11ProbHw1sol

11ProbHw1sol - c 2 SIEO 3658 Assignment#1 Solutions 4(a...

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SIEO 3658 Assignment #1 Solutions Probability October 4, 2011 Prof. Mariana Olvera-Cravioto Page 1 of 2 Assignment #1 Solutions 1. (a) False (b) Possibly true (c) False (d) True (e) True (f) Possibly true (g) True 2. (a) The Venn diagrams are: A B A B A B A B A B A B A B E F E F E F E F E F E F E F E F G E F G E F G E F E F c ( E F ) ( E F c ) We conclude that ( E F ) ( E F c ) = E . (b) The Venn diagrams are: A B A B A B A B A B A B A B E F E F E F E F E F E F E F E F G E F G E F G E F E c F E F c ( E F ) ( E c F ) ( E F c ) We conclude that ( E F ) ( E c F ) ( E F c ) = E F . (c) The Venn diagrams are: A B A B A B A B A B A B A B E F E F E F E F E F E F E F E F G E F G E F G E F F G ( E F ) ( F G ) We conclude that ( E F ) ( F G ) = F ( E G ). 3. (a) E c F G c (b) ( E F ) G c (c) E F G (d) E c F c G c (e) ( E F G )
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Unformatted text preview: c 2 SIEO 3658, Assignment #1 Solutions 4. (a) True (b) Possibly true (c) True (d) False 5. 10% 6. 6 21 = 0 . 286 7. S = { RR,RG,RB,GR,GG,GB,BR,BG,BB } The probability of each point is 1 9 8. (a) Yes. A and B can be mutually exclusive. (b) No. Assume B and C are mutually exclusive. Then P ( B ∪ C ) = P ( B )+ P ( C ) = 1 . 1 > 1. This is a contradiction. 9. (a) 0 (b) 1 (c) 1 2 = 0 . 5 (d) 1 2 = 0 . 5 (e) 1 10 = 0 . 1...
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11ProbHw1sol - c 2 SIEO 3658 Assignment#1 Solutions 4(a...

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