Finding Probability of identical and distinct objects












1












$begingroup$


10 objects are to be distributed among 10 people randomly. What is the probability that at least one of them will get nothing?



Now, I had thought of two ways of doing it:




  1. I assumed here objects to be identical.
    Hence total number of ways in which 10 objects would be distributed among 10 people is $frac{19!}{9!10!}$.
    Further, the total number of ways equals every one gets something + at least one gets nothing.
    And, there is only one way in which every one gets something.
    Hence the number of ways in which at least one person gets nothing is $frac{19!}{9!10!} - 1$.
    The required probability thus equals:
    $frac{frac{19!}{9!10!} - 1}{frac{19!}{9!10!}}$.


  2. I assumed all objects to be distinct.
    Hence number of total ways is $10^{10}$.
    Also, total ways = every one gets something + at least one gets nothing.
    Further, every one gets something in 10! Ways.
    Hence, number of ways in which at least one gets nothing = 10^10-10!.
    Hence the required probability is:
    $frac{10^{10}-10!}{10^{10}}$.



The problem is that my answers are different for these two methods.



I strongly feel that the answer should be same irrespective of whether they are distinct or identical. How and where am I going wrong? Why is the probability of quantity dependent on the quality of the objects?






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    10 objects are to be distributed among 10 people randomly. What is the probability that at least one of them will get nothing?



    Now, I had thought of two ways of doing it:




    1. I assumed here objects to be identical.
      Hence total number of ways in which 10 objects would be distributed among 10 people is $frac{19!}{9!10!}$.
      Further, the total number of ways equals every one gets something + at least one gets nothing.
      And, there is only one way in which every one gets something.
      Hence the number of ways in which at least one person gets nothing is $frac{19!}{9!10!} - 1$.
      The required probability thus equals:
      $frac{frac{19!}{9!10!} - 1}{frac{19!}{9!10!}}$.


    2. I assumed all objects to be distinct.
      Hence number of total ways is $10^{10}$.
      Also, total ways = every one gets something + at least one gets nothing.
      Further, every one gets something in 10! Ways.
      Hence, number of ways in which at least one gets nothing = 10^10-10!.
      Hence the required probability is:
      $frac{10^{10}-10!}{10^{10}}$.



    The problem is that my answers are different for these two methods.



    I strongly feel that the answer should be same irrespective of whether they are distinct or identical. How and where am I going wrong? Why is the probability of quantity dependent on the quality of the objects?






    probability probability-theory probability-distributions







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      10 objects are to be distributed among 10 people randomly. What is the probability that at least one of them will get nothing?



      Now, I had thought of two ways of doing it:




      1. I assumed here objects to be identical.
        Hence total number of ways in which 10 objects would be distributed among 10 people is $frac{19!}{9!10!}$.
        Further, the total number of ways equals every one gets something + at least one gets nothing.
        And, there is only one way in which every one gets something.
        Hence the number of ways in which at least one person gets nothing is $frac{19!}{9!10!} - 1$.
        The required probability thus equals:
        $frac{frac{19!}{9!10!} - 1}{frac{19!}{9!10!}}$.


      2. I assumed all objects to be distinct.
        Hence number of total ways is $10^{10}$.
        Also, total ways = every one gets something + at least one gets nothing.
        Further, every one gets something in 10! Ways.
        Hence, number of ways in which at least one gets nothing = 10^10-10!.
        Hence the required probability is:
        $frac{10^{10}-10!}{10^{10}}$.



      The problem is that my answers are different for these two methods.



      I strongly feel that the answer should be same irrespective of whether they are distinct or identical. How and where am I going wrong? Why is the probability of quantity dependent on the quality of the objects?






      probability probability-theory probability-distributions







      share|cite|improve this question











      $endgroup$




      10 objects are to be distributed among 10 people randomly. What is the probability that at least one of them will get nothing?



      Now, I had thought of two ways of doing it:




      1. I assumed here objects to be identical.
        Hence total number of ways in which 10 objects would be distributed among 10 people is $frac{19!}{9!10!}$.
        Further, the total number of ways equals every one gets something + at least one gets nothing.
        And, there is only one way in which every one gets something.
        Hence the number of ways in which at least one person gets nothing is $frac{19!}{9!10!} - 1$.
        The required probability thus equals:
        $frac{frac{19!}{9!10!} - 1}{frac{19!}{9!10!}}$.


      2. I assumed all objects to be distinct.
        Hence number of total ways is $10^{10}$.
        Also, total ways = every one gets something + at least one gets nothing.
        Further, every one gets something in 10! Ways.
        Hence, number of ways in which at least one gets nothing = 10^10-10!.
        Hence the required probability is:
        $frac{10^{10}-10!}{10^{10}}$.



      The problem is that my answers are different for these two methods.



      I strongly feel that the answer should be same irrespective of whether they are distinct or identical. How and where am I going wrong? Why is the probability of quantity dependent on the quality of the objects?






      probability probability-theory probability-distributions




      probability probability-theory probability-distributions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 10:40









      jvdhooft

      5,65961641




      5,65961641










      asked Feb 2 at 10:04









      Yashkalp SharmaYashkalp Sharma

      82




      82






















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          $begingroup$

          The probability of at least one of the people getting nothing, equals 1 minus the probability of all of them getting something. There are $10^{10}$ equally likely ways to distribute the objects, and only $10!$ valid ways in which this can be done. We indeed find:



          $$1 - frac{10!}{10^{10}}$$



          The problem with your other approach, is that the probabilities of the ${19 choose 9}$ possible combinations are not the same. For instance, the probability of all gifts being given to the first person equals $frac{1}{10^{10}}$, while the probability of each person getting something equals $frac{10!}{10^{10}}$.



          To make this more tangible, try doing the same for two people and two objects. The probability of the first person getting both items equals $left(frac{1}{2}right)^2 = frac{1}{4}$, while the probability of both people getting one object equals $1 cdot frac{1}{2} = frac{1}{2}$. Indeed, the probabilities of the three possible combinations are not the same.






          share|cite|improve this answer











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            $begingroup$

            The probability of at least one of the people getting nothing, equals 1 minus the probability of all of them getting something. There are $10^{10}$ equally likely ways to distribute the objects, and only $10!$ valid ways in which this can be done. We indeed find:



            $$1 - frac{10!}{10^{10}}$$



            The problem with your other approach, is that the probabilities of the ${19 choose 9}$ possible combinations are not the same. For instance, the probability of all gifts being given to the first person equals $frac{1}{10^{10}}$, while the probability of each person getting something equals $frac{10!}{10^{10}}$.



            To make this more tangible, try doing the same for two people and two objects. The probability of the first person getting both items equals $left(frac{1}{2}right)^2 = frac{1}{4}$, while the probability of both people getting one object equals $1 cdot frac{1}{2} = frac{1}{2}$. Indeed, the probabilities of the three possible combinations are not the same.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              The probability of at least one of the people getting nothing, equals 1 minus the probability of all of them getting something. There are $10^{10}$ equally likely ways to distribute the objects, and only $10!$ valid ways in which this can be done. We indeed find:



              $$1 - frac{10!}{10^{10}}$$



              The problem with your other approach, is that the probabilities of the ${19 choose 9}$ possible combinations are not the same. For instance, the probability of all gifts being given to the first person equals $frac{1}{10^{10}}$, while the probability of each person getting something equals $frac{10!}{10^{10}}$.



              To make this more tangible, try doing the same for two people and two objects. The probability of the first person getting both items equals $left(frac{1}{2}right)^2 = frac{1}{4}$, while the probability of both people getting one object equals $1 cdot frac{1}{2} = frac{1}{2}$. Indeed, the probabilities of the three possible combinations are not the same.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                The probability of at least one of the people getting nothing, equals 1 minus the probability of all of them getting something. There are $10^{10}$ equally likely ways to distribute the objects, and only $10!$ valid ways in which this can be done. We indeed find:



                $$1 - frac{10!}{10^{10}}$$



                The problem with your other approach, is that the probabilities of the ${19 choose 9}$ possible combinations are not the same. For instance, the probability of all gifts being given to the first person equals $frac{1}{10^{10}}$, while the probability of each person getting something equals $frac{10!}{10^{10}}$.



                To make this more tangible, try doing the same for two people and two objects. The probability of the first person getting both items equals $left(frac{1}{2}right)^2 = frac{1}{4}$, while the probability of both people getting one object equals $1 cdot frac{1}{2} = frac{1}{2}$. Indeed, the probabilities of the three possible combinations are not the same.






                share|cite|improve this answer











                $endgroup$



                The probability of at least one of the people getting nothing, equals 1 minus the probability of all of them getting something. There are $10^{10}$ equally likely ways to distribute the objects, and only $10!$ valid ways in which this can be done. We indeed find:



                $$1 - frac{10!}{10^{10}}$$



                The problem with your other approach, is that the probabilities of the ${19 choose 9}$ possible combinations are not the same. For instance, the probability of all gifts being given to the first person equals $frac{1}{10^{10}}$, while the probability of each person getting something equals $frac{10!}{10^{10}}$.



                To make this more tangible, try doing the same for two people and two objects. The probability of the first person getting both items equals $left(frac{1}{2}right)^2 = frac{1}{4}$, while the probability of both people getting one object equals $1 cdot frac{1}{2} = frac{1}{2}$. Indeed, the probabilities of the three possible combinations are not the same.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 2 at 10:42

























                answered Feb 2 at 10:36









                jvdhooftjvdhooft

                5,65961641




                5,65961641






























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