# Probability of a random variable

Shackman

## Homework Statement

5 men and 5 women are ranked according to exam scores. Assume no two scores are the same and each 10! rankings are equally likely. Let random variable X denote the highest ranking achieved by a woman e.g. X=2 means the highest test score was achieved by 1 of the 5 men and the second highest by 1 of the 5 women. Find P{X=i}, i=1,2,3,...,10.

## The Attempt at a Solution

I am quite lost on this problem, I have a very poor background in statistics so I'm not sure if there is a specific formula I should be using or what the most efficient way of doing this is, but P{X=i}, i=7,8,9,10 is 0 because there are only 5 men so the lowest a woman can be ranked is 6th place. I think P(X=1) = 5* 9!/10!, P(X=2)=5*8!/10!, P(X=3)=5*7!/10!, P(X=4)=5*6!/10!, P(X=5)=5*5!/10! and P(X=6)=5*4!/10! since there are 5*9! different rankings (9! combinations times 5 different possible women ranked at that position) with a woman ranked first and so on. So is P(X=i), i=1,2,...,10 the summation of these probabilities or P(X=i)=.567 (assuming they are correct)?

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## Answers and Replies

PingPong
Hi Shackman,

P(X=1): Choose a woman and place her at the top (there are nCr(5,1)=5 choices possible here). Then there are 9! permutations of the remaining people, for a total of 5*9! configurations.

P(X=2): Choose one male (again 5 choices here), place him at the top, and one woman (5 choices again) and place her at second. There are then 8! permutations of the remaining people, for a total of 5^2*8! combinations.

So it looks like you were on the right track for the first few. Repeat this type of procedure for the rest choosing (X-1) men and placing them above the woman, choosing one woman and placing her at the X spot, then permuting the leftovers. However, you must take into account that you must also permute the number of men that you place before the women!

If you want to check, the probabilities I get are

X=1: 50%
X=2: 27.78%
X=3: 13.9%
X=4: 5.95%
X=5: 1.98%
X=6: 0.397%
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Total probability: 100.07% (I'm off due to rounding...)