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#26 2013-03-23 21:51:19

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,247

Re: probability question help me

Look at the formula for no one getting drunk. The (k-1)! is the number of them.

Now if you tell me what you are trying to do I can suggest a good means of computing the numbers.

Last edited by bobbym (2013-03-23 22:33:11)


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#27 2013-03-23 23:13:58

NobodyButYou
Member
Registered: 2013-03-23
Posts: 8

Re: probability question help me

I am trying to understand the logic behind the terms that's all.

Summation from j= 0 to n of (-1)^j/j! in the numerator is not getting into my head.

As far as my understanding goes, it computes the number of ways you can give out n names to n people without any of them having their own name.

Why do we need to multiply the summation in the numerator with P(no one getting drunk)?

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#28 2013-03-23 23:21:29

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,247

Re: probability question help me

All the forms there were obtained empirically. By looking at the data. There just was no other way that I could solve the problem,

As far as my understanding goes, it computes the number of ways you can give out n names to n people without any of them having their own name.

That is the number of derangements which is always an integer. Your sum there is not always an integer.

Last edited by bobbym (2013-03-24 01:21:52)


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Offline

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