hw3_sol - IEOR 172: Probability and Risk Analysis for...

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IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007 Homework 3 Solution Chapter 3 Question 58 (a) P(head) = P(head | HH )P( HH ) + P(head | HT )P( HT ) +P(head | TH )P( TH ) + P(head | TT )P( TT ) = P(head) p 2 + 0 · p (1 - p ) + 1 · (1 - p ) p + P(head)(1 - p ) 2 then P(head) = p (1 - p ) 1 - p 2 - (1 - p ) 2 = 1 2 , and P(tail) = 1 - P(head) = . 5. (b) If the first flip is H , then the procedure continues until we get a T , and the result is T ; simi- larly, if the first flip is T , then the result is H . Therefore, P(head) = P( T ) = 1 - p and P(tail) = p , so the result is not equally likely to be either heads or tail. Theoretical exercise Question 21 (a) For n = 2, m = 1, voting order AAB : P 2 , 1 = 1 3 . For n = 3, m = 1, voting order AAAB , AABA : P 3 , 1 = 2 4 = 1 2 . For n = 3, m = 1, voting order AAAB , AABA : P 3 , 1 = 2 4 = 1 2 . For n = 3, m = 2, voting order AAABB , AABAB : P 3 , 2 = 2 10 = 1 5 . For
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This note was uploaded on 04/02/2008 for the course IEOR 172 taught by Professor Righter during the Fall '07 term at University of California, Berkeley.

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hw3_sol - IEOR 172: Probability and Risk Analysis for...

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