# Random variable and probability distribution: (5 men and 5 women are ranked)

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?

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 $\frac{5}{10}\cdot\frac{4}{9}\cdot \frac{5}{8}$.
• 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? – roob1n Nov 11 '15 at 21:39