3
$\begingroup$

My sister posed this question to me and I'm a little surprised to be struggling so much with it.

8 men and 8 women choose balls from a bin that contains 8 blue balls and 8 red balls. After the selection is done, what is the probability of that 4 men have a red ball and 4 men have a blue ball?

I was able to brute force the problem for 2 men, 2 women, 2 red and 2 blue balls (answer is 2/3) and I wrote a little script to brute force the answer for 4 men, 4 women, 4 red and 4 blue balls (answer is 20736 / 40320).

I can't figure out how to generalize the count though. Any help would be much appreciated.

$\endgroup$
3
$\begingroup$

You are distributing $8$ red balls among $8$ men and $8$ women (you can consider the blue ones as "not getting a red ball"). So the question is to check in how many ways you can select $4$ men and $4$ women, i e.:

$$\binom{8}{4} \cdot \binom{8}{4} = 4900$$

$\endgroup$
3
$\begingroup$

Imagine the $8$ balls to be distributed among the men are chosen at random, with all choices equally likely. There are $\binom{16}{8}$ ways to choose these $8$ balls. And there are $\binom{8}{4}\binom{8}{4}$ ways for the men to choose $4$ red and $4$ blue. Divide.

$\endgroup$
  • $\begingroup$ That gives a fraction, so is clearly wrong. $\endgroup$ – vonbrand Nov 8 '15 at 0:53
  • 2
    $\begingroup$ @vonbrand: OP asked for the probability. $\endgroup$ – André Nicolas Nov 8 '15 at 0:54
  • $\begingroup$ My bad.$$$$$$$$ $\endgroup$ – vonbrand Nov 8 '15 at 0:55
  • $\begingroup$ He did also ask for the count, although they're obviously very closely related. ;-) $\endgroup$ – Brian Tung Nov 8 '15 at 0:56
  • $\begingroup$ The count is in a sense not well defined, for we might for example in the numerator and denominator count the number of actual ways to distribute the balls among the people. Thar would blow up both counts by equal amounts. $\endgroup$ – André Nicolas Nov 8 '15 at 1:00
2
$\begingroup$

Suppose that there are $2n$ balls of each color, $2n$ men, and $2n$ women. There are $\binom{4n}{2n}$ sets of balls that the $2n$ men might collectively pick, and $\binom{2n}n^2$ of them consist of $n$ balls of each color. Each set of $2n$ balls is equally likely to be chosen by the men, so the desired probability is

$$\frac{\binom{2n}n^2}{\binom{4n}{2n}}\;.$$

When $n=4$, this is

$$\frac{\binom84^2}{\binom{16}8}=\frac{70^2}{12870}=\frac{4900}{12870}=\frac{490}{1287}\;.$$

When $n=2$ it’s

$$\frac{\binom42^2}{\binom84}=\frac{36}{70}=\frac{18}{35}\;.$$

$\endgroup$
  • $\begingroup$ No worries; I make many trivial errors like that. $\endgroup$ – Brian Tung Nov 8 '15 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.