Solutions03

# Solutions03 - ENGRD 2700 Spring 09 Homework 3 Solutions...

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ENGRD 2700, Spring ’09 Homework 3 Solutions Homework 3 Solutions Problem 1 When two dice are rolled, there are 36 possible outcomes, which can be repre- sented as ordered integer pairs ( a, b ) with 1 a, b 6. (a) We need to check if P ( A B 1 ) = P ( A ) P ( B 1 ). We have A = { (4 , 1) , (4 , 2) , (4 , 3) , (4 , 4) , (4 , 5) , (4 , 6) } , B 1 = { (1 , 2) , (2 , 2) , (3 , 2) , (4 , 2) , (5 , 2) , (6 , 2) } , A B 1 = { (4 , 2) } . Since each of the 36 outcomes is equally likely to be realized, we have P ( A ) = 6 × 1 / 36 = 1 / 6 , P ( B 1 ) = 6 × 1 / 36 = 1 / 6 , P ( A B 1 ) = 1 / 36 . Thus we see that P ( A B 1 ) = 1 / 36 = P ( A ) P ( B 1 ), which means that A and B 1 are independent. (b) B 2 = { (1 , 2) , (2 , 1) } . Clearly, A B 2 = , which means that A and B 2 are disjoint. (c) We have P ( A ) = 1 / 6 , P ( B 2 ) = 2 / 36 = 1 / 18 , P ( A B 2 ) = P ( ) = 0. Therefore, P ( A B 2 ) = 0 6 = 1 / 108 = P ( A ) P ( B 2 ), so A and B 2 are not inde- pendent. (d) B 3 = { (3 , 6) , (6 , 3) , (4 , 5) , (5 , 4) } . So A B 3 = { (4 , 5) } 6 = , which means that A and B 3 are not disjoint. (e) We have P ( A ) = 1 / 6 , P ( B 3 ) = 4 / 36 = 1 / 9 , P ( A B 3 ) = 1 / 36. There- fore, P ( A B 3 ) = 1 / 36 6 = 1 / 54 = P ( A ) P ( B 3 ), so A and B 3 are not independent. (f) B 4 = { (1 , 6) , (6 , 1) , (2 , 5) , (5 , 2) , (3 , 4) , (4 , 3) } . So A B 4 = { (4 , 3) } 6 = , which means that A and B 4 are not disjoint. (g) We have P ( A ) = 1 / 6 , P ( B 4 ) = 6 / 36 = 1 / 6 , P ( A B 4 ) = 1 / 36. Therefore, P ( A B 4 ) = 1 / 36 = P ( A ) P ( B 4 ), so A and B 4 are independent. (h) If two events are not disjoint, then they may or may not be independent. If two events are not independent, then they may or may not be disjoint. In short, independence and disjointness are different things. Problem 2 When a fair coin is flipped three times, there are 2 3 = 8 possible outcomes, 1

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ENGRD 2700, Spring ’09 Homework 3 Solutions which can be represented as three-letter sequences consisting of H ’s and T ’s. (a) The given events are A = { HHT, HHH, TTH, TTT } , B = { THH, HHH, HTT, TTT } , C = { HTH, HHH, THT, TTT } . Since each of the 8 outcomes is equally likely to be realized, we have P ( A ) = P ( B ) = P ( C ) = 4 / 8 = 1 / 2.
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• Spring '05
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• Probability theory, possible values, route

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