Probability distribution and floor functions: is this the right approach?












0












$begingroup$


The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.



(a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?



(b) Find the limits of these probabilities as $n rightarrowinfty$.



(c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?



MY ATTEMPT



The book gives the following answers to each case:



(a) It uses the floor function in order to solve this problem. Precisely, we have:
begin{align*}
textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}



From whence we obtain that



begin{align*}
textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}



(b) I do not know how to handle limits with floor function. But it says that it should be
begin{align*}
lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
end{align*}



(c) It is the same answer as given at (a).



Is there another approach to this problem? Thanks in advance.






probability probability-theory proof-verification







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    0












    $begingroup$


    The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.



    (a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?



    (b) Find the limits of these probabilities as $n rightarrowinfty$.



    (c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?



    MY ATTEMPT



    The book gives the following answers to each case:



    (a) It uses the floor function in order to solve this problem. Precisely, we have:
    begin{align*}
    textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
    end{align*}



    From whence we obtain that



    begin{align*}
    textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
    end{align*}



    (b) I do not know how to handle limits with floor function. But it says that it should be
    begin{align*}
    lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
    end{align*}



    (c) It is the same answer as given at (a).



    Is there another approach to this problem? Thanks in advance.






    probability probability-theory proof-verification







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.



      (a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?



      (b) Find the limits of these probabilities as $n rightarrowinfty$.



      (c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?



      MY ATTEMPT



      The book gives the following answers to each case:



      (a) It uses the floor function in order to solve this problem. Precisely, we have:
      begin{align*}
      textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
      end{align*}



      From whence we obtain that



      begin{align*}
      textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
      end{align*}



      (b) I do not know how to handle limits with floor function. But it says that it should be
      begin{align*}
      lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
      end{align*}



      (c) It is the same answer as given at (a).



      Is there another approach to this problem? Thanks in advance.






      probability probability-theory proof-verification







      share|cite|improve this question











      $endgroup$




      The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.



      (a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?



      (b) Find the limits of these probabilities as $n rightarrowinfty$.



      (c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?



      MY ATTEMPT



      The book gives the following answers to each case:



      (a) It uses the floor function in order to solve this problem. Precisely, we have:
      begin{align*}
      textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
      end{align*}



      From whence we obtain that



      begin{align*}
      textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
      end{align*}



      (b) I do not know how to handle limits with floor function. But it says that it should be
      begin{align*}
      lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
      end{align*}



      (c) It is the same answer as given at (a).



      Is there another approach to this problem? Thanks in advance.






      probability probability-theory proof-verification




      probability probability-theory proof-verification






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      edited Jan 14 at 2:26







      user1337

















      asked Jan 11 at 19:48









      user1337user1337

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          $begingroup$

          Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
          $$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
          and if $n=3k+2$
          for some $k$, then
          $$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
          Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
          Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
          $$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
          as you suspected. The argument works the same way for $D_{4}$.



          And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.






          share|cite|improve this answer









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            $begingroup$

            Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
            $$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
            and if $n=3k+2$
            for some $k$, then
            $$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
            Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
            Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
            $$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
            as you suspected. The argument works the same way for $D_{4}$.



            And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
              $$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
              and if $n=3k+2$
              for some $k$, then
              $$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
              Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
              Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
              $$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
              as you suspected. The argument works the same way for $D_{4}$.



              And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
                $$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
                and if $n=3k+2$
                for some $k$, then
                $$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
                Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
                Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
                $$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
                as you suspected. The argument works the same way for $D_{4}$.



                And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.






                share|cite|improve this answer









                $endgroup$



                Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
                $$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
                and if $n=3k+2$
                for some $k$, then
                $$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
                Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
                Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
                $$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
                as you suspected. The argument works the same way for $D_{4}$.



                And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 11 at 20:04









                pwerthpwerth

                3,233417




                3,233417






























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