Let $A$ and $B$ be events. Is it necessarily true that $frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)}$?












-1












$begingroup$


Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?



If so, explain why. If not, give a counterexample.




I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!






probability combinatorics conditional-probability







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
    $endgroup$
    – Simply Beautiful Art
    Oct 3 '16 at 22:28










  • $begingroup$
    @SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
    $endgroup$
    – Yuna Kun
    Oct 3 '16 at 22:32










  • $begingroup$
    what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
    $endgroup$
    – Cato
    Oct 4 '16 at 9:38












  • $begingroup$
    A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
    $endgroup$
    – Cato
    Oct 4 '16 at 9:43
















-1












$begingroup$


Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?



If so, explain why. If not, give a counterexample.




I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!






probability combinatorics conditional-probability







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
    $endgroup$
    – Simply Beautiful Art
    Oct 3 '16 at 22:28










  • $begingroup$
    @SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
    $endgroup$
    – Yuna Kun
    Oct 3 '16 at 22:32










  • $begingroup$
    what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
    $endgroup$
    – Cato
    Oct 4 '16 at 9:38












  • $begingroup$
    A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
    $endgroup$
    – Cato
    Oct 4 '16 at 9:43














-1












-1








-1





$begingroup$


Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?



If so, explain why. If not, give a counterexample.




I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!






probability combinatorics conditional-probability







share|cite|improve this question











$endgroup$




Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?



If so, explain why. If not, give a counterexample.




I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!






probability combinatorics conditional-probability




probability combinatorics conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 16:12









Martin Sleziak

44.8k10119272




44.8k10119272










asked Oct 3 '16 at 22:26









Yuna KunYuna Kun

631218




631218








  • 3




    $begingroup$
    Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
    $endgroup$
    – Simply Beautiful Art
    Oct 3 '16 at 22:28










  • $begingroup$
    @SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
    $endgroup$
    – Yuna Kun
    Oct 3 '16 at 22:32










  • $begingroup$
    what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
    $endgroup$
    – Cato
    Oct 4 '16 at 9:38












  • $begingroup$
    A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
    $endgroup$
    – Cato
    Oct 4 '16 at 9:43














  • 3




    $begingroup$
    Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
    $endgroup$
    – Simply Beautiful Art
    Oct 3 '16 at 22:28










  • $begingroup$
    @SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
    $endgroup$
    – Yuna Kun
    Oct 3 '16 at 22:32










  • $begingroup$
    what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
    $endgroup$
    – Cato
    Oct 4 '16 at 9:38












  • $begingroup$
    A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
    $endgroup$
    – Cato
    Oct 4 '16 at 9:43








3




3




$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28




$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28












$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32




$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32












$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38






$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38














$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43




$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43










4 Answers
4






active

oldest

votes


















3












$begingroup$

Hint:



$P(A|B)$ need not be equal to $P(B|A)$.



Let $A$ be the event that you get number $2$ from a dice toss.



Let $B$ be the event that you get an even number from the same dice toss.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.



    Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".



    Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
    $$
    frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
    $$






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Since
      $$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
      We can see
      $$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
      but
      $$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
      Hence,
      $$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$



      The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        Honestly, i think it does not(i might be wrong, but here we go).
        Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
        Now, think about it:
        Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
        You have:
        $$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
        Obviously, this equality only holds if
        $$P(A) = P(B) $$.
        An example:
        I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
        I throw the first dice, so i have:
        $$P(A) = 1/6 $$
        And then:
        $$P(B) = 1/6 $$
        Since both of the events are independent, we have that
        $$P(A cap B) = P(A)P(B) $$
        So:
        $$P(A/B) = P(A) = P(B/A) = P(B) $$.
        This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Hint:



          $P(A|B)$ need not be equal to $P(B|A)$.



          Let $A$ be the event that you get number $2$ from a dice toss.



          Let $B$ be the event that you get an even number from the same dice toss.






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            Hint:



            $P(A|B)$ need not be equal to $P(B|A)$.



            Let $A$ be the event that you get number $2$ from a dice toss.



            Let $B$ be the event that you get an even number from the same dice toss.






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              Hint:



              $P(A|B)$ need not be equal to $P(B|A)$.



              Let $A$ be the event that you get number $2$ from a dice toss.



              Let $B$ be the event that you get an even number from the same dice toss.






              share|cite|improve this answer









              $endgroup$



              Hint:



              $P(A|B)$ need not be equal to $P(B|A)$.



              Let $A$ be the event that you get number $2$ from a dice toss.



              Let $B$ be the event that you get an even number from the same dice toss.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Oct 3 '16 at 22:33









              Siong Thye GohSiong Thye Goh

              102k1466118




              102k1466118























                  3












                  $begingroup$

                  Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.



                  Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".



                  Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
                  $$
                  frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
                  $$






                  share|cite|improve this answer









                  $endgroup$


















                    3












                    $begingroup$

                    Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.



                    Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".



                    Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
                    $$
                    frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
                    $$






                    share|cite|improve this answer









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.



                      Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".



                      Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
                      $$
                      frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
                      $$






                      share|cite|improve this answer









                      $endgroup$



                      Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.



                      Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".



                      Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
                      $$
                      frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
                      $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Oct 3 '16 at 22:32









                      ZubzubZubzub

                      3,8891922




                      3,8891922























                          3












                          $begingroup$

                          Since
                          $$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
                          We can see
                          $$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
                          but
                          $$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
                          Hence,
                          $$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$



                          The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$






                          share|cite|improve this answer











                          $endgroup$


















                            3












                            $begingroup$

                            Since
                            $$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
                            We can see
                            $$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
                            but
                            $$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
                            Hence,
                            $$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$



                            The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$






                            share|cite|improve this answer











                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              Since
                              $$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
                              We can see
                              $$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
                              but
                              $$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
                              Hence,
                              $$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$



                              The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$






                              share|cite|improve this answer











                              $endgroup$



                              Since
                              $$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
                              We can see
                              $$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
                              but
                              $$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
                              Hence,
                              $$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$



                              The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Oct 3 '16 at 22:40

























                              answered Oct 3 '16 at 22:33









                              msmmsm

                              6,2292829




                              6,2292829























                                  1












                                  $begingroup$

                                  Honestly, i think it does not(i might be wrong, but here we go).
                                  Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
                                  Now, think about it:
                                  Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
                                  You have:
                                  $$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
                                  Obviously, this equality only holds if
                                  $$P(A) = P(B) $$.
                                  An example:
                                  I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
                                  I throw the first dice, so i have:
                                  $$P(A) = 1/6 $$
                                  And then:
                                  $$P(B) = 1/6 $$
                                  Since both of the events are independent, we have that
                                  $$P(A cap B) = P(A)P(B) $$
                                  So:
                                  $$P(A/B) = P(A) = P(B/A) = P(B) $$.
                                  This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    Honestly, i think it does not(i might be wrong, but here we go).
                                    Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
                                    Now, think about it:
                                    Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
                                    You have:
                                    $$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
                                    Obviously, this equality only holds if
                                    $$P(A) = P(B) $$.
                                    An example:
                                    I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
                                    I throw the first dice, so i have:
                                    $$P(A) = 1/6 $$
                                    And then:
                                    $$P(B) = 1/6 $$
                                    Since both of the events are independent, we have that
                                    $$P(A cap B) = P(A)P(B) $$
                                    So:
                                    $$P(A/B) = P(A) = P(B/A) = P(B) $$.
                                    This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      Honestly, i think it does not(i might be wrong, but here we go).
                                      Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
                                      Now, think about it:
                                      Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
                                      You have:
                                      $$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
                                      Obviously, this equality only holds if
                                      $$P(A) = P(B) $$.
                                      An example:
                                      I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
                                      I throw the first dice, so i have:
                                      $$P(A) = 1/6 $$
                                      And then:
                                      $$P(B) = 1/6 $$
                                      Since both of the events are independent, we have that
                                      $$P(A cap B) = P(A)P(B) $$
                                      So:
                                      $$P(A/B) = P(A) = P(B/A) = P(B) $$.
                                      This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.






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                                      $endgroup$



                                      Honestly, i think it does not(i might be wrong, but here we go).
                                      Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
                                      Now, think about it:
                                      Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
                                      You have:
                                      $$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
                                      Obviously, this equality only holds if
                                      $$P(A) = P(B) $$.
                                      An example:
                                      I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
                                      I throw the first dice, so i have:
                                      $$P(A) = 1/6 $$
                                      And then:
                                      $$P(B) = 1/6 $$
                                      Since both of the events are independent, we have that
                                      $$P(A cap B) = P(A)P(B) $$
                                      So:
                                      $$P(A/B) = P(A) = P(B/A) = P(B) $$.
                                      This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.







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                                      answered Oct 3 '16 at 22:42









                                      Vitor C GoergenVitor C Goergen

                                      8261416




                                      8261416






























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