hw09sol

# hw09sol - Math 55 Discrete Mathematics Fall 2008 Homework 9...

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Math 55: Discrete Mathematics, Fall 2008 Homework 9 Solutions * 6.3: 4. Let F be the event that Ann selects an orange ball, E the event that she chooses box 2. Then p ( E ) = 1 / 2, p ( F | E ) = 5 / 11, p ( F | E ) = 3 / 7. Bayes’ Theorem gives p ( E | F ) = (5 / 11)(1 / 2) / ((5 / 11)(1 / 2) + (3 / 7)(1 / 2)) = 35 / 68. 10. (a) ( . 97)( . 04) / (( . 97)( . 04) + ( . 02)( . 96)) 0 . 669 (b) ( . 02)( . 96) / (( . 97)( . 04) + ( . 02)( . 96)) . 331 (c) ( . 03)( . 04) / (( . 03)( . 04) + ( . 98)( . 96)) . 00127 (d) ( . 98)( . 96) / (( . 03)( . 04) + ( . 98)( . 96)) . 9987 12. (a) ( . 9)(2 / 3) + ( . 2)(1 / 3) = 2 / 3 (b) p (send 0 | receive 0) = p (receive 0 | send 0) p (send 0) /p (receive 0) = . 9(2 / 3) / (2 / 3) = . 9 14. (3 / 8)(1 / 2) / ((2 / 7)(1 / 6) + (3 / 8)(1 / 2) + (1 / 2)(1 / 3)) = 7 / 15 6.4: 4. 10( . 6) = 6 8. 3(7 / 2) = 21 / 2 16. p ( X = 0) = 1 / 4, p ( Y = 0) = 1 / 4, but p ( X = 0 | Y = 0) = 0 6 = (1 / 4)(1 / 4). You could also use other values of these random variables to give a counterexample to independence. * 30. Let X be the number of tails on n flips. Let X i be the indicator variable for tails on the i -th flip. Then E ( X i ) = . 4, hence E ( X ) = ( . 4) n . The flips are independent, and V ( X i ) = . 4 - ( . 4) 2 = . 24, so V ( X ) = ( . 24) n . By Chebyshev, p ( | X - ( . 4) n | ≥ n ) ( . 24) n/ ( n ) 2 = . 24. * 42. m/n . There are two ways to solve this problem using linearity of expectation: (i)
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