Line 14: Line 14:
 
 
 
         P(A=1,B=0) = P(A=1) . P(B=0) = p.(1-p)                 (2)
 
         P(A=1,B=0) = P(A=1) . P(B=0) = p.(1-p)                 (2)
 
+
       
 +
Since, (1) & (2) are not equal to each other, A & C are not independent of each other when bits are biased.
 
 
       
 
        Since, (1) & (2) are not equal to each other, A & C are not
 
independent of each other when bits are biased.
 
  
 
                             "OR"         
 
                             "OR"         
Line 25: Line 23:
 
 
 
         P(A=0,B=0) = P(A=1) . P(B=0) = (1-p).(1-p)                 (4)
 
         P(A=0,B=0) = P(A=1) . P(B=0) = (1-p).(1-p)                 (4)
 
+
         
       
+
Again, since (3) & (4) are not equal to each other, A & C are not independent of each other when bits are biased.
       
+
        Again, since (3) & (4) are not equal to each other, A & C are not
+
independent of each other when bits are biased.
+
  
 
         Hence, it is proved.
 
         Hence, it is proved.

Latest revision as of 14:42, 16 September 2008



       A		B		


       P(A=1) = p	P(B=1) = p
       P(A=0) = 1-p	P(B=0) = 1-p			


       P(A=1,C=1) = P(A=1) . P(C=1) = p.{P(A=1,B=0)+P(A=0,B=1)} = 2.p^2.(1-p)		(1)
       P(A=1,B=0) = P(A=1) . P(B=0) = p.(1-p)		                 		(2)
       

Since, (1) & (2) are not equal to each other, A & C are not independent of each other when bits are biased.


                            "OR"        
       P(A=0,C=0) = P(A=0) . P(C=0) = (1-p).{P(A=1,B=1)+P(A=0,B=0)} = (p^2 + (1-p)^2).(1-p)	(3)
       P(A=0,B=0) = P(A=1) . P(B=0) = (1-p).(1-p)		                 		(4)
         

Again, since (3) & (4) are not equal to each other, A & C are not independent of each other when bits are biased.

       Hence, it is proved.

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