Re: Probability question...
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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)]
f(x)=P(x<B)P(x<C)P(x<D)
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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?
Thanks,
gm
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