# Question: About 8 women in 100 000 have cervical cancer C so

About 8 women in 100,000 have cervical cancer (C), so P(C) = 0.00008 and P(no C) = 0.99992. The chance that a Pap smear will incorrectly indicate that a woman without cervical cancer has cervical cancer is 0.03. Therefore,

P(test pos | no C) = 0.03

What is the probability that a randomly chosen women who has this test will both be free of cervical cancer and test positive for cervical cancer (a false positive)?

P(test pos | no C) = 0.03

What is the probability that a randomly chosen women who has this test will both be free of cervical cancer and test positive for cervical cancer (a false positive)?

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