Bayesian Network Homework: Equations & Solutions

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The discussion focuses on solving a Bayesian network homework problem involving conditional probabilities. The participant outlines their approach for calculating P(B|JC), P(B|!JC, MC), and P(JC|MC), seeking validation on their equations. They express uncertainty about computing P(JC, MC) and detail the cases that contribute to this probability. The calculations include various scenarios related to burglary and alarm responses, emphasizing the need to consider all possible outcomes. Overall, the participant is working through complex probability concepts and seeks clarity on their methodology.
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Homework Statement



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Homework Equations

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)?
 
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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 won't =C1
Not Burgled --> No Alarm --> John might call, but Mary won't =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
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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