Prove that if events A and B are independent, then the complement events of A and B are also independent.












5












$begingroup$


I know that:

$$begin{gathered}Pleft(Acap Bright)=Pleft(Aright)Pleft(Bright)\
Pleft(A^{C}right)=1-Pleft(Aright)\
Pleft(B^{C}right)=1-Pleft(Bright)
end{gathered}
$$



My proof so far:

$$begin{gathered}Pleft(A^{C}cap B^{C}right)=left(1-Pleft(Aright)right)left(1-Pleft(Bright)right)=\
1-Pleft(Bright)-Pleft(Aright)+Pleft(Aright)Pleft(Bright)=1-Pleft(Bright)-Pleft(Aright)+Pleft(Acap Bright)
end{gathered}
$$



After that, I'm stuck. Any help would be appreciated.






probability







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$endgroup$








  • 1




    $begingroup$
    $=1-P(A cup B)$... Also your first line after "What I have so far:" is what you're trying to prove, not what you have.
    $endgroup$
    – Dustan Levenstein
    Oct 10 '16 at 19:14
















5












$begingroup$


I know that:

$$begin{gathered}Pleft(Acap Bright)=Pleft(Aright)Pleft(Bright)\
Pleft(A^{C}right)=1-Pleft(Aright)\
Pleft(B^{C}right)=1-Pleft(Bright)
end{gathered}
$$



My proof so far:

$$begin{gathered}Pleft(A^{C}cap B^{C}right)=left(1-Pleft(Aright)right)left(1-Pleft(Bright)right)=\
1-Pleft(Bright)-Pleft(Aright)+Pleft(Aright)Pleft(Bright)=1-Pleft(Bright)-Pleft(Aright)+Pleft(Acap Bright)
end{gathered}
$$



After that, I'm stuck. Any help would be appreciated.






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $=1-P(A cup B)$... Also your first line after "What I have so far:" is what you're trying to prove, not what you have.
    $endgroup$
    – Dustan Levenstein
    Oct 10 '16 at 19:14














5












5








5


3



$begingroup$


I know that:

$$begin{gathered}Pleft(Acap Bright)=Pleft(Aright)Pleft(Bright)\
Pleft(A^{C}right)=1-Pleft(Aright)\
Pleft(B^{C}right)=1-Pleft(Bright)
end{gathered}
$$



My proof so far:

$$begin{gathered}Pleft(A^{C}cap B^{C}right)=left(1-Pleft(Aright)right)left(1-Pleft(Bright)right)=\
1-Pleft(Bright)-Pleft(Aright)+Pleft(Aright)Pleft(Bright)=1-Pleft(Bright)-Pleft(Aright)+Pleft(Acap Bright)
end{gathered}
$$



After that, I'm stuck. Any help would be appreciated.






probability







share|cite|improve this question











$endgroup$




I know that:

$$begin{gathered}Pleft(Acap Bright)=Pleft(Aright)Pleft(Bright)\
Pleft(A^{C}right)=1-Pleft(Aright)\
Pleft(B^{C}right)=1-Pleft(Bright)
end{gathered}
$$



My proof so far:

$$begin{gathered}Pleft(A^{C}cap B^{C}right)=left(1-Pleft(Aright)right)left(1-Pleft(Bright)right)=\
1-Pleft(Bright)-Pleft(Aright)+Pleft(Aright)Pleft(Bright)=1-Pleft(Bright)-Pleft(Aright)+Pleft(Acap Bright)
end{gathered}
$$



After that, I'm stuck. Any help would be appreciated.






probability




probability






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share|cite|improve this question













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edited Sep 18 '18 at 20:59









Oren Milman

1747




1747










asked Oct 10 '16 at 19:10









Preston CharmellePreston Charmelle

26113




26113








  • 1




    $begingroup$
    $=1-P(A cup B)$... Also your first line after "What I have so far:" is what you're trying to prove, not what you have.
    $endgroup$
    – Dustan Levenstein
    Oct 10 '16 at 19:14














  • 1




    $begingroup$
    $=1-P(A cup B)$... Also your first line after "What I have so far:" is what you're trying to prove, not what you have.
    $endgroup$
    – Dustan Levenstein
    Oct 10 '16 at 19:14








1




1




$begingroup$
$=1-P(A cup B)$... Also your first line after "What I have so far:" is what you're trying to prove, not what you have.
$endgroup$
– Dustan Levenstein
Oct 10 '16 at 19:14




$begingroup$
$=1-P(A cup B)$... Also your first line after "What I have so far:" is what you're trying to prove, not what you have.
$endgroup$
– Dustan Levenstein
Oct 10 '16 at 19:14










4 Answers
4






active

oldest

votes


















12












$begingroup$

Assume $A$ and $B$ are independent. Then
begin{align}
P(A^c cap B^c)
&= 1 - P(A cup B) \
&= 1 - P(A) - P(B) + P(A cap B) \
&= 1 - P(A) - P(B) + P(A)P(B) \
&= (1-P(A))(1-P(B)) \
&= P(A^c)P(B^c).
end{align}






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    As you have found :



    P(A') = 1-P(A)



    P(B') = 1- P(B)



    Now clearly $P(A')P(B') =1 - [ P(A) + P(B)] + P(A cap B)$



    From set algebra we know that
    $P(A) + P(B) = P(A cup B) + P(Acap B)$
    Substituting, we have $P(A')P(B') = P([Acup B]')$



    Now from De morgans law we know that:



    $[Acup B]' = [A' cap B']$



    Substituting, we have $P(A')P(B') = P(A' cap B')$ , as required.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      gradient23's proof is great, in my opinion, but I would like to show another proof that seems more intuitive to me, though much less rigorous.



      The proof is based on a verbal definition of independence from wikipedia:




      two events are independent [...] if the occurrence of one does not affect the probability of occurrence of the other




      In addition, we use the fact that independence is symmetric.



      The (non-rigorous) proof:




      • We assume that $A$ and $B$ are independent.

      • By definition, the occurrence of $A$ doesn't affect the probability of $B$.

      • Thus, the occurrence of $A$ also doesn't affect the probability of $B^C$.

      • So by definition, $A$ and $B^C$ are also independent, which by definition again means that the occurrence of $B^C$ doesn't affect the probability of $A$.

        (Here we used the symmetry of independence.)

      • Therefore, the occurrence of $B^C$ also doesn't affect the probability of $A^C$.

      • So by definition, $B^C$ and $A^C$ are also independent.



      One could convert the proof to the language of math:
      $$begin{gathered}Atext{ and }B,text{are independent}\
      downarrow\
      Pleft(B|Aright)=Pleft(Bright)\
      downarrow\
      1-Pleft(B^{C}|Aright)=1-Pleft(B^{C}right)\
      downarrow\
      Pleft(B^{C}|Aright)=Pleft(B^{C}right)\
      downarrow\
      Atext{ and }B^{C},text{are independent}\
      downarrow\
      Pleft(A|B^{C}right)=Pleft(Aright)\
      downarrow\
      1-Pleft(A^{C}|B^{C}right)=1-Pleft(A^{C}right)\
      downarrow\
      Pleft(A^{C}|B^{C}right)=Pleft(A^{C}right)\
      downarrow\
      B^{C}text{ and }A^{C},text{are independent}
      end{gathered}
      $$



      But now we used conditional probabilities, which might be a problem in case $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).






      share|cite|improve this answer









      $endgroup$





















        -1












        $begingroup$

        By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.



        So, $P(A/B)=P(A cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A cap B)= P(A)P(B)$.



        Hence proved.






        share|cite|improve this answer











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          4 Answers
          4






          active

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          4 Answers
          4






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          12












          $begingroup$

          Assume $A$ and $B$ are independent. Then
          begin{align}
          P(A^c cap B^c)
          &= 1 - P(A cup B) \
          &= 1 - P(A) - P(B) + P(A cap B) \
          &= 1 - P(A) - P(B) + P(A)P(B) \
          &= (1-P(A))(1-P(B)) \
          &= P(A^c)P(B^c).
          end{align}






          share|cite|improve this answer









          $endgroup$


















            12












            $begingroup$

            Assume $A$ and $B$ are independent. Then
            begin{align}
            P(A^c cap B^c)
            &= 1 - P(A cup B) \
            &= 1 - P(A) - P(B) + P(A cap B) \
            &= 1 - P(A) - P(B) + P(A)P(B) \
            &= (1-P(A))(1-P(B)) \
            &= P(A^c)P(B^c).
            end{align}






            share|cite|improve this answer









            $endgroup$
















              12












              12








              12





              $begingroup$

              Assume $A$ and $B$ are independent. Then
              begin{align}
              P(A^c cap B^c)
              &= 1 - P(A cup B) \
              &= 1 - P(A) - P(B) + P(A cap B) \
              &= 1 - P(A) - P(B) + P(A)P(B) \
              &= (1-P(A))(1-P(B)) \
              &= P(A^c)P(B^c).
              end{align}






              share|cite|improve this answer









              $endgroup$



              Assume $A$ and $B$ are independent. Then
              begin{align}
              P(A^c cap B^c)
              &= 1 - P(A cup B) \
              &= 1 - P(A) - P(B) + P(A cap B) \
              &= 1 - P(A) - P(B) + P(A)P(B) \
              &= (1-P(A))(1-P(B)) \
              &= P(A^c)P(B^c).
              end{align}







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Oct 10 '16 at 19:39









              gradient23gradient23

              23319




              23319























                  1












                  $begingroup$

                  As you have found :



                  P(A') = 1-P(A)



                  P(B') = 1- P(B)



                  Now clearly $P(A')P(B') =1 - [ P(A) + P(B)] + P(A cap B)$



                  From set algebra we know that
                  $P(A) + P(B) = P(A cup B) + P(Acap B)$
                  Substituting, we have $P(A')P(B') = P([Acup B]')$



                  Now from De morgans law we know that:



                  $[Acup B]' = [A' cap B']$



                  Substituting, we have $P(A')P(B') = P(A' cap B')$ , as required.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    As you have found :



                    P(A') = 1-P(A)



                    P(B') = 1- P(B)



                    Now clearly $P(A')P(B') =1 - [ P(A) + P(B)] + P(A cap B)$



                    From set algebra we know that
                    $P(A) + P(B) = P(A cup B) + P(Acap B)$
                    Substituting, we have $P(A')P(B') = P([Acup B]')$



                    Now from De morgans law we know that:



                    $[Acup B]' = [A' cap B']$



                    Substituting, we have $P(A')P(B') = P(A' cap B')$ , as required.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      As you have found :



                      P(A') = 1-P(A)



                      P(B') = 1- P(B)



                      Now clearly $P(A')P(B') =1 - [ P(A) + P(B)] + P(A cap B)$



                      From set algebra we know that
                      $P(A) + P(B) = P(A cup B) + P(Acap B)$
                      Substituting, we have $P(A')P(B') = P([Acup B]')$



                      Now from De morgans law we know that:



                      $[Acup B]' = [A' cap B']$



                      Substituting, we have $P(A')P(B') = P(A' cap B')$ , as required.






                      share|cite|improve this answer









                      $endgroup$



                      As you have found :



                      P(A') = 1-P(A)



                      P(B') = 1- P(B)



                      Now clearly $P(A')P(B') =1 - [ P(A) + P(B)] + P(A cap B)$



                      From set algebra we know that
                      $P(A) + P(B) = P(A cup B) + P(Acap B)$
                      Substituting, we have $P(A')P(B') = P([Acup B]')$



                      Now from De morgans law we know that:



                      $[Acup B]' = [A' cap B']$



                      Substituting, we have $P(A')P(B') = P(A' cap B')$ , as required.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Oct 10 '16 at 19:43









                      LelouchLelouch

                      501313




                      501313























                          0












                          $begingroup$

                          gradient23's proof is great, in my opinion, but I would like to show another proof that seems more intuitive to me, though much less rigorous.



                          The proof is based on a verbal definition of independence from wikipedia:




                          two events are independent [...] if the occurrence of one does not affect the probability of occurrence of the other




                          In addition, we use the fact that independence is symmetric.



                          The (non-rigorous) proof:




                          • We assume that $A$ and $B$ are independent.

                          • By definition, the occurrence of $A$ doesn't affect the probability of $B$.

                          • Thus, the occurrence of $A$ also doesn't affect the probability of $B^C$.

                          • So by definition, $A$ and $B^C$ are also independent, which by definition again means that the occurrence of $B^C$ doesn't affect the probability of $A$.

                            (Here we used the symmetry of independence.)

                          • Therefore, the occurrence of $B^C$ also doesn't affect the probability of $A^C$.

                          • So by definition, $B^C$ and $A^C$ are also independent.



                          One could convert the proof to the language of math:
                          $$begin{gathered}Atext{ and }B,text{are independent}\
                          downarrow\
                          Pleft(B|Aright)=Pleft(Bright)\
                          downarrow\
                          1-Pleft(B^{C}|Aright)=1-Pleft(B^{C}right)\
                          downarrow\
                          Pleft(B^{C}|Aright)=Pleft(B^{C}right)\
                          downarrow\
                          Atext{ and }B^{C},text{are independent}\
                          downarrow\
                          Pleft(A|B^{C}right)=Pleft(Aright)\
                          downarrow\
                          1-Pleft(A^{C}|B^{C}right)=1-Pleft(A^{C}right)\
                          downarrow\
                          Pleft(A^{C}|B^{C}right)=Pleft(A^{C}right)\
                          downarrow\
                          B^{C}text{ and }A^{C},text{are independent}
                          end{gathered}
                          $$



                          But now we used conditional probabilities, which might be a problem in case $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            gradient23's proof is great, in my opinion, but I would like to show another proof that seems more intuitive to me, though much less rigorous.



                            The proof is based on a verbal definition of independence from wikipedia:




                            two events are independent [...] if the occurrence of one does not affect the probability of occurrence of the other




                            In addition, we use the fact that independence is symmetric.



                            The (non-rigorous) proof:




                            • We assume that $A$ and $B$ are independent.

                            • By definition, the occurrence of $A$ doesn't affect the probability of $B$.

                            • Thus, the occurrence of $A$ also doesn't affect the probability of $B^C$.

                            • So by definition, $A$ and $B^C$ are also independent, which by definition again means that the occurrence of $B^C$ doesn't affect the probability of $A$.

                              (Here we used the symmetry of independence.)

                            • Therefore, the occurrence of $B^C$ also doesn't affect the probability of $A^C$.

                            • So by definition, $B^C$ and $A^C$ are also independent.



                            One could convert the proof to the language of math:
                            $$begin{gathered}Atext{ and }B,text{are independent}\
                            downarrow\
                            Pleft(B|Aright)=Pleft(Bright)\
                            downarrow\
                            1-Pleft(B^{C}|Aright)=1-Pleft(B^{C}right)\
                            downarrow\
                            Pleft(B^{C}|Aright)=Pleft(B^{C}right)\
                            downarrow\
                            Atext{ and }B^{C},text{are independent}\
                            downarrow\
                            Pleft(A|B^{C}right)=Pleft(Aright)\
                            downarrow\
                            1-Pleft(A^{C}|B^{C}right)=1-Pleft(A^{C}right)\
                            downarrow\
                            Pleft(A^{C}|B^{C}right)=Pleft(A^{C}right)\
                            downarrow\
                            B^{C}text{ and }A^{C},text{are independent}
                            end{gathered}
                            $$



                            But now we used conditional probabilities, which might be a problem in case $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              gradient23's proof is great, in my opinion, but I would like to show another proof that seems more intuitive to me, though much less rigorous.



                              The proof is based on a verbal definition of independence from wikipedia:




                              two events are independent [...] if the occurrence of one does not affect the probability of occurrence of the other




                              In addition, we use the fact that independence is symmetric.



                              The (non-rigorous) proof:




                              • We assume that $A$ and $B$ are independent.

                              • By definition, the occurrence of $A$ doesn't affect the probability of $B$.

                              • Thus, the occurrence of $A$ also doesn't affect the probability of $B^C$.

                              • So by definition, $A$ and $B^C$ are also independent, which by definition again means that the occurrence of $B^C$ doesn't affect the probability of $A$.

                                (Here we used the symmetry of independence.)

                              • Therefore, the occurrence of $B^C$ also doesn't affect the probability of $A^C$.

                              • So by definition, $B^C$ and $A^C$ are also independent.



                              One could convert the proof to the language of math:
                              $$begin{gathered}Atext{ and }B,text{are independent}\
                              downarrow\
                              Pleft(B|Aright)=Pleft(Bright)\
                              downarrow\
                              1-Pleft(B^{C}|Aright)=1-Pleft(B^{C}right)\
                              downarrow\
                              Pleft(B^{C}|Aright)=Pleft(B^{C}right)\
                              downarrow\
                              Atext{ and }B^{C},text{are independent}\
                              downarrow\
                              Pleft(A|B^{C}right)=Pleft(Aright)\
                              downarrow\
                              1-Pleft(A^{C}|B^{C}right)=1-Pleft(A^{C}right)\
                              downarrow\
                              Pleft(A^{C}|B^{C}right)=Pleft(A^{C}right)\
                              downarrow\
                              B^{C}text{ and }A^{C},text{are independent}
                              end{gathered}
                              $$



                              But now we used conditional probabilities, which might be a problem in case $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).






                              share|cite|improve this answer









                              $endgroup$



                              gradient23's proof is great, in my opinion, but I would like to show another proof that seems more intuitive to me, though much less rigorous.



                              The proof is based on a verbal definition of independence from wikipedia:




                              two events are independent [...] if the occurrence of one does not affect the probability of occurrence of the other




                              In addition, we use the fact that independence is symmetric.



                              The (non-rigorous) proof:




                              • We assume that $A$ and $B$ are independent.

                              • By definition, the occurrence of $A$ doesn't affect the probability of $B$.

                              • Thus, the occurrence of $A$ also doesn't affect the probability of $B^C$.

                              • So by definition, $A$ and $B^C$ are also independent, which by definition again means that the occurrence of $B^C$ doesn't affect the probability of $A$.

                                (Here we used the symmetry of independence.)

                              • Therefore, the occurrence of $B^C$ also doesn't affect the probability of $A^C$.

                              • So by definition, $B^C$ and $A^C$ are also independent.



                              One could convert the proof to the language of math:
                              $$begin{gathered}Atext{ and }B,text{are independent}\
                              downarrow\
                              Pleft(B|Aright)=Pleft(Bright)\
                              downarrow\
                              1-Pleft(B^{C}|Aright)=1-Pleft(B^{C}right)\
                              downarrow\
                              Pleft(B^{C}|Aright)=Pleft(B^{C}right)\
                              downarrow\
                              Atext{ and }B^{C},text{are independent}\
                              downarrow\
                              Pleft(A|B^{C}right)=Pleft(Aright)\
                              downarrow\
                              1-Pleft(A^{C}|B^{C}right)=1-Pleft(A^{C}right)\
                              downarrow\
                              Pleft(A^{C}|B^{C}right)=Pleft(A^{C}right)\
                              downarrow\
                              B^{C}text{ and }A^{C},text{are independent}
                              end{gathered}
                              $$



                              But now we used conditional probabilities, which might be a problem in case $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Sep 19 '18 at 8:17









                              Oren MilmanOren Milman

                              1747




                              1747























                                  -1












                                  $begingroup$

                                  By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.



                                  So, $P(A/B)=P(A cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A cap B)= P(A)P(B)$.



                                  Hence proved.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    -1












                                    $begingroup$

                                    By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.



                                    So, $P(A/B)=P(A cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A cap B)= P(A)P(B)$.



                                    Hence proved.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      -1












                                      -1








                                      -1





                                      $begingroup$

                                      By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.



                                      So, $P(A/B)=P(A cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A cap B)= P(A)P(B)$.



                                      Hence proved.






                                      share|cite|improve this answer











                                      $endgroup$



                                      By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.



                                      So, $P(A/B)=P(A cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A cap B)= P(A)P(B)$.



                                      Hence proved.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Mar 25 at 20:11









                                      Brahadeesh

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                                      6,51642364










                                      answered Mar 25 at 18:32









                                      philip mutiaphilip mutia

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