Bayes network probability question











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I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
here is the diagram



so far i have gotten



Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48



Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56



not 100% sure or if these are correct or how to proceed when it comes to the next 2










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    up vote
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    down vote

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    I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
    here is the diagram



    so far i have gotten



    Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48



    Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56



    not 100% sure or if these are correct or how to proceed when it comes to the next 2










    share|cite|improve this question







    New contributor




    pog is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
      here is the diagram



      so far i have gotten



      Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48



      Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56



      not 100% sure or if these are correct or how to proceed when it comes to the next 2










      share|cite|improve this question







      New contributor




      pog is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
      here is the diagram



      so far i have gotten



      Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48



      Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56



      not 100% sure or if these are correct or how to proceed when it comes to the next 2







      probability bayesian bayes-theorem bayesian-network






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      asked Nov 16 at 22:16









      pog

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          What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:



          $$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$



          $$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$



          For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$



          Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$



          Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$



          We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$



          We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$



          Therefore, $p(c,d)=0.336$.




          • Note: double-check results in calculations.






          share|cite|improve this answer








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          • Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
            – pog
            Nov 18 at 17:42












          • $$p(neg b|a)=1-p(b|a)$$. It's the complement.
            – DavidPM
            Nov 18 at 19:17











          Your Answer





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          1 Answer
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          up vote
          0
          down vote



          accepted










          What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:



          $$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$



          $$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$



          For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$



          Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$



          Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$



          We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$



          We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$



          Therefore, $p(c,d)=0.336$.




          • Note: double-check results in calculations.






          share|cite|improve this answer








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          DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          • Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
            – pog
            Nov 18 at 17:42












          • $$p(neg b|a)=1-p(b|a)$$. It's the complement.
            – DavidPM
            Nov 18 at 19:17















          up vote
          0
          down vote



          accepted










          What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:



          $$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$



          $$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$



          For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$



          Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$



          Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$



          We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$



          We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$



          Therefore, $p(c,d)=0.336$.




          • Note: double-check results in calculations.






          share|cite|improve this answer








          New contributor




          DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          • Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
            – pog
            Nov 18 at 17:42












          • $$p(neg b|a)=1-p(b|a)$$. It's the complement.
            – DavidPM
            Nov 18 at 19:17













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:



          $$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$



          $$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$



          For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$



          Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$



          Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$



          We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$



          We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$



          Therefore, $p(c,d)=0.336$.




          • Note: double-check results in calculations.






          share|cite|improve this answer








          New contributor




          DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:



          $$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$



          $$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$



          For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$



          Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$



          Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$



          We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$



          We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$



          Therefore, $p(c,d)=0.336$.




          • Note: double-check results in calculations.







          share|cite|improve this answer








          New contributor




          DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          answered Nov 17 at 0:41









          DavidPM

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          • Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
            – pog
            Nov 18 at 17:42












          • $$p(neg b|a)=1-p(b|a)$$. It's the complement.
            – DavidPM
            Nov 18 at 19:17


















          • Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
            – pog
            Nov 18 at 17:42












          • $$p(neg b|a)=1-p(b|a)$$. It's the complement.
            – DavidPM
            Nov 18 at 19:17
















          Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
          – pog
          Nov 18 at 17:42






          Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
          – pog
          Nov 18 at 17:42














          $$p(neg b|a)=1-p(b|a)$$. It's the complement.
          – DavidPM
          Nov 18 at 19:17




          $$p(neg b|a)=1-p(b|a)$$. It's the complement.
          – DavidPM
          Nov 18 at 19:17










          pog is a new contributor. Be nice, and check out our Code of Conduct.










           

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