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hw5-sol
Course: CS 540, Fall 2002School: Wisconsin
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for Solution Written Part of Homework 5, CS540, Fall 2008 Question 1 (a) P(Y|X)=(0.70+0.015)/(0.70+0.015+0.10+0.02)=0.856 (b) P(Y|X,Z)=0.70/(0.70+0.10)=0.875 (c) P(Y)=(0.70+0.015+0.08+0.01)/(0.805+0.195)=0.805 (d) P(X,Z)=(0.70+0.10)/(0.70+0.10+0.015+0.02+0.08+0.07+0.01+0.005)=0.80 (e) Since P(X)=0.835, P(Z)=0.95, and P(X)*P(Z)!=P(X,Z), so the data is not consistent with X and Z being independent. Question 2 T:...
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for Solution Written Part of Homework 5, CS540, Fall 2008
Question 1 (a) P(Y|X)=(0.70+0.015)/(0.70+0.015+0.10+0.02)=0.856 (b) P(Y|X,Z)=0.70/(0.70+0.10)=0.875 (c) P(Y)=(0.70+0.015+0.08+0.01)/(0.805+0.195)=0.805 (d) P(X,Z)=(0.70+0.10)/(0.70+0.10+0.015+0.02+0.08+0.07+0.01+0.005)=0.80 (e) Since P(X)=0.835, P(Z)=0.95, and P(X)*P(Z)!=P(X,Z), so the data is not consistent with X and Z being independent.
Question 2 T: take the homework F: finish the homework Given: P(F)=0.9, P(!T|F)=0.01, P(!T|!F)=0.5 So, P(F|!T)=P(F,!T)/P(!T)=(0.01*0.9)/(0.01*0.9+0.5*0.1)=0.153
Question 3 (a) P(E,B,C)=sigma_A(sigma_D(P(A,B,C,D,E))) = sigma_B(sigma_C(P(A)*P(B|A)*P(C|B)*P(D|A,B)*P(E|D))) =0.8*0.2*0.95*0.85*0.60+0.8*0.2*0.95*0.15*0.50+0.2*0.9*0.95*0.15*0.60+0.2*0.9*0.95*0.85* 0.50=0.177 P(!E,B,C)= sigma_A(sigma_D(P(A,B,C,D,!E))) = sigma_B(sigma_C(P(A)*P(B|A)*P(C|B)*P(D|A,B)*P(!E|D))) =0.8*0.2*0.95*0.85*0.40+0.8*0.2*0.95*0.15*0.50+0.2*0.9*0.95*0.15*0.40+0.2*0.9*0.95*0.85* So, 0.50=0.146 P(E|B,C)=0.177/(0.177+0.146)=0.548 (b) P(D|E,A)=sigma_B(sigma_C(P(A,B,C,D,E))) = sigma_B(sigma_C(P(A)*P(B|A)*P(C|B)*P(D|A,B)*P(E|D))) =0.8*0.2*0.95*0.85*0.60+0.8*0.2*0.05*0.85*0.60+0.8*0.8*0.5*0.25*0.60+0.8*0.8*0.5*0.25*0. 60=0.178 P(!D|E,A)=sigma_B(sigma_C(P(A,B,C,!D,E))) = sigma_B(sigma_C(P(A)*P(B|A)*P(C|B)*P(!D|A,B)*P(E|!D))) =0.8*0.2*0.95*0.15*0.50+0.8*0.2*0.05*0.15*0.50+0.8*0.8*0.5*0.75*0.50+0.8*0.8*0.5*0.75*0. 50=0.2...
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