hw4_sol

hw4_sol - IEOR 172 Probability and Risk Analysis for...

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Unformatted text preview: IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007 Homework 4 Solution Chapter 4 Question 7 (a) { 1 , 2 , 3 , 4 , 5 , 6 } (b) { 1 , 2 , 3 , 4 , 5 , 6 } (c) { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } (d) {- 5 ,- 4 ,- 3 ,- 2 ,- 1 , , 1 , 2 , 3 , 4 , 5 } Question 8 (a) P( X = 1) = 1 36 , P( X = 2) = 1 12 , P( X = 3) = 5 36 , P( X = 4) = 7 36 , P( X = 5) = 1 4 , P( X = 6) = 11 36 . (b) P( X = 1) = 11 36 , P( X = 2) = 1 4 , P( X = 3) = 7 36 , P( X = 4) = 5 36 , P( X = 5) = 1 12 , P( X = 6) = 1 36 . (c) P( X = 2) = 1 36 , P( X = 3) = 1 18 , P( X = 4) = 1 12 , P( X = 5) = 1 9 , P( X = 6) = 5 36 , P( X = 7) = 1 6 , P( X = 8) = 5 36 , P( X = 9) = 1 9 , P( X = 10) = 1 12 , P( X = 11) = 1 18 , P( X = 12) = 1 36 . (d) P( X =- 5) = 1 36 , P( X =- 4) = 1 18 , P( X =- 3) = 1 12 , P( X =- 2) = 1 9 , P( X =- 1) = 5 36 , P( X = 0) = 1 6 , P( X = 1) = 5 36 , P( X = 2) = 1 9 , P( X = 3) = 1 12 , P( X = 4) = 1 18 , P( X = 5) = 1 36 . Question 15 p (1) = P { X = 1 } = 11 66 = . 1667 p (2) = P { X = 2 } = 10 66 · 11 56 + 9 66 · 11 57 + 8 66 · 11 58 + 7 66 · 11 59 + 6 66 · 11 60 + 5 66 · 11 61 1 + 4 66 · 11 62 + 3 66 · 11 63 + 2 66 · 11 64 + 1 66 · 11 65 = . 1556 p (3) = P { X = 3 } = 10 66 9 56 · 11 47 + ··· + 1 56 · 11 55 + ··· + 1 66 10 65 · 11 55 + · + 2 65 · 11 63 = . 1434 p (4) = P { X = 4 } = 1...
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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 Berkeley.

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hw4_sol - IEOR 172 Probability and Risk Analysis for...

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