MATH 211 JAN 2018 9002(LEE Test: Final Exam Part l This Question: 4 pts too Cai

Question

MATH 211 JAN 2018 9002(LEE Test: Final Exam Part l This Question: 4 pts too Cai 3/8/18 1:01 PM Time 5 of 17 (3 This Test: 100 pts According to a study published in a reputable science magazine, about 7 women in 100,000 have cervical cancer (C), so PIC) present is 0.89. Therefore, Pttest posjc) Prave CAND test posive)«O(Round to six de mal places as needed.) 0.00007. Suppose the chance that a Pap smear will detect cervical cancer when ti )-0.89. What is the probability that a randomly chosen woman who has this test wil both have cervical cancer AND test positive for it? 8

Explanation / Answer

P(C and test positive) = P(C) * P(test positive) [as both the events are not related to each other]

P(a n b) = P(a) * P(b) [if a and b are not related to each other]

=> P(C and test positive) = 0.00007 * 0.89 = 0.0000623

Answer = 0.000062 [round to six decimal places]

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