ways of selecting consecutive persons sitting at a table












9












$begingroup$


I'm trying to solve the following problem:




Ten people are sitting around a round table. Three of them are chosen
at random to give a presentation. What is the probability that the
three chosen people were sitting in consecutive seats?




I got the wrong answer but cannot see the error in my reasoning. This is how I see it:



1) the selection of the first person is unconstrained.



2) the next person must be selected from the 2 spots adjacent to the first. So this choice is limited to 2/9 of the possible choices.



3) the third choice must be taken from the one free spot next to the first person chosen, or the one free spot next to the 2nd person chosen. So this choice is limited to 2/8 of the possible choices.



4) multiplying these we get:



2/9 * 2/8 = 1/18


However, the official answer is:




Let's count as our outcomes the ways to select 3 people without regard
to order. There are $binom{10}{3} = 120$ ways to select any 3 people.
The number of successful outcomes is the number of ways to select 3
consecutive people. There are only 10 ways to do this -- think of
first selecting the middle person, then we take his or her two
neighbors. Therefore, the probability is $frac{10}{120} =
> boxed{frac{1}{12}}$.







combinatorics combinations







share|cite|improve this question







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David J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

















    9












    $begingroup$


    I'm trying to solve the following problem:




    Ten people are sitting around a round table. Three of them are chosen
    at random to give a presentation. What is the probability that the
    three chosen people were sitting in consecutive seats?




    I got the wrong answer but cannot see the error in my reasoning. This is how I see it:



    1) the selection of the first person is unconstrained.



    2) the next person must be selected from the 2 spots adjacent to the first. So this choice is limited to 2/9 of the possible choices.



    3) the third choice must be taken from the one free spot next to the first person chosen, or the one free spot next to the 2nd person chosen. So this choice is limited to 2/8 of the possible choices.



    4) multiplying these we get:



    2/9 * 2/8 = 1/18


    However, the official answer is:




    Let's count as our outcomes the ways to select 3 people without regard
    to order. There are $binom{10}{3} = 120$ ways to select any 3 people.
    The number of successful outcomes is the number of ways to select 3
    consecutive people. There are only 10 ways to do this -- think of
    first selecting the middle person, then we take his or her two
    neighbors. Therefore, the probability is $frac{10}{120} =
    > boxed{frac{1}{12}}$    .







    combinatorics combinations







    share|cite|improve this question







    New contributor




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







    $endgroup$















      9












      9








      9





      $begingroup$


      I'm trying to solve the following problem:




      Ten people are sitting around a round table. Three of them are chosen
      at random to give a presentation. What is the probability that the
      three chosen people were sitting in consecutive seats?




      I got the wrong answer but cannot see the error in my reasoning. This is how I see it:



      1) the selection of the first person is unconstrained.



      2) the next person must be selected from the 2 spots adjacent to the first. So this choice is limited to 2/9 of the possible choices.



      3) the third choice must be taken from the one free spot next to the first person chosen, or the one free spot next to the 2nd person chosen. So this choice is limited to 2/8 of the possible choices.



      4) multiplying these we get:



      2/9 * 2/8 = 1/18


      However, the official answer is:




      Let's count as our outcomes the ways to select 3 people without regard
      to order. There are $binom{10}{3} = 120$ ways to select any 3 people.
      The number of successful outcomes is the number of ways to select 3
      consecutive people. There are only 10 ways to do this -- think of
      first selecting the middle person, then we take his or her two
      neighbors. Therefore, the probability is $frac{10}{120} =
      > boxed{frac{1}{12}}$      .







      combinatorics combinations







      share|cite|improve this question







      New contributor




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







      $endgroup$




      I'm trying to solve the following problem:




      Ten people are sitting around a round table. Three of them are chosen
      at random to give a presentation. What is the probability that the
      three chosen people were sitting in consecutive seats?




      I got the wrong answer but cannot see the error in my reasoning. This is how I see it:



      1) the selection of the first person is unconstrained.



      2) the next person must be selected from the 2 spots adjacent to the first. So this choice is limited to 2/9 of the possible choices.



      3) the third choice must be taken from the one free spot next to the first person chosen, or the one free spot next to the 2nd person chosen. So this choice is limited to 2/8 of the possible choices.



      4) multiplying these we get:



      2/9 * 2/8 = 1/18


      However, the official answer is:




      Let's count as our outcomes the ways to select 3 people without regard
      to order. There are $binom{10}{3} = 120$ ways to select any 3 people.
      The number of successful outcomes is the number of ways to select 3
      consecutive people. There are only 10 ways to do this -- think of
      first selecting the middle person, then we take his or her two
      neighbors. Therefore, the probability is $frac{10}{120} =
      > boxed{frac{1}{12}}$      .







      combinatorics combinations




      combinatorics combinations






      share|cite|improve this question







      New contributor




      David J. 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 question







      New contributor




      David J. 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 question




      share|cite|improve this question






      New contributor




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









      asked 22 hours ago









      David J.David J.

      1634




      1634




      New contributor




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





      New contributor





      David J. 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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      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

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          18












          $begingroup$

          You forgot the possibility that the second person can be chosen to sit two seats away from the first, and then the third person is chosen to be the one between the first and the second. This gives an additional $frac29cdot frac18 = frac1{36}$, bringing the total up to $frac1{18}+frac1{36} = frac1{12}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            great answer, thanks!
            $endgroup$
            – David J.
            22 hours ago



















          9












          $begingroup$

          For your mistake see the answer of Arthur.



          A bit more concise solution:



          If the first person has been chosen then yet $2$ out of $9$ must be chosen.



          In $3$ of these cases the three chosen persons will sit consecutively so the probability on that is:$$frac3{binom92}=frac1{12}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            A nice middle ground between the OP's approach and the given official answer.
            $endgroup$
            – Arthur
            21 hours ago













          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          18












          $begingroup$

          You forgot the possibility that the second person can be chosen to sit two seats away from the first, and then the third person is chosen to be the one between the first and the second. This gives an additional $frac29cdot frac18 = frac1{36}$, bringing the total up to $frac1{18}+frac1{36} = frac1{12}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            great answer, thanks!
            $endgroup$
            – David J.
            22 hours ago
















          18












          $begingroup$

          You forgot the possibility that the second person can be chosen to sit two seats away from the first, and then the third person is chosen to be the one between the first and the second. This gives an additional $frac29cdot frac18 = frac1{36}$, bringing the total up to $frac1{18}+frac1{36} = frac1{12}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            great answer, thanks!
            $endgroup$
            – David J.
            22 hours ago














          18












          18








          18





          $begingroup$

          You forgot the possibility that the second person can be chosen to sit two seats away from the first, and then the third person is chosen to be the one between the first and the second. This gives an additional $frac29cdot frac18 = frac1{36}$, bringing the total up to $frac1{18}+frac1{36} = frac1{12}$.






          share|cite|improve this answer









          $endgroup$



          You forgot the possibility that the second person can be chosen to sit two seats away from the first, and then the third person is chosen to be the one between the first and the second. This gives an additional $frac29cdot frac18 = frac1{36}$, bringing the total up to $frac1{18}+frac1{36} = frac1{12}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 22 hours ago









          ArthurArthur

          114k7116198




          114k7116198












          • $begingroup$
            great answer, thanks!
            $endgroup$
            – David J.
            22 hours ago


















          • $begingroup$
            great answer, thanks!
            $endgroup$
            – David J.
            22 hours ago
















          $begingroup$
          great answer, thanks!
          $endgroup$
          – David J.
          22 hours ago




          $begingroup$
          great answer, thanks!
          $endgroup$
          – David J.
          22 hours ago











          9












          $begingroup$

          For your mistake see the answer of Arthur.



          A bit more concise solution:



          If the first person has been chosen then yet $2$ out of $9$ must be chosen.



          In $3$ of these cases the three chosen persons will sit consecutively so the probability on that is:$$frac3{binom92}=frac1{12}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            A nice middle ground between the OP's approach and the given official answer.
            $endgroup$
            – Arthur
            21 hours ago


















          9












          $begingroup$

          For your mistake see the answer of Arthur.



          A bit more concise solution:



          If the first person has been chosen then yet $2$ out of $9$ must be chosen.



          In $3$ of these cases the three chosen persons will sit consecutively so the probability on that is:$$frac3{binom92}=frac1{12}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            A nice middle ground between the OP's approach and the given official answer.
            $endgroup$
            – Arthur
            21 hours ago
















          9












          9








          9





          $begingroup$

          For your mistake see the answer of Arthur.



          A bit more concise solution:



          If the first person has been chosen then yet $2$ out of $9$ must be chosen.



          In $3$ of these cases the three chosen persons will sit consecutively so the probability on that is:$$frac3{binom92}=frac1{12}$$






          share|cite|improve this answer









          $endgroup$



          For your mistake see the answer of Arthur.



          A bit more concise solution:



          If the first person has been chosen then yet $2$ out of $9$ must be chosen.



          In $3$ of these cases the three chosen persons will sit consecutively so the probability on that is:$$frac3{binom92}=frac1{12}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 22 hours ago









          drhabdrhab

          101k544130




          101k544130












          • $begingroup$
            A nice middle ground between the OP's approach and the given official answer.
            $endgroup$
            – Arthur
            21 hours ago




















          • $begingroup$
            A nice middle ground between the OP's approach and the given official answer.
            $endgroup$
            – Arthur
            21 hours ago


















          $begingroup$
          A nice middle ground between the OP's approach and the given official answer.
          $endgroup$
          – Arthur
          21 hours ago






          $begingroup$
          A nice middle ground between the OP's approach and the given official answer.
          $endgroup$
          – Arthur
          21 hours ago












          David J. is a new contributor. Be nice, and check out our Code of Conduct.










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          David J. is a new contributor. Be nice, and check out our Code of Conduct.
















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