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Unformatted text preview: H [ X j ( Y;Z ) = (0 ; 3))] = 0 H [ X j ( Y;Z ) = ( 1 ; 1))] = 0 H [ X j ( Y;Z ) = ( 1 ; 2))] = 0 H [ X j ( Y;Z ) = ( 1 ; 3))] = 0 H [ X j ( Y;Z ) = (1 ; 1))] = 0 H [ X j ( Y;Z ) = (1 ; 2))] = 0 H [ X j Y;Z ] = 0 : H [ X j ( Y;Z ) = (1 ; 3))] = 0 5. H [ Z j Y ] & 1 : 5850 H [ Z j Y ] = P ( Y = 1) H [ Z j Y = 1]+ P ( Y = 0) H [ Z j Y = 0]+ P ( Y = 1) H [ Z j Y = 1] : Now H [ Z j Y = 1] = 3 X k =1 P [ Z = k j Y = 1] log 2 & 1 P [ Z = k j Y = 1] = 3 X k =1 1 3 log 2 3 & 1 : 585 : H [ Z j Y = 1] & 1 : 585 H [ Z j Y = 0] & 1 : 585 So H [ Z j Y ] & 1 : 585 2...
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This note was uploaded on 05/22/2010 for the course MATH 187 taught by Professor Staff during the Spring '08 term at UCSD.
 Spring '08
 staff
 Math, Cryptography

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