hypergeometric/permutation problem

br00547

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Nov 7, 2011
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Hello,

I am totally stumped, and hope that someone can point me in the right direction.

Imagine that there are 70 applicants for a job. Four committee members independently create a top-ten list from the applicant pool.

I believe there are 70! / (70 - 10)! possible lists. But what I am stumped by is the question of the probability of there being differing amounts of overlap between the lists, assuming that they were created randomly.

For example, what is the probability that only one of the 70 applicants will appear on all fours lists, assuming they were created independently and randomly? Or what is the probability that no applicants will appear on all four lists (independently and randomly created)? I've been drawing multinomial trees all morning for scaled-down versions of this situation to see if I can generate a solution that I can scale up, but to no avail. So I'm hoping that someone out there can either solve this for me or at least point me in the right direction.

Thank you!
 
Imagine that there are 70 applicants for a job. Four committee members independently create a top-ten list from the applicant pool.

I believe there are 70! / (70 - 10)! possible lists. But what I am stumped by is the question of the probability of there being differing amounts of overlap between the lists, assuming that they were created randomly.

For example, what is the probability that only one of the 70 applicants will appear on all fours lists, assuming they were created independently and randomly? Or what is the probability that no applicants will appear on all four lists (independently and randomly created)? I've been drawing multinomial trees all morning for scaled-down versions of this situation to see if I can generate a solution that I can scale up, but to no avail. So I'm hoping that someone out there can either solve this for me or at least point me in the right direction.
I would need to know the exact question to be really confident of an answer.

However as written, it does not say that the committee members were to 'rank' each applicant, only make a list of the top ten.
So that makes this a combination as opposed to a permutation.
If that is correct then there are \(\displaystyle \dfrac{70!}{(10!)^4(30!)}\) ways to make four disjoint lists.
Whereas there are \(\displaystyle \left( {\frac{{70!}}{{10! \cdot 60!}}} \right)^4 \) to make four lists.

AGAIN! these are not ordered list, just content lists.
 
Thank you!

The top-ten lists would be ranked, but even this answer to the unranked version of the question gives me some insight into how to solve the problem.

I would need to know the exact question to be really confident of an answer.

However as written, it does not say that the committee members were to 'rank' each applicant, only make a list of the top ten.
So that makes this a combination as opposed to a permutation.
If that is correct then there are \(\displaystyle \dfrac{70!}{(10!)^4(30!)}\) ways to make four disjoint lists.
Whereas there are \(\displaystyle \left( {\frac{{70!}}{{10! \cdot 60!}}} \right)^4 \) to make four lists.

AGAIN! these are not ordered list, just content lists.
 
The top-ten lists would be ranked, but even this answer to the unranked version of the question gives me some insight into how to solve the problem.
But do you not see that it makes no difference.
If the question is "What if the same person is at the top of each list?" is quite different from the question "What if the same person is on the each of the lists?"

Thus you must post the exact wording of the question.
Otherwise, there is no exact solution.
 
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