From: stromhau@stud.ntnu.no   
      
   Hi,   
      
   Given tha classical alarm, earthquake, burglary, marycalls and johncalls.   
   Some sort of drawing below.   
      
   My problem is that i want to calculate P(A|M).   
      
   I use bayes rule to get ;   
      
   P(A|M) = P(M|A)*P(A)/P(M). Because i dont know P(A) i get a problem. I dont   
   know p(M), but this i could have avoided if i had known P(A).   
      
   The first thing on my mind was to calculate P(A) as:   
      
   sum P(A,M,J,B,E) where i use all the combinations of(negatet or positive) of   
   M,J,B,E.   
      
   Then i found one solution where P(A) was calculated as   
      
   p(A) = p(A | ~B, ~E) p(~B, ~E) + p(A | ~B, E) p(~B, E) + p(A | B, ~E) p(B,   
   ~E) + p(A | B, E) p(B, E)   
      
   My big question is why?   
      
   It seems like P(A) is calculated as P(A,E,B)+P(A,-B,-E)+P(A,-B,E)+.......   
      
   just what i would have done except that i belived that i had to use P(J) and   
   P(M) also.   
      
   How is it that this is true ?   
      
          B            E   
           \           /   
            \         /   
              \     /   
                 A   
               /   \   
              /     \   
            /        \   
          J           M   
      
   arrows from B to A and from E to A   
   arrows from A to J and from A to M   
      
      
      
   Tommy Stromhaug   
      
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