using Bayes’ Rule to calculate conditional probability












1












$begingroup$


i have following problem,



"Students who party before an exam are twice as likely to fail the exam as those who don't party (and presumably study). Of 20% of students partied before the exam, what percentage of students who failed the exam went partying?"



i believe that this problem related to conditional probability, but i couldn't find all necessary elements for answer. appreciate your help.






conditional-probability







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$endgroup$












  • $begingroup$
    Do you understand Bayes’ theorem?
    $endgroup$
    – gHem
    Dec 20 '18 at 4:38










  • $begingroup$
    Is that supposed to read If 20% of students... instead of Of 20% of students... ?
    $endgroup$
    – David Diaz
    Dec 20 '18 at 4:53








  • 1




    $begingroup$
    An alternative approach to Aaron's answer is to first write down the given info as: begin{align}P[fail : mid : party] &= 2P[ fail : mid : party^c]\ P[party] &= 0.2end{align} and you want to compute $P[party : mid : fail]$.
    $endgroup$
    – Michael
    Dec 20 '18 at 5:18


















1












$begingroup$


i have following problem,



"Students who party before an exam are twice as likely to fail the exam as those who don't party (and presumably study). Of 20% of students partied before the exam, what percentage of students who failed the exam went partying?"



i believe that this problem related to conditional probability, but i couldn't find all necessary elements for answer. appreciate your help.






conditional-probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you understand Bayes’ theorem?
    $endgroup$
    – gHem
    Dec 20 '18 at 4:38










  • $begingroup$
    Is that supposed to read If 20% of students... instead of Of 20% of students... ?
    $endgroup$
    – David Diaz
    Dec 20 '18 at 4:53








  • 1




    $begingroup$
    An alternative approach to Aaron's answer is to first write down the given info as: begin{align}P[fail : mid : party] &= 2P[ fail : mid : party^c]\ P[party] &= 0.2end{align} and you want to compute $P[party : mid : fail]$.
    $endgroup$
    – Michael
    Dec 20 '18 at 5:18
















1












1








1


1



$begingroup$


i have following problem,



"Students who party before an exam are twice as likely to fail the exam as those who don't party (and presumably study). Of 20% of students partied before the exam, what percentage of students who failed the exam went partying?"



i believe that this problem related to conditional probability, but i couldn't find all necessary elements for answer. appreciate your help.






conditional-probability







share|cite|improve this question









$endgroup$




i have following problem,



"Students who party before an exam are twice as likely to fail the exam as those who don't party (and presumably study). Of 20% of students partied before the exam, what percentage of students who failed the exam went partying?"



i believe that this problem related to conditional probability, but i couldn't find all necessary elements for answer. appreciate your help.






conditional-probability




conditional-probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 20 '18 at 4:30









NourNour

254




254












  • $begingroup$
    Do you understand Bayes’ theorem?
    $endgroup$
    – gHem
    Dec 20 '18 at 4:38










  • $begingroup$
    Is that supposed to read If 20% of students... instead of Of 20% of students... ?
    $endgroup$
    – David Diaz
    Dec 20 '18 at 4:53








  • 1




    $begingroup$
    An alternative approach to Aaron's answer is to first write down the given info as: begin{align}P[fail : mid : party] &= 2P[ fail : mid : party^c]\ P[party] &= 0.2end{align} and you want to compute $P[party : mid : fail]$.
    $endgroup$
    – Michael
    Dec 20 '18 at 5:18




















  • $begingroup$
    Do you understand Bayes’ theorem?
    $endgroup$
    – gHem
    Dec 20 '18 at 4:38










  • $begingroup$
    Is that supposed to read If 20% of students... instead of Of 20% of students... ?
    $endgroup$
    – David Diaz
    Dec 20 '18 at 4:53








  • 1




    $begingroup$
    An alternative approach to Aaron's answer is to first write down the given info as: begin{align}P[fail : mid : party] &= 2P[ fail : mid : party^c]\ P[party] &= 0.2end{align} and you want to compute $P[party : mid : fail]$.
    $endgroup$
    – Michael
    Dec 20 '18 at 5:18


















$begingroup$
Do you understand Bayes’ theorem?
$endgroup$
– gHem
Dec 20 '18 at 4:38




$begingroup$
Do you understand Bayes’ theorem?
$endgroup$
– gHem
Dec 20 '18 at 4:38












$begingroup$
Is that supposed to read If 20% of students... instead of Of 20% of students... ?
$endgroup$
– David Diaz
Dec 20 '18 at 4:53






$begingroup$
Is that supposed to read If 20% of students... instead of Of 20% of students... ?
$endgroup$
– David Diaz
Dec 20 '18 at 4:53






1




1




$begingroup$
An alternative approach to Aaron's answer is to first write down the given info as: begin{align}P[fail : mid : party] &= 2P[ fail : mid : party^c]\ P[party] &= 0.2end{align} and you want to compute $P[party : mid : fail]$.
$endgroup$
– Michael
Dec 20 '18 at 5:18






$begingroup$
An alternative approach to Aaron's answer is to first write down the given info as: begin{align}P[fail : mid : party] &= 2P[ fail : mid : party^c]\ P[party] &= 0.2end{align} and you want to compute $P[party : mid : fail]$.
$endgroup$
– Michael
Dec 20 '18 at 5:18












2 Answers
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$begingroup$

Let $x$ be the total number of students and $p$ be the probability of a student who didn't party failing the exam. The probability of a student who partied before the exam failing the exam is then $2p$.



$x/5$ students partied, out of which $2px/5$ failed. Out of the $4x/5$ who didn't party, $4px/5$ failed the exam. The total students who failed is $2px/5+4px/5=6px/5$, out of which $2px/5$ partied. The required probability is $2/6=1/3$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I would recommend drawing a tree diagram to start with. Look at https://www.mathsisfun.com/data/probability-events-conditional.html for more information.



    In your case, I assigned percentages making sure to fulfill your condition.



    20% partied - Of them, 60% of them failed and 40% of them passed, while

    80% no party - Of them, 30% of them failed and 70% of them passed.



    Using these numbers, you can that if taking a sample out of $100$ for example, you will see that 12 people failed who went partying and 24 failed who didn't party. So the percentage who failed (and went partying) is $33.3$%.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      Let $x$ be the total number of students and $p$ be the probability of a student who didn't party failing the exam. The probability of a student who partied before the exam failing the exam is then $2p$.



      $x/5$ students partied, out of which $2px/5$ failed. Out of the $4x/5$ who didn't party, $4px/5$ failed the exam. The total students who failed is $2px/5+4px/5=6px/5$, out of which $2px/5$ partied. The required probability is $2/6=1/3$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $x$ be the total number of students and $p$ be the probability of a student who didn't party failing the exam. The probability of a student who partied before the exam failing the exam is then $2p$.



        $x/5$ students partied, out of which $2px/5$ failed. Out of the $4x/5$ who didn't party, $4px/5$ failed the exam. The total students who failed is $2px/5+4px/5=6px/5$, out of which $2px/5$ partied. The required probability is $2/6=1/3$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $x$ be the total number of students and $p$ be the probability of a student who didn't party failing the exam. The probability of a student who partied before the exam failing the exam is then $2p$.



          $x/5$ students partied, out of which $2px/5$ failed. Out of the $4x/5$ who didn't party, $4px/5$ failed the exam. The total students who failed is $2px/5+4px/5=6px/5$, out of which $2px/5$ partied. The required probability is $2/6=1/3$.






          share|cite|improve this answer









          $endgroup$



          Let $x$ be the total number of students and $p$ be the probability of a student who didn't party failing the exam. The probability of a student who partied before the exam failing the exam is then $2p$.



          $x/5$ students partied, out of which $2px/5$ failed. Out of the $4x/5$ who didn't party, $4px/5$ failed the exam. The total students who failed is $2px/5+4px/5=6px/5$, out of which $2px/5$ partied. The required probability is $2/6=1/3$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 6:42









          Shubham JohriShubham Johri

          5,525818




          5,525818























              1












              $begingroup$

              I would recommend drawing a tree diagram to start with. Look at https://www.mathsisfun.com/data/probability-events-conditional.html for more information.



              In your case, I assigned percentages making sure to fulfill your condition.



              20% partied - Of them, 60% of them failed and 40% of them passed, while

              80% no party - Of them, 30% of them failed and 70% of them passed.



              Using these numbers, you can that if taking a sample out of $100$ for example, you will see that 12 people failed who went partying and 24 failed who didn't party. So the percentage who failed (and went partying) is $33.3$%.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                I would recommend drawing a tree diagram to start with. Look at https://www.mathsisfun.com/data/probability-events-conditional.html for more information.



                In your case, I assigned percentages making sure to fulfill your condition.



                20% partied - Of them, 60% of them failed and 40% of them passed, while

                80% no party - Of them, 30% of them failed and 70% of them passed.



                Using these numbers, you can that if taking a sample out of $100$ for example, you will see that 12 people failed who went partying and 24 failed who didn't party. So the percentage who failed (and went partying) is $33.3$%.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I would recommend drawing a tree diagram to start with. Look at https://www.mathsisfun.com/data/probability-events-conditional.html for more information.



                  In your case, I assigned percentages making sure to fulfill your condition.



                  20% partied - Of them, 60% of them failed and 40% of them passed, while

                  80% no party - Of them, 30% of them failed and 70% of them passed.



                  Using these numbers, you can that if taking a sample out of $100$ for example, you will see that 12 people failed who went partying and 24 failed who didn't party. So the percentage who failed (and went partying) is $33.3$%.






                  share|cite|improve this answer









                  $endgroup$



                  I would recommend drawing a tree diagram to start with. Look at https://www.mathsisfun.com/data/probability-events-conditional.html for more information.



                  In your case, I assigned percentages making sure to fulfill your condition.



                  20% partied - Of them, 60% of them failed and 40% of them passed, while

                  80% no party - Of them, 30% of them failed and 70% of them passed.



                  Using these numbers, you can that if taking a sample out of $100$ for example, you will see that 12 people failed who went partying and 24 failed who didn't party. So the percentage who failed (and went partying) is $33.3$%.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 4:58









                  AfronPieAfronPie

                  322213




                  322213






























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