How do you multiply two conditional probabilities?












4












$begingroup$


I have to multiply this:



$P(a|b,c)·P(b|c)$



How do you multiply those two expressions?



It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Intuitively you could compare this with P(a|b)P(b) = P(a,b)
    $endgroup$
    – Martijn Weterings
    Jan 8 at 20:48










  • $begingroup$
    Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
    $endgroup$
    – Alexis
    Jan 8 at 20:52










  • $begingroup$
    The nomenglature.
    $endgroup$
    – VansFannel
    Jan 8 at 20:55










  • $begingroup$
    Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
    $endgroup$
    – StatsStudent
    Jan 8 at 21:08










  • $begingroup$
    @VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
    $endgroup$
    – StatsStudent
    Jan 8 at 22:40


















4












$begingroup$


I have to multiply this:



$P(a|b,c)·P(b|c)$



How do you multiply those two expressions?



It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Intuitively you could compare this with P(a|b)P(b) = P(a,b)
    $endgroup$
    – Martijn Weterings
    Jan 8 at 20:48










  • $begingroup$
    Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
    $endgroup$
    – Alexis
    Jan 8 at 20:52










  • $begingroup$
    The nomenglature.
    $endgroup$
    – VansFannel
    Jan 8 at 20:55










  • $begingroup$
    Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
    $endgroup$
    – StatsStudent
    Jan 8 at 21:08










  • $begingroup$
    @VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
    $endgroup$
    – StatsStudent
    Jan 8 at 22:40
















4












4








4


1



$begingroup$


I have to multiply this:



$P(a|b,c)·P(b|c)$



How do you multiply those two expressions?



It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.






probability







share|cite|improve this question











$endgroup$




I have to multiply this:



$P(a|b,c)·P(b|c)$



How do you multiply those two expressions?



It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 20:44







VansFannel

















asked Jan 8 at 19:58









VansFannelVansFannel

15118




15118












  • $begingroup$
    Intuitively you could compare this with P(a|b)P(b) = P(a,b)
    $endgroup$
    – Martijn Weterings
    Jan 8 at 20:48










  • $begingroup$
    Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
    $endgroup$
    – Alexis
    Jan 8 at 20:52










  • $begingroup$
    The nomenglature.
    $endgroup$
    – VansFannel
    Jan 8 at 20:55










  • $begingroup$
    Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
    $endgroup$
    – StatsStudent
    Jan 8 at 21:08










  • $begingroup$
    @VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
    $endgroup$
    – StatsStudent
    Jan 8 at 22:40




















  • $begingroup$
    Intuitively you could compare this with P(a|b)P(b) = P(a,b)
    $endgroup$
    – Martijn Weterings
    Jan 8 at 20:48










  • $begingroup$
    Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
    $endgroup$
    – Alexis
    Jan 8 at 20:52










  • $begingroup$
    The nomenglature.
    $endgroup$
    – VansFannel
    Jan 8 at 20:55










  • $begingroup$
    Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
    $endgroup$
    – StatsStudent
    Jan 8 at 21:08










  • $begingroup$
    @VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
    $endgroup$
    – StatsStudent
    Jan 8 at 22:40


















$begingroup$
Intuitively you could compare this with P(a|b)P(b) = P(a,b)
$endgroup$
– Martijn Weterings
Jan 8 at 20:48




$begingroup$
Intuitively you could compare this with P(a|b)P(b) = P(a,b)
$endgroup$
– Martijn Weterings
Jan 8 at 20:48












$begingroup$
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
$endgroup$
– Alexis
Jan 8 at 20:52




$begingroup$
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
$endgroup$
– Alexis
Jan 8 at 20:52












$begingroup$
The nomenglature.
$endgroup$
– VansFannel
Jan 8 at 20:55




$begingroup$
The nomenglature.
$endgroup$
– VansFannel
Jan 8 at 20:55












$begingroup$
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
$endgroup$
– StatsStudent
Jan 8 at 21:08




$begingroup$
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
$endgroup$
– StatsStudent
Jan 8 at 21:08












$begingroup$
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
$endgroup$
– StatsStudent
Jan 8 at 22:40






$begingroup$
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
$endgroup$
– StatsStudent
Jan 8 at 22:40












2 Answers
2






active

oldest

votes


















8












$begingroup$

Those two expressions simply mean:



(the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).



The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:



$P(a|b,c)·P(b|c)=0.40times0.70=0.28$



UPDATED BASED ON YOUR EDITED QUESTION:



The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$



begin{eqnarray*}
P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(c)}\
& = & frac{P[(ab)c]}{P(c)}\
& = & P(ab|c)
end{eqnarray*}



So, in all, we have:



begin{eqnarray*}
P(a|bc)P(b|c) & = & P(ab|c)
end{eqnarray*}






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
    $$
    P(a,b,c) = P(a|b,c) P(b|c) P(c)
    $$
    which would imply:
    $$
    P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
    $$
    so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
      $endgroup$
      – VansFannel
      Jan 8 at 20:45












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    Those two expressions simply mean:



    (the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).



    The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:



    $P(a|b,c)·P(b|c)=0.40times0.70=0.28$



    UPDATED BASED ON YOUR EDITED QUESTION:



    The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$



    begin{eqnarray*}
    P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
    & = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
    & = & frac{P(abc)}{P(c)}\
    & = & frac{P[(ab)c]}{P(c)}\
    & = & P(ab|c)
    end{eqnarray*}



    So, in all, we have:



    begin{eqnarray*}
    P(a|bc)P(b|c) & = & P(ab|c)
    end{eqnarray*}






    share|cite|improve this answer











    $endgroup$


















      8












      $begingroup$

      Those two expressions simply mean:



      (the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).



      The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:



      $P(a|b,c)·P(b|c)=0.40times0.70=0.28$



      UPDATED BASED ON YOUR EDITED QUESTION:



      The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$



      begin{eqnarray*}
      P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
      & = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
      & = & frac{P(abc)}{P(c)}\
      & = & frac{P[(ab)c]}{P(c)}\
      & = & P(ab|c)
      end{eqnarray*}



      So, in all, we have:



      begin{eqnarray*}
      P(a|bc)P(b|c) & = & P(ab|c)
      end{eqnarray*}






      share|cite|improve this answer











      $endgroup$
















        8












        8








        8





        $begingroup$

        Those two expressions simply mean:



        (the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).



        The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:



        $P(a|b,c)·P(b|c)=0.40times0.70=0.28$



        UPDATED BASED ON YOUR EDITED QUESTION:



        The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$



        begin{eqnarray*}
        P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
        & = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
        & = & frac{P(abc)}{P(c)}\
        & = & frac{P[(ab)c]}{P(c)}\
        & = & P(ab|c)
        end{eqnarray*}



        So, in all, we have:



        begin{eqnarray*}
        P(a|bc)P(b|c) & = & P(ab|c)
        end{eqnarray*}






        share|cite|improve this answer











        $endgroup$



        Those two expressions simply mean:



        (the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).



        The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:



        $P(a|b,c)·P(b|c)=0.40times0.70=0.28$



        UPDATED BASED ON YOUR EDITED QUESTION:



        The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$



        begin{eqnarray*}
        P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
        & = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
        & = & frac{P(abc)}{P(c)}\
        & = & frac{P[(ab)c]}{P(c)}\
        & = & P(ab|c)
        end{eqnarray*}



        So, in all, we have:



        begin{eqnarray*}
        P(a|bc)P(b|c) & = & P(ab|c)
        end{eqnarray*}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 9 at 0:00

























        answered Jan 8 at 20:16









        StatsStudentStatsStudent

        6,05532044




        6,05532044

























            3












            $begingroup$

            If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
            $$
            P(a,b,c) = P(a|b,c) P(b|c) P(c)
            $$
            which would imply:
            $$
            P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
            $$
            so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
              $endgroup$
              – VansFannel
              Jan 8 at 20:45
















            3












            $begingroup$

            If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
            $$
            P(a,b,c) = P(a|b,c) P(b|c) P(c)
            $$
            which would imply:
            $$
            P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
            $$
            so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
              $endgroup$
              – VansFannel
              Jan 8 at 20:45














            3












            3








            3





            $begingroup$

            If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
            $$
            P(a,b,c) = P(a|b,c) P(b|c) P(c)
            $$
            which would imply:
            $$
            P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
            $$
            so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.






            share|cite|improve this answer









            $endgroup$



            If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
            $$
            P(a,b,c) = P(a|b,c) P(b|c) P(c)
            $$
            which would imply:
            $$
            P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
            $$
            so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 20:28









            Stat_ProgrammerStat_Programmer

            3411




            3411












            • $begingroup$
              I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
              $endgroup$
              – VansFannel
              Jan 8 at 20:45


















            • $begingroup$
              I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
              $endgroup$
              – VansFannel
              Jan 8 at 20:45
















            $begingroup$
            I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
            $endgroup$
            – VansFannel
            Jan 8 at 20:45




            $begingroup$
            I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
            $endgroup$
            – VansFannel
            Jan 8 at 20:45


















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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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            Gyllenstierna

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