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# HW12Answers(4.1, 4.2) - Math331 Spring 2008 Instructor...

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Math331, Spring 2008 Instructor: David Anderson Section 4.2 Homework Answers Homework: pgs. 150-158 #’s 1, 4, 5, 6, 7, 16. 1. The possible values of X are { 0 , 1 , 2 , 3 , 4 , 5 } . To find the probabilities of each, we note that all combinations of rolls in the sample space S = { ( a,b ) | a,b ∈ { 1 , 2 , 3 , 4 , 5 , 6 }} are equiprob- able. So we need to count the number of ways to get each of the values in { 0 , 1 , 2 , 3 , 4 , 5 } , and divide by 36. To get 0, both die must be the same. Therefore, there are exactly 6 ways of doing this. So P { X = 0 } = 6 / 36 = 1 / 6. To have X = 1, we think in the following way. If the first roll is a 2, 3, 4, or 5, there are always two rolls possible on the second roll that gives us X = 1. However, if the first roll is 1 or 6, there is only one possibility for the second roll. Thus, P { X = 1 } = 10 / 36. Similarly, if the first roll is a 3 or 4, there are two possibilities for the second roll that gives X = 2, and a 1, 2, 5, or 6 gives only one possibility for the second roll. Thus, P { X = 2 } = 8 / 36. Similarly, P { X = 3 } = 6 / 36, P { X = 4 } = 4 / 36 (a roll of 3 or 4 on the first roll makes this impossible), and P

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