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SilasBarta comments on Case study: abuse of frequentist statistics - Less Wrong
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P(E) = P(E|H) P(H) + P(E|~H)P(~H)
The quantities P(H), P(E|H), and P(E|~H) are in general independent of each other, in the sense that you can move any one of them without changing the others -- provided you adjust P(E) accordingly.
Thanks, that helps. See how I apply that point in my reply to Psy-Kosh here.