HW3-sol (Math 425) - Math 425 Section 007 Homework 3...

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Math 425 – Section 007 Homework 3 – Chapter 3, problems 7, 10, 15, 26, 30, 39, 57, 62, 70 Problem 15. An ectopic pregnancy is twice as likely to develop when the preg- nant woman is a smoker as it is when she is a nonsmoker. If 32% of women of childbearing age are smokers, what percentage of women having ectopic preg- nancies are smokers? Solution. Let E denote the event that a woman has ectopic pregnancy and S be the event that the woman is a smoker. By Bayes’ formula: P ( S | E ) = P ( E | S ) P ( S ) P ( E | S ) P ( S ) + P ( E | S c ) P ( S c ) Now, we don’t know P ( E | S ) or P ( E | S c ), but we know the that P ( E | S ) P ( E | S c ) = 2. Dividing the numerator and denominator by P ( E | S c ), we get: P ( S | E ) = P ( S ) P ( S ) + P ( E | S c ) P ( E | S ) P ( S c ) = 0 . 32 0 . 32 + 1 2 × (1 - 0 . 32) 0 . 484 Thus, 48.4% of women having ectopic pregnancies are smokers. Problem 39. An insurance company believes that people can be divided into two classes: accident prone and non-accident prone. An AP person will have an
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This note was uploaded on 04/02/2008 for the course MATH 425 taught by Professor Buckingham during the Winter '08 term at University of Michigan.

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HW3-sol (Math 425) - Math 425 Section 007 Homework 3...

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