probability of a board containing a certain number of accountants












1














A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?



cant figure this one out






probability statistics







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  • 2




    binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
    – Lost1
    Apr 14 '13 at 0:11


















1














A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?



cant figure this one out






probability statistics







share|cite|improve this question


















  • 2




    binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
    – Lost1
    Apr 14 '13 at 0:11
















1












1








1







A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?



cant figure this one out






probability statistics







share|cite|improve this question













A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?



cant figure this one out






probability statistics




probability statistics






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Apr 14 '13 at 0:08









notamathwiz

50341330




50341330








  • 2




    binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
    – Lost1
    Apr 14 '13 at 0:11
















  • 2




    binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
    – Lost1
    Apr 14 '13 at 0:11










2




2




binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11






binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11












2 Answers
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oldest

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1














There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.



But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?






share|cite|improve this answer





















  • so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
    – notamathwiz
    Apr 14 '13 at 0:34












  • You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
    – André Nicolas
    Apr 14 '13 at 0:37










  • this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
    – notamathwiz
    Apr 14 '13 at 0:40










  • The denominator is right. The numerator isn't.
    – André Nicolas
    Apr 14 '13 at 1:09






  • 1




    Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
    – André Nicolas
    Apr 14 '13 at 1:37



















-1














Using the Hypergeometric distribution formular,
$$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$



$N=12$ , $n=5$ ,
$k=4$ ,
$x=2$ ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$






share|cite|improve this answer










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    2 Answers
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    2 Answers
    2






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    active

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    active

    oldest

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    1














    There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.



    But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?






    share|cite|improve this answer





















    • so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
      – notamathwiz
      Apr 14 '13 at 0:34












    • You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
      – André Nicolas
      Apr 14 '13 at 0:37










    • this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
      – notamathwiz
      Apr 14 '13 at 0:40










    • The denominator is right. The numerator isn't.
      – André Nicolas
      Apr 14 '13 at 1:09






    • 1




      Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
      – André Nicolas
      Apr 14 '13 at 1:37
















    1














    There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.



    But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?






    share|cite|improve this answer





















    • so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
      – notamathwiz
      Apr 14 '13 at 0:34












    • You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
      – André Nicolas
      Apr 14 '13 at 0:37










    • this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
      – notamathwiz
      Apr 14 '13 at 0:40










    • The denominator is right. The numerator isn't.
      – André Nicolas
      Apr 14 '13 at 1:09






    • 1




      Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
      – André Nicolas
      Apr 14 '13 at 1:37














    1












    1








    1






    There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.



    But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?






    share|cite|improve this answer












    There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.



    But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 14 '13 at 0:27









    André Nicolas

    451k36422806




    451k36422806












    • so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
      – notamathwiz
      Apr 14 '13 at 0:34












    • You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
      – André Nicolas
      Apr 14 '13 at 0:37










    • this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
      – notamathwiz
      Apr 14 '13 at 0:40










    • The denominator is right. The numerator isn't.
      – André Nicolas
      Apr 14 '13 at 1:09






    • 1




      Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
      – André Nicolas
      Apr 14 '13 at 1:37


















    • so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
      – notamathwiz
      Apr 14 '13 at 0:34












    • You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
      – André Nicolas
      Apr 14 '13 at 0:37










    • this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
      – notamathwiz
      Apr 14 '13 at 0:40










    • The denominator is right. The numerator isn't.
      – André Nicolas
      Apr 14 '13 at 1:09






    • 1




      Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
      – André Nicolas
      Apr 14 '13 at 1:37
















    so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
    – notamathwiz
    Apr 14 '13 at 0:34






    so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
    – notamathwiz
    Apr 14 '13 at 0:34














    You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
    – André Nicolas
    Apr 14 '13 at 0:37




    You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
    – André Nicolas
    Apr 14 '13 at 0:37












    this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
    – notamathwiz
    Apr 14 '13 at 0:40




    this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
    – notamathwiz
    Apr 14 '13 at 0:40












    The denominator is right. The numerator isn't.
    – André Nicolas
    Apr 14 '13 at 1:09




    The denominator is right. The numerator isn't.
    – André Nicolas
    Apr 14 '13 at 1:09




    1




    1




    Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
    – André Nicolas
    Apr 14 '13 at 1:37




    Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
    – André Nicolas
    Apr 14 '13 at 1:37











    -1














    Using the Hypergeometric distribution formular,
    $$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$



    $N=12$ , $n=5$ ,
    $k=4$ ,
    $x=2$ ,
    $n-x=3$ (non-accountants) , $p(x) = 14/33$






    share|cite|improve this answer










    New contributor




    Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      -1














      Using the Hypergeometric distribution formular,
      $$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$



      $N=12$ , $n=5$ ,
      $k=4$ ,
      $x=2$ ,
      $n-x=3$ (non-accountants) , $p(x) = 14/33$






      share|cite|improve this answer










      New contributor




      Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















        -1












        -1








        -1






        Using the Hypergeometric distribution formular,
        $$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$



        $N=12$ , $n=5$ ,
        $k=4$ ,
        $x=2$ ,
        $n-x=3$ (non-accountants) , $p(x) = 14/33$






        share|cite|improve this answer










        New contributor




        Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        Using the Hypergeometric distribution formular,
        $$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$



        $N=12$ , $n=5$ ,
        $k=4$ ,
        $x=2$ ,
        $n-x=3$ (non-accountants) , $p(x) = 14/33$







        share|cite|improve this answer










        New contributor




        Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday









        Davide Giraudo

        125k16150260




        125k16150260






        New contributor




        Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered yesterday









        Janet Anagli

        1




        1




        New contributor




        Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























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