Conditional probabilities summing to one












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Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.



Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$






probability conditional-probability







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    $begingroup$


    Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.



    Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$






    probability conditional-probability







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      1












      1








      1





      $begingroup$


      Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.



      Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$






      probability conditional-probability







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      $endgroup$




      Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.



      Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$






      probability conditional-probability




      probability conditional-probability






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      edited Dec 22 '18 at 11:21









      Saad

      20.6k92452




      20.6k92452










      asked Dec 22 '18 at 11:11









      machinerymachinery

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          $begingroup$

          No it's incorrect.
          $$P(A=true|B=true)+P(A=false|B=true)$$
          $$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
          $$=frac{P(B=true)}{P(B=true)}=1$$
          Similarly,
          $$P(A=true|B=false)+P(A=false|B=false)=1$$
          Thus, yes you can calculate the variables required, but the equation you wrote was incorrect






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            $begingroup$

            Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$



            but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$






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              1












              $begingroup$

              The summation you mention gives sum $2$ (not $1$).



              This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$



              In both cases if you know one of terms then you also know the other one.






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                3 Answers
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                active

                oldest

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                3 Answers
                3






                active

                oldest

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                active

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                active

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                2












                $begingroup$

                No it's incorrect.
                $$P(A=true|B=true)+P(A=false|B=true)$$
                $$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
                $$=frac{P(B=true)}{P(B=true)}=1$$
                Similarly,
                $$P(A=true|B=false)+P(A=false|B=false)=1$$
                Thus, yes you can calculate the variables required, but the equation you wrote was incorrect






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  No it's incorrect.
                  $$P(A=true|B=true)+P(A=false|B=true)$$
                  $$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
                  $$=frac{P(B=true)}{P(B=true)}=1$$
                  Similarly,
                  $$P(A=true|B=false)+P(A=false|B=false)=1$$
                  Thus, yes you can calculate the variables required, but the equation you wrote was incorrect






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    No it's incorrect.
                    $$P(A=true|B=true)+P(A=false|B=true)$$
                    $$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
                    $$=frac{P(B=true)}{P(B=true)}=1$$
                    Similarly,
                    $$P(A=true|B=false)+P(A=false|B=false)=1$$
                    Thus, yes you can calculate the variables required, but the equation you wrote was incorrect






                    share|cite|improve this answer









                    $endgroup$



                    No it's incorrect.
                    $$P(A=true|B=true)+P(A=false|B=true)$$
                    $$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
                    $$=frac{P(B=true)}{P(B=true)}=1$$
                    Similarly,
                    $$P(A=true|B=false)+P(A=false|B=false)=1$$
                    Thus, yes you can calculate the variables required, but the equation you wrote was incorrect







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 22 '18 at 11:25









                    Ankit KumarAnkit Kumar

                    1,542221




                    1,542221























                        2












                        $begingroup$

                        Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$



                        but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$



                          but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$



                            but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$






                            share|cite|improve this answer









                            $endgroup$



                            Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$



                            but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 22 '18 at 11:27









                            HenryHenry

                            101k482170




                            101k482170























                                1












                                $begingroup$

                                The summation you mention gives sum $2$ (not $1$).



                                This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$



                                In both cases if you know one of terms then you also know the other one.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  The summation you mention gives sum $2$ (not $1$).



                                  This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$



                                  In both cases if you know one of terms then you also know the other one.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    The summation you mention gives sum $2$ (not $1$).



                                    This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$



                                    In both cases if you know one of terms then you also know the other one.






                                    share|cite|improve this answer









                                    $endgroup$



                                    The summation you mention gives sum $2$ (not $1$).



                                    This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$



                                    In both cases if you know one of terms then you also know the other one.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 22 '18 at 11:33









                                    drhabdrhab

                                    104k545136




                                    104k545136






























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