How can I calculate the probability of two independent events with only the union and the intersection?












2












$begingroup$


Suppose two events A and B are two independent events with $P(A) > P(B)$ and $P(A ∪ B) = 0.626$ and $P(A ∩ B) = 0.144$, determine the values of P(A) and P(B).



So far I have:



$$P(A ∩ B) = P(A) + P(B) - P(A U B)$$
$$0.144 = P(A) + P(B) - 0.626$$






probability statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You haven't used the independence of $A$ and $B$, which will give you another equation so you can solve $P(A)$ and $P(B)$.
    $endgroup$
    – Eclipse Sun
    Jan 28 at 19:02
















2












$begingroup$


Suppose two events A and B are two independent events with $P(A) > P(B)$ and $P(A ∪ B) = 0.626$ and $P(A ∩ B) = 0.144$, determine the values of P(A) and P(B).



So far I have:



$$P(A ∩ B) = P(A) + P(B) - P(A U B)$$
$$0.144 = P(A) + P(B) - 0.626$$






probability statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You haven't used the independence of $A$ and $B$, which will give you another equation so you can solve $P(A)$ and $P(B)$.
    $endgroup$
    – Eclipse Sun
    Jan 28 at 19:02














2












2








2





$begingroup$


Suppose two events A and B are two independent events with $P(A) > P(B)$ and $P(A ∪ B) = 0.626$ and $P(A ∩ B) = 0.144$, determine the values of P(A) and P(B).



So far I have:



$$P(A ∩ B) = P(A) + P(B) - P(A U B)$$
$$0.144 = P(A) + P(B) - 0.626$$






probability statistics







share|cite|improve this question











$endgroup$




Suppose two events A and B are two independent events with $P(A) > P(B)$ and $P(A ∪ B) = 0.626$ and $P(A ∩ B) = 0.144$, determine the values of P(A) and P(B).



So far I have:



$$P(A ∩ B) = P(A) + P(B) - P(A U B)$$
$$0.144 = P(A) + P(B) - 0.626$$






probability statistics




probability statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 19:01









Maria Mazur

48.9k1260122




48.9k1260122










asked Jan 28 at 18:56









Elena TorreElena Torre

596




596








  • 1




    $begingroup$
    You haven't used the independence of $A$ and $B$, which will give you another equation so you can solve $P(A)$ and $P(B)$.
    $endgroup$
    – Eclipse Sun
    Jan 28 at 19:02














  • 1




    $begingroup$
    You haven't used the independence of $A$ and $B$, which will give you another equation so you can solve $P(A)$ and $P(B)$.
    $endgroup$
    – Eclipse Sun
    Jan 28 at 19:02








1




1




$begingroup$
You haven't used the independence of $A$ and $B$, which will give you another equation so you can solve $P(A)$ and $P(B)$.
$endgroup$
– Eclipse Sun
Jan 28 at 19:02




$begingroup$
You haven't used the independence of $A$ and $B$, which will give you another equation so you can solve $P(A)$ and $P(B)$.
$endgroup$
– Eclipse Sun
Jan 28 at 19:02










2 Answers
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$begingroup$

For independent events you have
$$ P(Acap B) = P(A)P(B) \
P(Acup B) + P(Acap B) = P(A) + P(B) $$
If you know the product and sum of two numbers, finding the numbers themselves is a matter of solving a quadratic equation (and in fact one of the classical formulation of quadratic problems).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint: Let $$P(A) = pimplies P(B) = {P(Acap B)over P(A)} = {0,144over p}$$ then by PIE we have $$ 0.626 = p+{0,144over p}-0,144$$



    Now solve for $p$...






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

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      active

      oldest

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      1












      $begingroup$

      For independent events you have
      $$ P(Acap B) = P(A)P(B) \
      P(Acup B) + P(Acap B) = P(A) + P(B) $$
      If you know the product and sum of two numbers, finding the numbers themselves is a matter of solving a quadratic equation (and in fact one of the classical formulation of quadratic problems).






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        For independent events you have
        $$ P(Acap B) = P(A)P(B) \
        P(Acup B) + P(Acap B) = P(A) + P(B) $$
        If you know the product and sum of two numbers, finding the numbers themselves is a matter of solving a quadratic equation (and in fact one of the classical formulation of quadratic problems).






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          For independent events you have
          $$ P(Acap B) = P(A)P(B) \
          P(Acup B) + P(Acap B) = P(A) + P(B) $$
          If you know the product and sum of two numbers, finding the numbers themselves is a matter of solving a quadratic equation (and in fact one of the classical formulation of quadratic problems).






          share|cite|improve this answer









          $endgroup$



          For independent events you have
          $$ P(Acap B) = P(A)P(B) \
          P(Acup B) + P(Acap B) = P(A) + P(B) $$
          If you know the product and sum of two numbers, finding the numbers themselves is a matter of solving a quadratic equation (and in fact one of the classical formulation of quadratic problems).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 28 at 19:06









          Henning MakholmHenning Makholm

          243k17308553




          243k17308553























              1












              $begingroup$

              Hint: Let $$P(A) = pimplies P(B) = {P(Acap B)over P(A)} = {0,144over p}$$ then by PIE we have $$ 0.626 = p+{0,144over p}-0,144$$



              Now solve for $p$...






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint: Let $$P(A) = pimplies P(B) = {P(Acap B)over P(A)} = {0,144over p}$$ then by PIE we have $$ 0.626 = p+{0,144over p}-0,144$$



                Now solve for $p$...






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint: Let $$P(A) = pimplies P(B) = {P(Acap B)over P(A)} = {0,144over p}$$ then by PIE we have $$ 0.626 = p+{0,144over p}-0,144$$



                  Now solve for $p$...






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Let $$P(A) = pimplies P(B) = {P(Acap B)over P(A)} = {0,144over p}$$ then by PIE we have $$ 0.626 = p+{0,144over p}-0,144$$



                  Now solve for $p$...







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 28 at 19:00









                  Maria MazurMaria Mazur

                  48.9k1260122




                  48.9k1260122






























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