
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
September 26th, 2018, 12:42 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2  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: 6,641 Thanks: 625 
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: 200 Thanks: 2 
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) 

Tags 
independence, probability 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
By The Independence......  WWRtelescoping  Advanced Statistics  1  February 22nd, 2016 03:02 PM 
Independence in probability  creatingembla  Probability and Statistics  1  November 1st, 2015 04:26 AM 
Understanding Dependence/Independence in Probability  supernerd707  Probability and Statistics  1  July 9th, 2012 01:52 PM 
probability independence question  gbux  Probability and Statistics  1  February 16th, 2012 06:57 PM 
Probability: independence, disjoint.  DQ  Advanced Statistics  2  October 9th, 2010 12:17 PM 