# hw2-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...

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Unformatted text preview: AMS 311 (Fall, 2010) Joe Mitchell PROBABILITY THEORY Homework Set # 2 – Solution Notes (1). (14 points) Consider an experiment in which three fair dice are rolled. What is the conditional probability that at least one lands on 6, given that the sum of the three dice is at least 15? We consider the sample space, S = { (1 , 1 , 1) , (1 , 1 , 2) ,... } , to be the set of all 6 3 equally likely outcomes (triples of integers between 1 and 6), where, e.g., (2,4,4) is the outcome in which the first die lands 2, the second lands 4, and the third lands 4. Let E = { (6 , 1 , 1) , (6 , 1 , 2) ,..., (6 , 6 , 6) , (1 , 6 , 1) , (1 , 6 , 2) ,... } be the event that at least one of the two dice lands on 6. Note that | E | = 1 + 3 · 5 + 3 · 25 = 91, so P ( E ) = 91 / 216. (Reasoning: there is 1 outcome with three 6’s, 3 · 5 outcomes of the form “66x” (3 choices for position of the “x”, and 5 choices for “x”), 3 · 5 2 outcomes of the form “6xy” (3 choices for position of the “6”, and 5 2 = 25 choices for the digits “x” and “y”, each between 1 and 5).) Let F = { (4 , 5 , 6) , (4 , 6 , 5) , (5 , 5 , 5) , (5 , 5 , 6) , (5 , 6 , 5) , (5 , 6 , 6) , (6 , 4 , 5) ,... , (6 , 6 , 6) } be the event that the sum of the three dice is at least 15. Note that | F | = 20, so P ( F ) = 20 / 216 = 5 / 54. (To do the counting, either list the entries explicitly, or note that there are 3 outcomes using the digits “366”, 3! outcomes using the digits “456”, 3 with digits “466”, 1 with “555”, 3 with “556”, 3 with “566”, and 1 with “666”.) Now, E ∩ F = F \ { (5 , 5 , 5) } , since all elements of F have at least one 6, except the one outcome (5,5,5). Thus, P ( E ∩ F ) = (20 − 1) / 216 = 19 / 216. We want to compute P ( E | F ) = P ( E ∩ F ) P ( F ) = P ( E ∩ F ) P ( F ) = 19 / 216 20 / 216 = 19 20 (2). (14 points) A poll shows that 41 percent of the women and 17 percent of the men that took a weight-loss class kept the weight off for at least one year after completing the class. These people then attended a success party at the end of a year. If 72 percent of the original class were male, (a) what percentage of those attending the party were women? (b) what percentage of the original class attended the party? Select a random member of the class. Let A be the event that this member attends the party (i.e., this member kept the weight off for at least a year). Let W be the event that this member is a woman. We are told that P ( A | W ) = 0 . 41 and P ( A | W c ) = 0 . 17, and that P ( W ) = 1 − . 72 = 0 . 28. (a). We want to compute P ( W | A ) = P ( W ∩ A ) P ( A ) = P ( A | W ) P ( W ) P ( A | W ) P ( W ) + P ( A | W c ) P ( W c ) = ( . 41)( . 28) ( . 41)( . 28) + ( . 17)( . 72) ≈ . 484 Thus, 48.4% of the attendees were women....
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## This note was uploaded on 07/15/2011 for the course AMS 311 taught by Professor Tucker,a during the Spring '08 term at SUNY Stony Brook.

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hw2-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...

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