Bayesial+network

P(M|B)=P(M^B) P(B) =∑P(M                B  e  a  j)  ∑P(m  B  e  a  j) ∑ P(M Λ B Λ e Λ a Λ j): 1) P(M Λ B Λ E Λ A Λ J)=P(M|A) ×P(B) ×P(E) ×P(J|A) ×P(A|B Λ E) =0.70×0.001×0.002×0.90×0.95 =1.197×10-6 2) P(M Λ B Λ  E Λ A Λ J)=P(M|A) ×P(B) ×P(  E) ×P(J|A) ×P(A|B Λ  E)  =0.70×0.001×0.998×0.9×0.94 =5.910156×10-4 3) P(M Λ B Λ E Λ A Λ J)=P(M|  A) ×P(B) ×P(E) ×P(J|  A) ×P(  A|B Λ E)   =0.01×0.001×0.002×0.05×0.05  =5×10-11 4) P(M Λ B Λ E Λ A Λ  J)=P(M|A) ×P(B) ×P(E) ×P(  J|A) ×P(A|B Λ E)  =0.70×0.001×0.002×0.1×0.95 =1.33×10-7 5) P(M Λ B Λ E Λ A Λ  J)=P(M|A) ×P(B) ×P(  E) ×P(  J|A) ×P(A|B Λ  E)  =0.70×0.001×0.998×0.1×0.94  =6.56684×10-5 6) P(M Λ B Λ E Λ  A Λ  J)=P(M|  A) ×P(B) ×P(E) ×P(  J|  A) ×P(  A|B Λ E)  =0.01×0.001×0.002×0.95×0.05 =9.5×10-10 7) P(M Λ B Λ E Λ  A Λ J)=P(M|  A) ×P(B) ×P(  E) ×P(J|  A) ×P(  A|B Λ  E)  =0.01×0.001×0.998×0.05×0.06  =2.994×10-8 8) P(M Λ B Λ  E Λ  A Λ  J)=P(M|  A) ×P(B) ×P(  E) ×P(  J|  A) ×P(  A|B Λ  E)  =0.01×0.001×0.998×0.95×0.06 =5.6886×10-7 Sum=6.5861326×10-4

P(B)=P(j Λ B Λ e Λ a Λ m) 1) P(J Λ B Λ E Λ A Λ M)=P(J|A) ×P(B) ×P(E) ×P(M|A) ×P(A|B Λ E) =0.9×0.001×0.002×0.70×0.95  =1.197×10-6 2) P(J Λ B Λ  E Λ A Λ M)=P(J|A) ×P(B) ×P(  E)  × P(M|A)  × P(A|B Λ  E)  = 0.90×0.001×0.998×0.70×0.94 =5.9101×10-4 3) P(J Λ B Λ E Λ A Λ M)= P(J|  A) ×P(B) ×P(E) ×P(M|  A) ×P(  A|B Λ E)  =    = 5×10-11 4) P(J Λ B Λ E Λ A Λ  M)= P(J|A) ×P(B) ×P(E) ×P(  M|A) ×P(A|B Λ E)  = 0.90×0.001×0.002×0.30×0.95 = 5.13×10-7 5) P(J Λ B Λ E Λ A Λ  M )=P(J|A) ×P(B) ×P(  E) ×P(  M|A) ×P(A|B Λ  E)  =0.90×0.001×0.998×0.30×0.94  =2.532924×10-4 6) P(J Λ B Λ E Λ  A Λ  M)=P(J|  A) ×P(B) ×P(E) ×P(  M|  A) ×P(  A|B Λ E)  =0.05×0.001×0.002×0.99×0.05 =4.95×10-9 7) P(J Λ B Λ E Λ  A Λ M)=P(J|  A) ×P(B) ×P(  E) ×P(M|  A) ×P(  A|B Λ  E)  = 0.05×0.001×0.998×0.01×0.06  =2.994×10-8 8) P(J Λ B Λ  E Λ  A Λ  M )=P(J|  A) ×P(B) ×P(  E) ×P(  M|  A) ×P(  A|B Λ  E)  =0.05×0.001×0.998×0.99×0.06 =2.96406×10-6 9) P( J Λ B Λ E Λ A Λ M )=P(  J|A) ×P(B) ×P(E) ×P(M|A) ×P(A|B Λ E)  =0.10×0.001×0.002×0.70×0.95 =1.33×10-7 10) P(  J Λ B Λ  E Λ A Λ M )=P(  J |A) ×P(B) ×P(  E) ×P(M|A) ×P(A|B Λ  E)   =0.10×0.001×0.998×0.7×0.94 =6.56684×10-5 11) P( J Λ B Λ  E Λ  A Λ M)=P(  J|  A) ×P(B) ×P(  E) ×P(M|  A) ×P(  A|B Λ  E)  =0.95×0.001×0.998×0.01×0.06  =5.586×10-7 12) P(  J Λ B Λ  E Λ  A Λ  M )=P(  J|  A) ×P(B) ×P(  E) ×P(  M|  A) ×P(  A|B Λ  E)   =0.95×0.001×0.998×0.99×0.06 =5.6317×10-5 13) P( J Λ B Λ E Λ  A Λ M)=P(  J|  A) ×P(B) ×P(E) ×P(M|  A) ×P(  A|B Λ E)  =0.95×0.001×0.002×0.01×0.05  =9.5×10-10 14) P(  J Λ B Λ E Λ  A Λ  M)=P(  J|  A) ×P(B) ×P(E) ×P(  M|  A) ×P(  A|B Λ E)  =0.95×0.001×0.002×0.99×0.05 =9.405×10-8 15) P( J Λ B Λ E Λ A Λ  M)= P(  J|A) ×P(B) ×P(E) ×P(  M|A) ×P(A|B Λ E)  = 0.10×0.001×0.002×0.30×0.95  = 5.7×10-8 16) P(  J Λ B Λ  E Λ A Λ  M )=P(  J|A) ×P(B) ×P(  E) ×P(  M|A) ×P(A|B Λ  E)  =0.10×0.001×0.998×0.30×0.94 =2.81436×10-5

Sum=9.9998346×10-4 ≈0.001 ---

P(M|B)=P(M^B) P(B) __=∑P(M B  e  a  j)__ ∑P(m B  e  a  j) = 6.5861326×10-4 0.001 =0.6586 __=0.66__

so, the probability for Mary to make call when there is Burgary is 0.66. However, the probability for Jonh to make a call when there is burgary is 0.84 So, we can conclude that John is more reliable to make a call when there is burgary.