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January 31

Curious Sample Spaces

Hello again; I was wondering if somebody could help me figure out via extrapolation from a previous reference desk question how to define a finite probabilistic sample space in terms of a multiset so that I may create a piecewise probability mass function that would evaluate either the classical probability $\left.P\right|_{S}\!\left(E\right)=\!\;^{\left|E\right|}\!/_{\left|S\right|}$ derived here or the empirical probability $\left.P\right|_{S}\!\left(E\right)=\!\;^{f_{E}}\!/_{\sum _{E\in S}f_{E}}$ based on whether the multiplicity of any of this probabilistic sample space's member events is to 1 or any larger integer, respectively. I ask this not only out of my own curiosity but also because I anticipate that the resulting probability space would make the most mathematical sense to me both in its implementation and in the operations that such an implementation would generate even though defining it is not explicitly required in the context of the homework which I am finishing independently of the Probability and Statistics class that I was supposed to have finished last semester. I would be very grateful if somebody could explain to me the material necessary for me to figure this out.
Many thanks in advance,
RandomDSdevel (talk) 20:42, 31 January 2013 (UTC)[reply]

Probability question

I'm looking for a probability formula. I asked someone in a chat room and they pointed me towards binomial probability, but it doesn't work when you don't replace the objects.

The problem is this: if you have U cards, V of which help you, if you draw X of them without replacement, what's the chance that at least Y of the ones you drew will help you.

If you do this with binomial probability it doesn't take into account that the cards aren't replaced once they are drawn. Help! :( 70.173.177.190 (talk) 21:46, 31 January 2013 (UTC)[reply]

I think this might be helpful to you Hypergeometric distribution ---- _nonsense ferret 22:36, 31 January 2013 (UTC)[reply]

Exactly what I was after. Thank you so much! 70.173.177.190 (talk) 23:24, 31 January 2013 (UTC)[reply]