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September 26th, 2018, 12:42 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  Probability and Independence
Let A, B, C, D, E be five boolean random variables. Assume that you are give the independence assumptions: * A and B are independent absolutely. * D is conditionally independent of both A and C, given B. * E is conditionally independent of A, B and D, given C. Assume you also are give the following probabilities: P(A=T) = 0.9 P(B=T) = 0.6 P(A=F) = 0.1 P(B=F) = 0.4 P(C=TA=T,B=T) = 0.9 P(C=TA=T,B=F) = 0.8 P(C=TA=F,B=T) = 0.4 P(C=TA=F,B=F) = 0.1 P(C=FA=T,B=T) = 0.1 P(C=FA=T,B=F) = 0.2 P(C=FA=F,B=T) = 0.6 P(C=FA=F,B=F) = 0.9 P(D=TB=T) = 0.8 P(D=TB=F) = 0.2 P(E=TC=T) = 0.1 P(E=TC=F) = 0.8 P(D=FB=T) = 0.2 P(D=FB=F) = 0.8 P(E=FC=T) = 0.9 P(E=FC=F) = 0.2 Compute the following probabilties: 1. P(A=T and B=T) = P(A) * P(B) = 0.54 2. P(A=T or B=T) = P(A) + P(B) = 1.5 3. P(C=T) = ????????? 4. P(C=TA=T) = ???????? 5. P(C=T and D=T) = ??????? I am totally lost of getting the answers of 3,4 5. I would be much appreciate if anyone able to show me how to solve these. Thanks. 
September 26th, 2018, 02:38 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,683 Thanks: 658 
General approach: P(C=T)=P(C=TA=T,B=T)P(A=T)P(B=T)+P(C=TA=T,B=F)P( A=T)P(B=F)+P(C=TA=F,B=T)P(A=F)P(B=T)+P(C=TA=F,B= F)P(A=F)P(B=F). This uses the independence of A and B. P(C=TA=T)=P(C=TA=T,B=T)P(B=T)+P(C=TA=T,B=F)P(B= F) P(D=T)=1 Exercise for you  why? Therefore answer for 5. P(C=T) 
September 27th, 2018, 11:15 AM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
It took me a while but I finally got it!!!!!!!!!! P(D=T) = P(D=T, B=T) + P(D=T, B=F) P(D=T) = P(D=TB=T)P(B=T) + P(D=TB=F)P(B=F) P(D=T) = ( 0.8 )( 0.6 ) + ( 0.2 )( 0.4 ) = 0.56 P(C = T and D = T) = P(C) * P (D) 

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independence, probability 
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