Line 15: Line 15:
 
         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
+
 +
        Since, (1) & (2) are not equal to each other, A & C are not
 
independent of each other when bits are biased.
 
independent of each other when bits are biased.
  
Line 24: Line 25:
 
         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
 
         Again, since (3) & (4) are not equal to each other, A & C are not
 
independent of each other when bits are biased.
 
independent of each other when bits are biased.
  
 
         Hence, it is proved.
 
         Hence, it is proved.

Revision as of 15:38, 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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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang