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 September 26th, 2018, 11:42 AM #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=T|A=T,B=T) = 0.9 P(C=T|A=T,B=F) = 0.8 P(C=T|A=F,B=T) = 0.4 P(C=T|A=F,B=F) = 0.1 P(C=F|A=T,B=T) = 0.1 P(C=F|A=T,B=F) = 0.2 P(C=F|A=F,B=T) = 0.6 P(C=F|A=F,B=F) = 0.9 P(D=T|B=T) = 0.8 P(D=T|B=F) = 0.2 P(E=T|C=T) = 0.1 P(E=T|C=F) = 0.8 P(D=F|B=T) = 0.2 P(D=F|B=F) = 0.8 P(E=F|C=T) = 0.9 P(E=F|C=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=T|A=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, 01:38 PM #2 Global Moderator   Joined: May 2007 Posts: 6,805 Thanks: 716 General approach: P(C=T)=P(C=T|A=T,B=T)P(A=T)P(B=T)+P(C=T|A=T,B=F)P( A=T)P(B=F)+P(C=T|A=F,B=T)P(A=F)P(B=T)+P(C=T|A=F,B= F)P(A=F)P(B=F). This uses the independence of A and B. P(C=T|A=T)=P(C=T|A=T,B=T)P(B=T)+P(C=T|A=T,B=F)P(B= F) P(D=T)=1 Exercise for you - why? Therefore answer for 5. P(C=T) Thanks from zollen September 27th, 2018, 10:15 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=T|B=T)P(B=T) + P(D=T|B=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) Tags independence, probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post WWRtelescoping Advanced Statistics 1 February 22nd, 2016 02:02 PM creatingembla Probability and Statistics 1 November 1st, 2015 03:26 AM supernerd707 Probability and Statistics 1 July 9th, 2012 12:52 PM gbux Probability and Statistics 1 February 16th, 2012 05:57 PM -DQ- Advanced Statistics 2 October 9th, 2010 11:17 AM

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