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HW3_Solutions

# HW3_Solutions - PSTAT 120A Spring 2010 HW 3 Solutions...

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Unformatted text preview: PSTAT 120A Spring 2010 HW 3 Solutions 1. (Ross 3.15) Note that percentages represented as a decimal between 0 and 1 will obey essentially the same properties as probabilities, because if a person is picked randomly, then the probability that they have some property A is the percentage of people who have property A . Let M be the event that a randomly picked person is a smoker, and E that that person has an ectopic pregnancy. The percentage of smokers amongst the ectopic pregnancies is then obtained by using Bayes’ formula P ( M | E ) = P ( E | M ) P ( M ) P ( E | M ) P ( M ) + P ( E | M c ) P ( M c ) = (2 x )( . 32) (2 x )( . 32) + x (1- . 32) = . 64 1 . 32 = 0 . 48 , where x := P ( E | M c ). Note that there is no way to figure out the value of x given the information, but it’s value is not needed to solve for P ( M | E ). 2. (Ross 3.20) Let F be the event that a randomly selected student is female, C that the student is a computer science major....
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HW3_Solutions - PSTAT 120A Spring 2010 HW 3 Solutions...

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