose 2}{2 \choose 2} 4!$ because once the 4 people are chosen so that each ethnic group is represented, they can be placed in various positions in $4!$ ways. For the same reason, the denominator is ${12 \choose 4} 4!$. The $4!$'s cancel out and we get the same probability.

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