Probability of exclusive and exhaustive events [closed]

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If $A, B$ and $C$ are three mutually exclusive and exhaustive events and $$2Pr(A) = 3 Pr(B) = 4 Pr(C).$$ Find $Pr(A cup B)$.




probability elementary-set-theory




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closed as off-topic by drhab, Arnaud Mortier, Graham Kemp, Siong Thye Goh, José Carlos Santos Aug 22 at 8:14


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up vote
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down vote

favorite
1












If $A, B$ and $C$ are three mutually exclusive and exhaustive events and $$2Pr(A) = 3 Pr(B) = 4 Pr(C).$$ Find $Pr(A cup B)$.




probability elementary-set-theory




share|cite|improve this question














closed as off-topic by drhab, Arnaud Mortier, Graham Kemp, Siong Thye Goh, José Carlos Santos Aug 22 at 8:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – drhab, Arnaud Mortier, Graham Kemp, Siong Thye Goh, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.












  • To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck. Our aim is to improve your mathematical skills.
    – drhab
    Aug 22 at 8:01










  • On top of your own efforts you should add your background and knowledge, including the definition of the keywords used here. If you don't know them, you should look them up.
    – Arnaud Mortier
    Aug 22 at 8:05













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If $A, B$ and $C$ are three mutually exclusive and exhaustive events and $$2Pr(A) = 3 Pr(B) = 4 Pr(C).$$ Find $Pr(A cup B)$.




probability elementary-set-theory




share|cite|improve this question














If $A, B$ and $C$ are three mutually exclusive and exhaustive events and $$2Pr(A) = 3 Pr(B) = 4 Pr(C).$$ Find $Pr(A cup B)$.





probability elementary-set-theory





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edited Aug 22 at 11:04









Nash J.

1,094315




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asked Aug 22 at 7:53









Aayush Mukharji

214




214




closed as off-topic by drhab, Arnaud Mortier, Graham Kemp, Siong Thye Goh, José Carlos Santos Aug 22 at 8:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – drhab, Arnaud Mortier, Graham Kemp, Siong Thye Goh, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by drhab, Arnaud Mortier, Graham Kemp, Siong Thye Goh, José Carlos Santos Aug 22 at 8:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – drhab, Arnaud Mortier, Graham Kemp, Siong Thye Goh, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.











  • To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck. Our aim is to improve your mathematical skills.
    – drhab
    Aug 22 at 8:01










  • On top of your own efforts you should add your background and knowledge, including the definition of the keywords used here. If you don't know them, you should look them up.
    – Arnaud Mortier
    Aug 22 at 8:05

















  • To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck. Our aim is to improve your mathematical skills.
    – drhab
    Aug 22 at 8:01










  • On top of your own efforts you should add your background and knowledge, including the definition of the keywords used here. If you don't know them, you should look them up.
    – Arnaud Mortier
    Aug 22 at 8:05
















To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck. Our aim is to improve your mathematical skills.
– drhab
Aug 22 at 8:01




To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck. Our aim is to improve your mathematical skills.
– drhab
Aug 22 at 8:01












On top of your own efforts you should add your background and knowledge, including the definition of the keywords used here. If you don't know them, you should look them up.
– Arnaud Mortier
Aug 22 at 8:05





On top of your own efforts you should add your background and knowledge, including the definition of the keywords used here. If you don't know them, you should look them up.
– Arnaud Mortier
Aug 22 at 8:05











2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










If $x=P(A),y=P(B)$ and $z=P(C)$, you have to solve the linear system



$x+y+z=1$



$2x-3y=0$



$2x-4z=0$.



It is your turn to show that $x=frac613, y=frac413$ and $z=frac313$.



Hence $P(A cup B)=x+y=frac1013$.






share|cite|improve this answer



























    up vote
    1
    down vote













    Hint:



    Using the mutually exclusive and exhaustive information, you have three equations in three unknowns



    $$2 mathbb P (A) = 3 mathbb P (B)$$
    $$3 mathbb P (B) = 4 mathbb P (C)$$
    $$mathbb P (A) + mathbb P (B) + mathbb P (C) =mathbb P (A cup B cup C)= 1$$



    so you can solve these. (You also know $2 mathbb P (A) = 4 mathbb P (C)$ and $mathbb P (A) ge 0$, $mathbb P (B)ge 0$, $mathbb P (C)ge 0$ though these do not add information here)



    You then want $mathbb P (A cup B) =mathbb P (A) + mathbb P (B)$






    share|cite|improve this answer






















    • Is the answer 10/13?
      – Aayush Mukharji
      Aug 22 at 8:01











    • @AayushMukharji - I would have thought so, though you are supposed to show your working
      – Henry
      Aug 22 at 8:04










    • According to book answer is 7/3 and my answer doesn't match.
      – Aayush Mukharji
      Aug 22 at 8:04










    • $mathbb P (A cup B)$ is a probability and so cannot be greater than $1$. Not much help but $frac713=mathbb P (B cup C)$
      – Henry
      Aug 22 at 8:05










    • @AayushMukharji If the book says that the probability is $7/3$, then that is clearly an error because probabilities can only be between $0$ and $1$, and $7/3 > 1$.
      – Eff
      Aug 22 at 8:07

















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    If $x=P(A),y=P(B)$ and $z=P(C)$, you have to solve the linear system



    $x+y+z=1$



    $2x-3y=0$



    $2x-4z=0$.



    It is your turn to show that $x=frac613, y=frac413$ and $z=frac313$.



    Hence $P(A cup B)=x+y=frac1013$.






    share|cite|improve this answer
























      up vote
      0
      down vote



      accepted










      If $x=P(A),y=P(B)$ and $z=P(C)$, you have to solve the linear system



      $x+y+z=1$



      $2x-3y=0$



      $2x-4z=0$.



      It is your turn to show that $x=frac613, y=frac413$ and $z=frac313$.



      Hence $P(A cup B)=x+y=frac1013$.






      share|cite|improve this answer






















        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        If $x=P(A),y=P(B)$ and $z=P(C)$, you have to solve the linear system



        $x+y+z=1$



        $2x-3y=0$



        $2x-4z=0$.



        It is your turn to show that $x=frac613, y=frac413$ and $z=frac313$.



        Hence $P(A cup B)=x+y=frac1013$.






        share|cite|improve this answer












        If $x=P(A),y=P(B)$ and $z=P(C)$, you have to solve the linear system



        $x+y+z=1$



        $2x-3y=0$



        $2x-4z=0$.



        It is your turn to show that $x=frac613, y=frac413$ and $z=frac313$.



        Hence $P(A cup B)=x+y=frac1013$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 22 at 8:10









        Fred

        38.2k1238




        38.2k1238




















            up vote
            1
            down vote













            Hint:



            Using the mutually exclusive and exhaustive information, you have three equations in three unknowns



            $$2 mathbb P (A) = 3 mathbb P (B)$$
            $$3 mathbb P (B) = 4 mathbb P (C)$$
            $$mathbb P (A) + mathbb P (B) + mathbb P (C) =mathbb P (A cup B cup C)= 1$$



            so you can solve these. (You also know $2 mathbb P (A) = 4 mathbb P (C)$ and $mathbb P (A) ge 0$, $mathbb P (B)ge 0$, $mathbb P (C)ge 0$ though these do not add information here)



            You then want $mathbb P (A cup B) =mathbb P (A) + mathbb P (B)$






            share|cite|improve this answer






















            • Is the answer 10/13?
              – Aayush Mukharji
              Aug 22 at 8:01











            • @AayushMukharji - I would have thought so, though you are supposed to show your working
              – Henry
              Aug 22 at 8:04










            • According to book answer is 7/3 and my answer doesn't match.
              – Aayush Mukharji
              Aug 22 at 8:04










            • $mathbb P (A cup B)$ is a probability and so cannot be greater than $1$. Not much help but $frac713=mathbb P (B cup C)$
              – Henry
              Aug 22 at 8:05










            • @AayushMukharji If the book says that the probability is $7/3$, then that is clearly an error because probabilities can only be between $0$ and $1$, and $7/3 > 1$.
              – Eff
              Aug 22 at 8:07














            up vote
            1
            down vote













            Hint:



            Using the mutually exclusive and exhaustive information, you have three equations in three unknowns



            $$2 mathbb P (A) = 3 mathbb P (B)$$
            $$3 mathbb P (B) = 4 mathbb P (C)$$
            $$mathbb P (A) + mathbb P (B) + mathbb P (C) =mathbb P (A cup B cup C)= 1$$



            so you can solve these. (You also know $2 mathbb P (A) = 4 mathbb P (C)$ and $mathbb P (A) ge 0$, $mathbb P (B)ge 0$, $mathbb P (C)ge 0$ though these do not add information here)



            You then want $mathbb P (A cup B) =mathbb P (A) + mathbb P (B)$






            share|cite|improve this answer






















            • Is the answer 10/13?
              – Aayush Mukharji
              Aug 22 at 8:01











            • @AayushMukharji - I would have thought so, though you are supposed to show your working
              – Henry
              Aug 22 at 8:04










            • According to book answer is 7/3 and my answer doesn't match.
              – Aayush Mukharji
              Aug 22 at 8:04










            • $mathbb P (A cup B)$ is a probability and so cannot be greater than $1$. Not much help but $frac713=mathbb P (B cup C)$
              – Henry
              Aug 22 at 8:05










            • @AayushMukharji If the book says that the probability is $7/3$, then that is clearly an error because probabilities can only be between $0$ and $1$, and $7/3 > 1$.
              – Eff
              Aug 22 at 8:07












            up vote
            1
            down vote










            up vote
            1
            down vote









            Hint:



            Using the mutually exclusive and exhaustive information, you have three equations in three unknowns



            $$2 mathbb P (A) = 3 mathbb P (B)$$
            $$3 mathbb P (B) = 4 mathbb P (C)$$
            $$mathbb P (A) + mathbb P (B) + mathbb P (C) =mathbb P (A cup B cup C)= 1$$



            so you can solve these. (You also know $2 mathbb P (A) = 4 mathbb P (C)$ and $mathbb P (A) ge 0$, $mathbb P (B)ge 0$, $mathbb P (C)ge 0$ though these do not add information here)



            You then want $mathbb P (A cup B) =mathbb P (A) + mathbb P (B)$






            share|cite|improve this answer














            Hint:



            Using the mutually exclusive and exhaustive information, you have three equations in three unknowns



            $$2 mathbb P (A) = 3 mathbb P (B)$$
            $$3 mathbb P (B) = 4 mathbb P (C)$$
            $$mathbb P (A) + mathbb P (B) + mathbb P (C) =mathbb P (A cup B cup C)= 1$$



            so you can solve these. (You also know $2 mathbb P (A) = 4 mathbb P (C)$ and $mathbb P (A) ge 0$, $mathbb P (B)ge 0$, $mathbb P (C)ge 0$ though these do not add information here)



            You then want $mathbb P (A cup B) =mathbb P (A) + mathbb P (B)$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 22 at 8:02

























            answered Aug 22 at 7:59









            Henry

            93.5k471149




            93.5k471149











            • Is the answer 10/13?
              – Aayush Mukharji
              Aug 22 at 8:01











            • @AayushMukharji - I would have thought so, though you are supposed to show your working
              – Henry
              Aug 22 at 8:04










            • According to book answer is 7/3 and my answer doesn't match.
              – Aayush Mukharji
              Aug 22 at 8:04










            • $mathbb P (A cup B)$ is a probability and so cannot be greater than $1$. Not much help but $frac713=mathbb P (B cup C)$
              – Henry
              Aug 22 at 8:05










            • @AayushMukharji If the book says that the probability is $7/3$, then that is clearly an error because probabilities can only be between $0$ and $1$, and $7/3 > 1$.
              – Eff
              Aug 22 at 8:07
















            • Is the answer 10/13?
              – Aayush Mukharji
              Aug 22 at 8:01











            • @AayushMukharji - I would have thought so, though you are supposed to show your working
              – Henry
              Aug 22 at 8:04










            • According to book answer is 7/3 and my answer doesn't match.
              – Aayush Mukharji
              Aug 22 at 8:04










            • $mathbb P (A cup B)$ is a probability and so cannot be greater than $1$. Not much help but $frac713=mathbb P (B cup C)$
              – Henry
              Aug 22 at 8:05










            • @AayushMukharji If the book says that the probability is $7/3$, then that is clearly an error because probabilities can only be between $0$ and $1$, and $7/3 > 1$.
              – Eff
              Aug 22 at 8:07















            Is the answer 10/13?
            – Aayush Mukharji
            Aug 22 at 8:01





            Is the answer 10/13?
            – Aayush Mukharji
            Aug 22 at 8:01













            @AayushMukharji - I would have thought so, though you are supposed to show your working
            – Henry
            Aug 22 at 8:04




            @AayushMukharji - I would have thought so, though you are supposed to show your working
            – Henry
            Aug 22 at 8:04












            According to book answer is 7/3 and my answer doesn't match.
            – Aayush Mukharji
            Aug 22 at 8:04




            According to book answer is 7/3 and my answer doesn't match.
            – Aayush Mukharji
            Aug 22 at 8:04












            $mathbb P (A cup B)$ is a probability and so cannot be greater than $1$. Not much help but $frac713=mathbb P (B cup C)$
            – Henry
            Aug 22 at 8:05




            $mathbb P (A cup B)$ is a probability and so cannot be greater than $1$. Not much help but $frac713=mathbb P (B cup C)$
            – Henry
            Aug 22 at 8:05












            @AayushMukharji If the book says that the probability is $7/3$, then that is clearly an error because probabilities can only be between $0$ and $1$, and $7/3 > 1$.
            – Eff
            Aug 22 at 8:07




            @AayushMukharji If the book says that the probability is $7/3$, then that is clearly an error because probabilities can only be between $0$ and $1$, and $7/3 > 1$.
            – Eff
            Aug 22 at 8:07


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