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







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    Mar 24 at 5:49








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    Mar 24 at 5:51






  • 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
    Mar 24 at 10:01










  • $begingroup$
    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
    Mar 24 at 18:23
















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











$endgroup$








  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    Mar 24 at 5:49








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    Mar 24 at 5:51






  • 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
    Mar 24 at 10:01










  • $begingroup$
    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
    Mar 24 at 18:23














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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 24 at 18:22









Robert Howard

2,2933935




2,2933935










asked Mar 24 at 5:48









Gordon NgGordon Ng

292




292








  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    Mar 24 at 5:49








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    Mar 24 at 5:51






  • 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
    Mar 24 at 10:01










  • $begingroup$
    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
    Mar 24 at 18:23














  • 6




    $begingroup$
    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
    $endgroup$
    – Paras Khosla
    Mar 24 at 5:49








  • 3




    $begingroup$
    A common way to show that something doesn’t hold is to come up with a specific counterexample.
    $endgroup$
    – amd
    Mar 24 at 5:51






  • 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
    Mar 24 at 10:01










  • $begingroup$
    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
    Mar 24 at 18:23








6




6




$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49






$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49






3




3




$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51




$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51




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
Mar 24 at 10:01




$begingroup$
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
Mar 24 at 10:01












$begingroup$
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
Mar 24 at 18:23




$begingroup$
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
Mar 24 at 18:23










6 Answers
6






active

oldest

votes


















23












$begingroup$

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
    Mar 24 at 18:29



















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









    $endgroup$





















      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











        $endgroup$





















          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









          $endgroup$














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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            23












            $begingroup$

            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
              Mar 24 at 18:29
















            23












            $begingroup$

            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
              Mar 24 at 18:29














            23












            23








            23





            $begingroup$

            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 Mar 24 at 5:54









            Kavi Rama MurthyKavi Rama Murthy

            71.5k53170




            71.5k53170












            • $begingroup$
              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
              Mar 24 at 18:29


















            • $begingroup$
              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
              Mar 24 at 18:29
















            $begingroup$
            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
            Mar 24 at 18:29




            $begingroup$
            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
            Mar 24 at 18:29











            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 Mar 24 at 11:28









                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









                    $endgroup$


















                      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









                      $endgroup$
















                        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









                        $endgroup$



                        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












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Mar 24 at 5:55









                        TojrahTojrah

                        3436




                        3436























                            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 Mar 24 at 11:29









                                user21820user21820

                                39.9k544159




                                39.9k544159























                                    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











                                    $endgroup$


















                                      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











                                      $endgroup$
















                                        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.






                                        share|cite|improve this answer











                                        $endgroup$



                                        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.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Mar 24 at 18:08









                                        Robert Howard

                                        2,2933935




                                        2,2933935










                                        answered Mar 24 at 17:57









                                        Artemis FowlArtemis Fowl

                                        1114




                                        1114























                                            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









                                            $endgroup$


















                                              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









                                              $endgroup$
















                                                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









                                                $endgroup$



                                                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












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Mar 24 at 16:28









                                                BakuriuBakuriu

                                                11115




                                                11115






























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