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Let's say there are $5$ man and $5$ women running in a race. Let the variable X that indicates the highest position in ranking for a woman. So, for example, if $X = 1$, that means that a woman took first place(any woman) and if $X = 6$, that means that women took the last places on the ranking.

I guess, obviously there are $10!$ different outcomes for ranking. How can I calculate $p_X(x)$ with $x = 1, 2, 3, 4, 5, 6$?

Edit : Two or more people cannot take the same place in the ranking. All 10! different outcomes are equally possible.

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    $\begingroup$ Are we to suppose that each ranking is equiprobable? (which of course is completely unlikely for a real race with real people!) $\endgroup$ – Bram28 Mar 24 '18 at 20:58
  • $\begingroup$ The women are assumed to be indistinguishable. There are ${10 \choose 5}$ ways to place them in $10$ empty slots. You're correct that $6$ is highest value $X$ can take on. $\endgroup$ – Remy Mar 24 '18 at 20:59
  • $\begingroup$ You didn't really answer Bram28's question. Are all of the $10!$ possible outcomes considered equally likely? $\endgroup$ – saulspatz Mar 24 '18 at 21:01
  • $\begingroup$ I wasn't responding to Bram28 $\endgroup$ – Remy Mar 24 '18 at 21:02
  • $\begingroup$ @Bram28 Yes, all 10! outcomes are equally possible. I will edit the post. $\endgroup$ – Κωνσταντίνος Κορναράκης Mar 24 '18 at 21:06
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The women are assumed to be indistinguishable. There are ${10 \choose 5}$ ways to place them in $10$ empty slots. You're correct that $6$ is highest value $X$ can take on.

If we are assuming every one of the $10$ people are equally fast, then the probability that a woman is in the first position is $\frac{1}{2}$ by symmetry.

The probability that the best position by a woman is $X=2$ is $$\frac{5}{10}\cdot\frac{5}{9}=\frac{{5 \choose 1}}{{10 \choose 1}}\cdot\frac{{5 \choose 1}}{{9 \choose 1}}$$

since a man has to be in the first position and then a woman has to be in the second position

The probability that the best position by a woman is $X=3$ is $$\frac{5}{10}\cdot\frac{4}{9}\cdot\frac{5}{8}=\frac{{5 \choose 2}}{{10 \choose 2}}\cdot\frac{{5 \choose 1}}{{8 \choose 1}}$$

since a man has to be in the first and second position and then a woman has to be in the third position

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The probability that the best position by a woman is $X=6$ is $$\frac{5}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{1}{6}\cdot\frac{5}{5}=\frac{{5 \choose 5}}{{10 \choose 5}}\cdot\frac{{5 \choose 1}}{{5 \choose 1}}$$

Can you go from here to obtain $p_X(x)$?

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Here is a solution where order matters. $P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)$

$=\frac{{5 \choose 1}\times9!}{10!}+\frac{{5 \choose 1}\times{5 \choose 1}\times8!}{10!}+\frac{{5 \choose 2}\times 2! \times{5 \choose 1}\times7!}{10!}+\frac{{5 \choose 3}\times 3! \times{5 \choose 1}\times6!}{10!}+\frac{{5 \choose 4}\times 4! \times{5 \choose 1}\times5!}{10!}\frac{{5 \choose 5}\times 5! \times{5 \choose 1}\times4!}{10!}$

$=\frac{5}{10}+\frac{5\times5}{10\times9}+\frac{5\times4\times5}{10\times9\times8}+\frac{5\times4\times3\times5}{10\times9\times8\times7}+\frac{5\times4\times3\times2\times5}{10\times9\times8\times7\times6}+\frac{5\times4\times3\times2\times1\times5}{10\times9\times8\times7\times6\times5}$

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