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0.5 × 0.1 + 0.1 × 0.65 + 0.4 × 0.25 M. Chen ([email protected]) ENGG2430C lecture 4 5 / 19 Total Probability Law: Divide and Conquer (D. P. Bertsekas & J. N. Tsitsiklis, Introduction to Probability, Athena Scientiﬁc Publishers, 2002) Noticing B = B ∩ Ω = [A1 ∩ B ] ∪ [A2 ∩ B ] ∪ [A3 ∩ B ] and sets
Ai ∩ B (i = 1, 2, 3) are disjoint, we get
P (B ) = P (A1 ∩ B ) + P (A2 ∩ B ) + P (A3 ∩ B )
= P (A1 )P (B A1 ) + P (A2 )P (B A2 ) + P (A3 )P (B A3 )
M. Chen ([email protected]) ENGG2430C lecture 4 6 / 19 Binary Symmetric Channel (BSC) * Error probability ε : P (0r 1t ) = P (1r 0t ) = ε .
Question: Assume P (0t ) = P (1t ) = 0.5, P (1r ) =?
If I know what is transmitted, then it is easy:
P (1r 1t ) = 1 − ε , P (1r 0t ) = ε .
Without such knowledge, it is a weighted sum of these two conditional
probabilities:
P (1r ) = P (1r 1t )P (1t ) + P (1r 0t )P (0t ) = (1 − ε )0.5 + ε · 0.5 = 0.5.
The answer does not depend on ε . Intuitive explanation?
M. Chen ([email protected]) ENGG2430C lecture 4 7 / 19 Bayes’ Rule Law for combining evidence. Infer from observations. (D. P. Bertsekas & J. N. Tsitsiklis, Introduction to Probability, Athena Scientiﬁc Publishers, 2002) “Prior” probability: P (Ai ), i = 1, 2, 3
For each i , we know P (B Ai )
Compute P (A1 B )
P (A1 B ) = (Apply total probability theorem on P (B )) = M. Chen ([email protected]) ENGG2430C lecture 4 P (B A1 )P (A1 )
P (A1 ∩ B )
=
P (B )
P (B )
P (B A1 )P (A1 )
∑j P (Aj )P (B Aj )
8 / 19 Binary Symmetric Channel (BSC) * Error probability ε : P (0r 1t ) = P (1r 0t ) = ε .
Question: Assume P (0t ) = P (1t ) = 0.5. Given that we receive 1,
what is probability that 1 is transmitted?
P (1t 1r ) = M. Chen ([email protected]) P (1t , 1r ) P (1r 1t )P (1t )
=
= 1 − ε.
P (1r )
P (1r ) ENGG2430C lecture 4 9 / 19 Example: FalsePositive Puzzle * P (having the disease) = 0.001, P (healthy) = 0.999.
Question: Given that the person tested positive,...
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 Spring '14
 Conditional Probability, Probability theory, Law of total probability, M. Chen, Minghua Chen, engg2430c lecture

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