This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 8 (due March 30) 6.3, #2: We use Bayes Theorem: p ( E  F ) = p ( F  E ) p ( E ) p ( F ) = 5 9 . 6.3, #4: Let E be the event that Ann selects an orange ball, and F the event that she has chosen the second box. We are then given that p ( F ) = 1 2 , p ( E  F ) = 5 11 and p ( E  F ) = 3 7 , and we want to find p ( F  E ). Applying Bayes Theorem, we find p ( F  E ) = 35 68 . 6.3, #6: Let E be the event that the player tests positive, and F the event that he takes steroids. We want p ( F  E ), and we are given p ( F ) = . 05, p ( E  F ) = . 98 and p ( E  F ) = . 12. Applying Bayes Theorem, we find p ( F  E ) . 301, so 30 . 1% of the players who test positive are in fact taking steroids. 6.3, #10: Let E be the event that a patient tests positive, and F the event that they have the bird flu. We are given that p ( F ) = . 04, p ( E  F ) = . 97 and p ( E  F ) = . 02. a) We want to know p ( F  E ); by Bayes Theorem, this is 66 . 9%. b) We want p ( F  E ); since this and p ( F  E ) sum to 1, this is 33 . 1%. c) We want p ( F  E ); by Bayes Theorem, this is . 13%. d) We want p ( F  E ); this is 99 . 87%. 6.3, #11: Let E be the event that success is predicted, and F the event of an actual success. We see p ( F ) = . 6, p ( E  F ) = . 7 and p ( E  F ) =...
View Full
Document
 Spring '08
 STRAIN
 Math

Click to edit the document details