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23.AMCwithstates0,1,2hasP=0@1=21=31=601=32=31=201=21A:IfP(X0=0)=P(X0=1 23.AMCwithstates0,1,2hasP=0@1=21=31=601=32=31=201=21A:IfP(X0=0)=P(X0=1

23.AMCwithstates0,1,2hasP=0@1=21=31=601=32=31=201=21A:IfP(X0=0)=P(X0=1 - PDF document

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23.AMCwithstates0,1,2hasP=0@1=21=31=601=32=31=201=21A:IfP(X0=0)=P(X0=1 - PPT Presentation

AfteralittlealgebraweobtainP31 10803922474816444524391AThensince 131313wegetPX3j1Xi0P3ij i1 32Xi0P3ijforj012Inparticularwe ndthatPX30132324PX3162324andPX32 ID: 409696

Afteralittlealgebra weobtainP3=1 1080@3922474816444524391A:Thensince =(1=3;1=3;1=3) wegetP(X3=j)=1Xi=0P(3)ij i=1 32Xi=0P(3)ij;forj=0;1;2:Inparticular we ndthatP(X3=0)=132=324 P(X3=1)=62=324 andP(X3=2)

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