By means of an example, show that $P(A) + P(B) = 1$ does not mean that $B$ is the complement of $A$












5












$begingroup$


I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?



This is what I got to:



$P(A) + P(B) = 1$



$P(A) + P(A') = 1$



How could it be proven that $B$ isn't the complement of $A$?



Help would be greatly appreciated.






probability







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  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    yesterday








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    yesterday






  • 7




    $begingroup$
    If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
    $endgroup$
    – Eric Duminil
    yesterday










  • $begingroup$
    Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Robert Howard
    yesterday
















5












$begingroup$


I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?



This is what I got to:



$P(A) + P(B) = 1$



$P(A) + P(A') = 1$



How could it be proven that $B$ isn't the complement of $A$?



Help would be greatly appreciated.






probability







share|cite|improve this question









New contributor




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







$endgroup$








  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    yesterday








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    yesterday






  • 7




    $begingroup$
    If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
    $endgroup$
    – Eric Duminil
    yesterday










  • $begingroup$
    Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Robert Howard
    yesterday














5












5








5





$begingroup$


I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?



This is what I got to:



$P(A) + P(B) = 1$



$P(A) + P(A') = 1$



How could it be proven that $B$ isn't the complement of $A$?



Help would be greatly appreciated.






probability







share|cite|improve this question









New contributor




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







$endgroup$




I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?



This is what I got to:



$P(A) + P(B) = 1$



$P(A) + P(A') = 1$



How could it be proven that $B$ isn't the complement of $A$?



Help would be greatly appreciated.






probability




probability






share|cite|improve this question









New contributor




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











share|cite|improve this question









New contributor




Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




share|cite|improve this question








edited yesterday









Robert Howard

2,2383935




2,2383935






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asked yesterday









Gordon NgGordon Ng

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  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    yesterday








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    yesterday






  • 7




    $begingroup$
    If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
    $endgroup$
    – Eric Duminil
    yesterday










  • $begingroup$
    Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Robert Howard
    yesterday














  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    yesterday








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    yesterday






  • 7




    $begingroup$
    If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
    $endgroup$
    – Eric Duminil
    yesterday










  • $begingroup$
    Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Robert Howard
    yesterday








6




6




$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday






$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday






3




3




$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday




$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday




7




7




$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday




$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday












$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday




$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday










6 Answers
6






active

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22












$begingroup$

Take any event of probability $frac 1 2$ and take $B=A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
    $endgroup$
    – aschepler
    yesterday



















11












$begingroup$

A counterexample



Take a normal 6-sided die.



Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6



Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6



P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1



But B is not the complement of A.



The complement of A is the event "roll any of the numbers 5 or 6".



By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.



Further



It also seems you have misunderstood the question.



You wrote How could it be proven that B isn't the complement of A?



This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.






    share|cite|improve this answer








    New contributor




    Tojrah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$





















      1












      $begingroup$

      From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        Another example:
        $$A = text{Getting a head on coin }A$$
        $$B = text{Getting a head on coin }B$$



        Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.






        share|cite|improve this answer










        New contributor




        Shadow The Gohst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          0












          $begingroup$

          Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.



          Since $S = A cup B cup {e}$ we have:



          $$
          P(A) + P(B) + P({e}) = 1
          $$



          but also



          $$
          P({e}) = 0
          $$



          since it is finite, thus:



          $$
          P(A) + P(B) = 1
          $$



          For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.






          share|cite|improve this answer








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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            22












            $begingroup$

            Take any event of probability $frac 1 2$ and take $B=A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
              $endgroup$
              – aschepler
              yesterday
















            22












            $begingroup$

            Take any event of probability $frac 1 2$ and take $B=A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
              $endgroup$
              – aschepler
              yesterday














            22












            22








            22





            $begingroup$

            Take any event of probability $frac 1 2$ and take $B=A$.






            share|cite|improve this answer









            $endgroup$



            Take any event of probability $frac 1 2$ and take $B=A$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            Kavi Rama MurthyKavi Rama Murthy

            69.9k53170




            69.9k53170












            • $begingroup$
              And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
              $endgroup$
              – aschepler
              yesterday


















            • $begingroup$
              And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
              $endgroup$
              – aschepler
              yesterday
















            $begingroup$
            And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
            $endgroup$
            – aschepler
            yesterday




            $begingroup$
            And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
            $endgroup$
            – aschepler
            yesterday











            11












            $begingroup$

            A counterexample



            Take a normal 6-sided die.



            Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6



            Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6



            P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1



            But B is not the complement of A.



            The complement of A is the event "roll any of the numbers 5 or 6".



            By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.



            Further



            It also seems you have misunderstood the question.



            You wrote How could it be proven that B isn't the complement of A?



            This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.






            share|cite|improve this answer









            $endgroup$


















              11












              $begingroup$

              A counterexample



              Take a normal 6-sided die.



              Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6



              Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6



              P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1



              But B is not the complement of A.



              The complement of A is the event "roll any of the numbers 5 or 6".



              By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.



              Further



              It also seems you have misunderstood the question.



              You wrote How could it be proven that B isn't the complement of A?



              This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.






              share|cite|improve this answer









              $endgroup$
















                11












                11








                11





                $begingroup$

                A counterexample



                Take a normal 6-sided die.



                Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6



                Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6



                P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1



                But B is not the complement of A.



                The complement of A is the event "roll any of the numbers 5 or 6".



                By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.



                Further



                It also seems you have misunderstood the question.



                You wrote How could it be proven that B isn't the complement of A?



                This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.






                share|cite|improve this answer









                $endgroup$



                A counterexample



                Take a normal 6-sided die.



                Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6



                Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6



                P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1



                But B is not the complement of A.



                The complement of A is the event "roll any of the numbers 5 or 6".



                By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.



                Further



                It also seems you have misunderstood the question.



                You wrote How could it be proven that B isn't the complement of A?



                This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                user985366user985366

                432310




                432310























                    2












                    $begingroup$

                    As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.






                    share|cite|improve this answer








                    New contributor




                    Tojrah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    $endgroup$


















                      2












                      $begingroup$

                      As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.






                      share|cite|improve this answer








                      New contributor




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






                      $endgroup$
















                        2












                        2








                        2





                        $begingroup$

                        As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.






                        share|cite|improve this answer








                        New contributor




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






                        $endgroup$



                        As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.







                        share|cite|improve this answer








                        New contributor




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









                        share|cite|improve this answer



                        share|cite|improve this answer






                        New contributor




                        Tojrah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                        answered yesterday









                        TojrahTojrah

                        2455




                        2455




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                            1












                            $begingroup$

                            From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.






                            share|cite|improve this answer









                            $endgroup$


















                              1












                              $begingroup$

                              From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.






                              share|cite|improve this answer









                              $endgroup$
















                                1












                                1








                                1





                                $begingroup$

                                From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.






                                share|cite|improve this answer









                                $endgroup$



                                From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered yesterday









                                user21820user21820

                                39.7k544158




                                39.7k544158























                                    1












                                    $begingroup$

                                    Another example:
                                    $$A = text{Getting a head on coin }A$$
                                    $$B = text{Getting a head on coin }B$$



                                    Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.






                                    share|cite|improve this answer










                                    New contributor




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






                                    $endgroup$


















                                      1












                                      $begingroup$

                                      Another example:
                                      $$A = text{Getting a head on coin }A$$
                                      $$B = text{Getting a head on coin }B$$



                                      Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.






                                      share|cite|improve this answer










                                      New contributor




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






                                      $endgroup$
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        Another example:
                                        $$A = text{Getting a head on coin }A$$
                                        $$B = text{Getting a head on coin }B$$



                                        Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.






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                                        $endgroup$



                                        Another example:
                                        $$A = text{Getting a head on coin }A$$
                                        $$B = text{Getting a head on coin }B$$



                                        Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.







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                                        edited yesterday









                                        Robert Howard

                                        2,2383935




                                        2,2383935






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                                        answered yesterday









                                        Shadow The GohstShadow The Gohst

                                        111




                                        111




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                                            0












                                            $begingroup$

                                            Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.



                                            Since $S = A cup B cup {e}$ we have:



                                            $$
                                            P(A) + P(B) + P({e}) = 1
                                            $$



                                            but also



                                            $$
                                            P({e}) = 0
                                            $$



                                            since it is finite, thus:



                                            $$
                                            P(A) + P(B) = 1
                                            $$



                                            For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.






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                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.



                                              Since $S = A cup B cup {e}$ we have:



                                              $$
                                              P(A) + P(B) + P({e}) = 1
                                              $$



                                              but also



                                              $$
                                              P({e}) = 0
                                              $$



                                              since it is finite, thus:



                                              $$
                                              P(A) + P(B) = 1
                                              $$



                                              For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.






                                              share|cite|improve this answer








                                              New contributor




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                                              Check out our Code of Conduct.






                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.



                                                Since $S = A cup B cup {e}$ we have:



                                                $$
                                                P(A) + P(B) + P({e}) = 1
                                                $$



                                                but also



                                                $$
                                                P({e}) = 0
                                                $$



                                                since it is finite, thus:



                                                $$
                                                P(A) + P(B) = 1
                                                $$



                                                For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.






                                                share|cite|improve this answer








                                                New contributor




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






                                                $endgroup$



                                                Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.



                                                Since $S = A cup B cup {e}$ we have:



                                                $$
                                                P(A) + P(B) + P({e}) = 1
                                                $$



                                                but also



                                                $$
                                                P({e}) = 0
                                                $$



                                                since it is finite, thus:



                                                $$
                                                P(A) + P(B) = 1
                                                $$



                                                For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.







                                                share|cite|improve this answer








                                                New contributor




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









                                                share|cite|improve this answer



                                                share|cite|improve this answer






                                                New contributor




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                                                answered yesterday









                                                BakuriuBakuriu

                                                11115




                                                11115




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