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EL630 HW1 Solution

# EL630 HW1 Solution - Solution to Homework 1 1(Book:2 1 Show...

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1 Solution to Homework 1 1. (Book :2 1) Show that ( ) ; ( )( ) . a AB A b A B AB AB BA − +++= += + : ( )AccordingtoDe Morgan's Law , wehave a A B Solution ,and then, ( ) . A B AB AB A B AB A B A B AB A B B AS A = +++= + = + = ( ) () {} b A B ABAB A AA BB B AB BA + = + + =+++ = +++ = + (Also try to use the other identity of De Morgan's Law ) AB A B = + 2. If {2 5}and {3 6},find , ,and( ) . A x B x A B AB A B AB = ≤≤ = + + Solution : A B

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2 {2 5} {3 6} {2 6} {2 5} {3 6} {3 5} {3 } {5 } ( ) {2 6} ({ 3} { 5}) {2 3} {5 6} AB x x x AB x x x AB x x A B AB x x x x x + = ≤≤ + = = = = <+ > + = < + > = ≤< + <≤ 3. Show thatif { },then ( ) ( ). AB P A P B = ∅≤ : From Property 4 of probability in Lecture1,weknow that () () .N ow {} . Therefore, ( ) ( ). B A PA PB A B ⊂⇒ = += Solution Another way to show is 1 ( ) () () () 1 () ) () () ( ) () () () 1 () () PB B PB PB PB PA B PA PB PAB PA PB PA = + = + = + ≤− = A B B
3 4. Show that ( ) if ( ) ( ) ( ), then ( ) 0; ( ) if ( ) ( ) 1, then ( ) 1. a PA PB PAB PAB BA b PA PB PAB = = += = = = Solution: () a ( ) ( )( ) PA B PAB AB AB PAB PAB PAB + + = + + and ( ) ( ) . PA B PA PB PAB + These two identies imply that ( ) ( ) ( ) 0. PAB PAB PAB AB + = : ( ) a nd ( ) i m pl y ( ) 0 ; ( ) ( ) a ( ) i m y 0 ; { } ( ) 0 PA PAB AB PAB PAB PA PAB PAB PB PAB AB PAB PAB PB PAB PAB AB AB P AB AB P AB P AB = + = = = + =∅⇒ + = + =

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EL630 HW1 Solution - Solution to Homework 1 1(Book:2 1 Show...

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