Question

17. Consider the Bayesian network drawn below A P(A true) 0.4 P(B-true | A-false)0.9 D P(B-true | A-true0.3 P(C-true) 0.7 P(D true | B-false and C-false)0.8 P(D-true | B-false and C-true) 0.3 P(D-true | B-true and C-false) 0.5 P(D-true | B-true and C-true)0.1 Show your work for the following calculations a) Compute P(A true and B -false and C- true and D -false) b) Compute P(D true | A -false and B-true and C-false) cCompute P(A true | B -false and C tue and D -false). d) Compute P(B -false A true and C -false e) Compute P(B -false)

Answer 1

Joint distribution: P(A,B,C,D) = P(A)P(C)P(B/A)P(D/BC)

a) P(A=true, B=false, C=true, D=false)

= P(A=true)P(C=true)P(B=false/A=true)P(D=true/B=false, C=true)

= 0.7 * 0.4* (1-0.3) * (1-0.3)

= 0.1372

b) P(D=true/A=false, B=true, C=false)

= P(A=false)P(C=false)P(B=true/A=false)P(D=true/B=true, C=false)/ (summation of D over P(A=false)P(C=false)P(B=true/A=false)P(D/B=true, C=false))

= (0.6 * 0.3 * 0.9 * 0.5) /((0.6 * 0.3* 0.9* 0.5) + (0.6* 0.3* 0.9* 0.5))

= 0.5

e) P(B=false) = summation of ACD over (P(A)P(C)P(B=false/A)P(D/B=false,C))

= (0.6 * 0.3 * 0.1 * 0.2) + (0.6 * 0.7 * 0.1 * 0.2) + (0.6 * 0.3 * 0.1 * 0.8) + (0.6 * 0.7 * 0.1 * 0.8) + (0.4 * 0.3 * 0.7 * 0.2) + (0.4 * 0.7 * 0.7 * 0.2) + (0.4 * 0.3 * 0.7 * 0.8) + (0.4 * 0.7 * 0.7 * 0.8)

= 0.34

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