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Unformatted text preview: CSE 250A. Assignment 2 Out: Tue Oct 5 Due: Tue Oct 12 Reading : Russell & Norvig, Chapters 1314. 2.1 Probabilistic inference Recall the alarm belief network described in class. The directed acyclic graph (DAG) and conditional probability tables (CPTs) are shown below: Alarm Earthquake Burglar John Calls Mary Calls P(E=1) = 0.002 P(B=1) = 0.001 P(A=1E=0,B=0) = 0.001 P(A=1E=0,B=1) = 0.94 P(A=1E=1,B=0) = 0.29 P(A=1E=1,B=1) = 0.95 P(J=1A=0) = 0.05 P(J=1A=1) = 0.90 P(M=1A=0) = 0.01 P(M=1A=1) = 0.70 Compute numeric values for the following probabilities, exploiting relations of conditional independence as much as possible to simplify your calculations. Show your work. (a) P ( M =1) (d) P ( E =1  A =1 ,M =0) (g) P ( B =1  J =1 ,E =1) (b) P ( M =1  E =0 ,B =1) (e) P ( E =1  A =1 ,J =1) (h) P ( J =0  A =0 ,M =1) (c) P ( E =0  A =1) (f) P ( B =1  J =1) (i) P ( J =0  M =1 ,B =1) 2.2 Nonmonotonic reasoning A patient is known to have contracted a rare disease which comes in two forms, represented by the values of a binary random variable X ∈ { , 1 } . Symptoms of the disease are represented by the binary random variables Y k ∈ { , 1 } , and knowledge of the disease is summarized by the belief network: X Y 1 Y 2 Y 3 Y n ... 1 The conditional probability tables (CPTs) for this belief network are as follows. In the absence of evidence, both forms of the disease are equally likely, with prior probabilities: P ( X =0) = P ( X =1) = 1 2 . In the first form of the disease ( X =0) , the first symptom occurs with probability one, P ( Y 1 =1  X =0) = 1 , while the k th symptom (with k ≥ 2 ) occurs with probability P ( Y k =1  X =0) = f ( k 1) f ( k ) , where the function...
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 Spring '10
 lawrenCe
 Probability theory, conditional independence, belief network

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