Prove directly that P(E|F) = P(E|FG)P(G|F) + P(E|FGc)P(Gc|F)
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P(E|F) = P(E|FG)P(G|F) + P(E|FGc)P(Gc|F)
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PE F PEF PF PE FGPG F PEFG ...View the full answer
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