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Bayesian Network

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data

    Hrp3Htc.png
    2. Relevant equations


    3. The attempt at a solution

    For part A I solved for P(B|JC) = P(B,JC)/P(JC)
    For part B I am thinking P(B|!JC, MC) = P(B, !JC, MC) / P(!JC, MC)
    For part C I am thinking P(JC|MC) = P(JC, MC)/P(MC)

    Am I on track with these equations? Especially for part c? How do I compute P(JC, MC)?
     
  2. jcsd
  3. Mar 2, 2015 #2

    RUber

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    Homework Helper

    The probability that john calls and mary calls can be found by examining all cases:
    Burgled --> Alarm --> John calls =A1
    Burgled --> Alarm --> Mary calls=A2
    Not burgled --> Alarm --> John calls =B1
    Not burgled --> Alarm --> Mary calls =B2
    Burgled --> No Alarm --> John might call, but Mary wont =C1
    Not Burgled --> No Alarm --> John might call, but Mary wont =D1
    edit: P(JC & MC) = A1*A2/p(Burgled + Alarm) + B1*B2/p(Not burgled + alarm) + C1*0/p(burgled and no alarm) + D1*0/p(not burgled and no alarm)
    P(MC) = A2+B2+0+0
    P(JC) = A1+B1+C1+D1
     
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