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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) | ||
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| + | Since, (1) & (2) are not equal to each other, A & C are not independent of each other when bits are biased. | ||
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"OR" | "OR" | ||
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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) | ||
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| − | + | Again, since (3) & (4) are not equal to each other, A & C are not independent of each other when bits are biased. | |
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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.
