According to a study published in a reputable science magazi

According to a study published in a reputable science magazine, about 7 women in 100,000 have cervical cancer(C), so P(C) = 0.00007. 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? P(have C AND test positive = (Round to six decimal places as needed.)

Solution

Let P(A | B) denote P(A given B)

We know that P(A | B) =P(A,B) / P(B) => P(A,B)=P(A | B)*P(B)

P(C and positive)=P(C, positive) = P(positive | C)*P(C)

=0.84*0.00007=0.0000588=0.000059 (approx)