# Hw1_sol - EE 351K Probability Statistics and Random Processes Instructor Shakkottai/Vishwanath Homework 1 SPRING 2009

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EE 351K Probability, Statistics, and Random Processes SPRING 2009 Instructor: Shakkottai/Vishwanath { shakkott,sriram } @ece.utexas.edu Homework 1 Problem 1 We are given that P ( A ) = 0 . 3 , P ( B c ) = 0 . 45 , and P ( A B ) = 0 . 6 . Determine P ( B ) and P ( A B ) . Solution : We have P ( B ) = 1 - P ( B c ) = 1 - 0 . 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 . 3 + 0 . 55 - 0 . 6 = 0 . 25 . 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 a 3 or 4 comes up. Let B be the set of outcomes where a prime number comes up. Calculate the sets on both sides of the equalities in part (a), and verify that the equalities hold. Solution : (a) We ﬁrst prove De Morgan’s law. That is to show ( E F ) c = E c F c for any two sets E and F . Given ω E c F c , ω must be either in E c or F c . As E F E and E F F , ω must be not in E F . That is ω ( E F ) c . On the other hand, suppose ω ( E F ) c . As E F E , ω E c or ω E \ ( E F ) . Further, E \ ( E F ) F c . Therefore, ω E c F c . Similarly, we can also prove ( E F ) c = E c F c . To show ( A c B c ) c = A B . Note that by De Morgan’s law ( A c B c ) c = ( A c ) c ( B c ) c = A B. Similarly, ( A c B c ) c = ( A c ) c ( B c ) c = A B. (b) We have

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## This note was uploaded on 10/26/2009 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.

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Hw1_sol - EE 351K Probability Statistics and Random Processes Instructor Shakkottai/Vishwanath Homework 1 SPRING 2009

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