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Old 05-01-2005, 01:18 AM
gaming_mouse gaming_mouse is offline
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Join Date: Oct 2004
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Default Re: Probability question...

You could get away with a single integral here, if I'm interpreting the assumptions right:

P(A<B,A<C,A<D)=E[P(A<B,A<C,A<D | A)]=E[f(A)]


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Jason, I don't follow this. Are you saying you can just multiply the individual probabilities together? I could be wrong, but for some reason that didn't seem correct to me.

Also, would you mind taking a look at that other thread I started, down below?

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