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An exercise in my homework: 5 women and 5 men are ranked based on their results in a test. Assume that all results are different and every of 10! possible orders has the same probability.

X (random variable) is the place of the highest ranked woman on the test. So it's possible values are 1,2,3,4,5,6.

Find the probability distribution.

So my approach is: $$\frac{(x-1)! \cdot 5 \cdot \left({(10-x)!}\over{((10-x)-(5-x+1))!}\right) \cdot 4!}{10!}$$

My thoughts:

  • $10!$ : are the possible cases
  • $(x-1)!$ : are the possibilities to sort every man that is higher ranked than the first woman.
  • 5 : there are five woman so every woman can be the highest ranked.
  • $\left({(10-x)!}\over{((10-x)-(5-x+1))!}\right) $ : all men behind the best woman can be placed on all the left places (not taken by the best woman or better men).
  • $4!$ : are the possibilities to place the last 4 women on the left places.

To calculate each probability. Now my questions are:

  1. Am I right?

  2. Is there a simpler way to achieve this / can I simplify it?

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1 Answer 1

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The probability that $X=1$ is $1/2$ by symmetry.

The probability that the top ranked woman is second overall is the probability that the first is male and the second is female. This is $\frac{5}{10}\cdot\frac{5}{9}$.

The probability that the top ranked woman is third overall is the probability that the first two are male and the third is female. This is $\frac{5}{10}\cdot\frac{4}{9}\cdot \frac{5}{8}$.

Continue.

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  • $\begingroup$ I'm sorry, I had to ajust the expression. $\left({(10-x)!}\over{((10-x)-(5-x+1))!}\right)$ is a partial permutation, where I choose the lower ranked men out of the left places. And $4!$ is actually $4!\over0!$ and is the same, i have $4!$ possibilities to place 4 women on the left 4 places. Now it also works for $x=1$. Regarding your answer, that's a much simple way. Am I thinking too far? Is it a trivial problem? $\endgroup$
    – roob1n
    Commented Nov 11, 2015 at 21:39
  • $\begingroup$ Yours is a reasonable counting approach. If you are going to count, it may be simpler to count the number of ways to assign positions to the women. That makes the sample space smaller. $\endgroup$
    – André Nicolas
    Commented Nov 11, 2015 at 21:51

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