hw8sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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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 ) =...
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hw8sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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