What is the probability that both tubes are good?

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This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?




probability




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  • Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
    – Joffan
    Jul 14 at 23:35














up vote
1
down vote

favorite












enter image description here
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?




probability




share|cite|improve this question



















  • Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
    – Joffan
    Jul 14 at 23:35












up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?




probability




share|cite|improve this question











enter image description here
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?





probability





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asked Jul 14 at 23:03









Rayri

485




485











  • Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
    – Joffan
    Jul 14 at 23:35
















  • Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
    – Joffan
    Jul 14 at 23:35















Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35




Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35










2 Answers
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$P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.



$P(A)=frac13$, you're right.



To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.






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    You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$



    Similarly the probability both are bad is $$frac410timesfrac39=frac215$$



    And so the probability at least one is good $$1-frac215 = frac1315$$



    which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)



    so the answer to the question is $$fracfrac13,frac1315,=frac513$$






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      2 Answers
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      2 Answers
      2






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      up vote
      1
      down vote













      $P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.



      $P(A)=frac13$, you're right.



      To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.






      share|cite|improve this answer

























        up vote
        1
        down vote













        $P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.



        $P(A)=frac13$, you're right.



        To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          $P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.



          $P(A)=frac13$, you're right.



          To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.






          share|cite|improve this answer













          $P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.



          $P(A)=frac13$, you're right.



          To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 14 at 23:20









          Arnaud Mortier

          19.2k22159




          19.2k22159




















              up vote
              1
              down vote













              You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$



              Similarly the probability both are bad is $$frac410timesfrac39=frac215$$



              And so the probability at least one is good $$1-frac215 = frac1315$$



              which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)



              so the answer to the question is $$fracfrac13,frac1315,=frac513$$






              share|cite|improve this answer



























                up vote
                1
                down vote













                You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$



                Similarly the probability both are bad is $$frac410timesfrac39=frac215$$



                And so the probability at least one is good $$1-frac215 = frac1315$$



                which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)



                so the answer to the question is $$fracfrac13,frac1315,=frac513$$






                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$



                  Similarly the probability both are bad is $$frac410timesfrac39=frac215$$



                  And so the probability at least one is good $$1-frac215 = frac1315$$



                  which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)



                  so the answer to the question is $$fracfrac13,frac1315,=frac513$$






                  share|cite|improve this answer















                  You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$



                  Similarly the probability both are bad is $$frac410timesfrac39=frac215$$



                  And so the probability at least one is good $$1-frac215 = frac1315$$



                  which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)



                  so the answer to the question is $$fracfrac13,frac1315,=frac513$$







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 14 at 23:25


























                  answered Jul 14 at 23:20









                  Henry

                  93.1k469147




                  93.1k469147






















                       

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