According to a study published in a reputable science magazine, about 77 women i

Question

According to a study published in a reputable science magazine, about

77

women in 100,000 have cervical cancer (C), so

P(C)equals=0.000070.00007.

Suppose the chance that a Pap smear will detect cervical cancer when it is present is

0.830.83.

Therefore,

Upper P left parenthesis test pos|C right parenthesisP(test pos|C)equals=0.830.83.

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 0.83. Therefore, P(test posjC) 0.83. 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.)

Explanation / Answer

P(Have C and test Positive)

= P(Test positive | Have C) * P(Have C)

= 0.83 * 0.00007

= 0.000058

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