A large company employs 20 individuals as statisticians, 7 of whom are women and 13 of whom are men. No two people earn the same amount.
What is the probability that 6 of the women earn salaries below the median salary of the group?
If r is the number of "successes" in a set of N element, and x is the number of success in n elements draw, then p(x) = (r choose x) ( (n - r) choose (n - x) ) / (N choose n).
The Attempt at a Solution
I'm pretty sure that I just plug & chug with N = 20, n = 10, r = 7, x = 6. But I'm not 100% sure about the logic of all this. So, we look at the 10 persons below the median, who, for all we know, are just 10 random people from the group of 20. We know that there are 7 women in the group of 20, and we want to know the probability that 6 of our 10 random persons are women. Is that right? I feel a little thrown-off by the "median" thing.