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Unformatted text preview: EE 351K Probability, Statistics, and Random Processes SPRING 2012 Instructor:S.Shakkottai firstname.lastname@example.org Homework 1 Solutions Problem 1 We are given that P ( A ) = 0 . 6 , P ( B c ) = 0 . 45 , and P ( A B ) = 0 . 85 . Determine P ( B ) and P ( A B ) . Solution : We have P ( B ) = 1- P ( B c ) = 1- . 45 = 0 . 55 . Also, by rearranging the formula P ( A B ) = P ( A ) + P ( B )- P ( A B ) , we obtain P ( A B ) = P ( A ) + P ( B )- P ( A B ) = 0 . 6 + 0 . 55- . 85 = 0 . 3 . Problem 2 Let A and B be two sets. (a) Show that ( A c B c ) c = A B and ( A c B c ) c = A B . (b) Consider rolling a six-sided die once. Let A be the set of outcomes where an odd number comes up. Let B be the set of outcomes where a 1 or a 2 comes up. Calculate the sets on both sides of the equalities in part (a), and verify that the equalities hold. Solution : (a) Sketching Venn diagram or applying De Morgans law directly leads to the desired result. Let us here prove De Morgans law. That is, let us show ( E F ) c = E c F c for any two sets E and F . Given E c F c , must be either in E c or F c . As E F E and E F F , must be not in E F . Therefore, ( E F ) c . On the other hand, suppose ( E F ) c . As E F E ,...
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This note was uploaded on 02/20/2012 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.
- Spring '07