# EE 351K - EE 351K Probability, Statistics, and Random...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 351K Probability, Statistics, and Random Processes SPRING 2012 Instructor:S.Shakkottai shakkott@ece.utexas.edu 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 ,...
View Full Document

## 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.

### Page1 / 3

EE 351K - EE 351K Probability, Statistics, and Random...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online