Question: According to a study published in Scientific American, about 8 women in 100,000 have cervical cancer (which well call event C), so P(C) = 0.00008.

According to a study published in Scientific American, about 8 women in 100,000 have cervical cancer (which we’ll call event C), so P(C) = 0.00008. Suppose the chance that a Pap smear will detect cervical cancer when it is present is 0.84. Therefore,
P(test pos | C) = 0.84
What is the probability that a randomly chosen woman who has this test will both have cervical cancer AND test positive for it?

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