Probability-Bayes's rule












0














Here is the question:



There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.



We choose randomly a basket and take out a ball after another with returning.



a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?



b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?




ok,here is my approach:



a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $



I don't know how to get rid of the sigma



b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
$frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$



here too I could'nt get rid of i from the whole equation.any help please?






probability







share|cite|improve this question



























    0














    Here is the question:



    There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.



    We choose randomly a basket and take out a ball after another with returning.



    a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?



    b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?




    ok,here is my approach:



    a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $



    I don't know how to get rid of the sigma



    b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
    $frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$



    here too I could'nt get rid of i from the whole equation.any help please?






    probability







    share|cite|improve this question

























      0












      0








      0


      0





      Here is the question:



      There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.



      We choose randomly a basket and take out a ball after another with returning.



      a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?



      b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?




      ok,here is my approach:



      a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $



      I don't know how to get rid of the sigma



      b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
      $frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$



      here too I could'nt get rid of i from the whole equation.any help please?






      probability







      share|cite|improve this question













      Here is the question:



      There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.



      We choose randomly a basket and take out a ball after another with returning.



      a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?



      b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?




      ok,here is my approach:



      a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $



      I don't know how to get rid of the sigma



      b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
      $frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$



      here too I could'nt get rid of i from the whole equation.any help please?






      probability




      probability






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      share|cite|improve this question











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      asked Nov 25 '18 at 16:56









      Adddison

      846




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          For (a), you've mixed up the summation index -- and what is $a_n$? The denominator should be
          $$sum_{k=0}^N 2frac{k}{N}frac{N-k}{N}frac{1}{N+1}=
          frac{2}{N^2(N+1)}sum_{k=0}^{N}(kN-k^2)=frac{2}{N^2(N+1)}
          left(Nsum_{k=0}^{N}k-sum_{k=0}^{N}k^2right).
          $$
          The last two sums are all known; finally you should arrive in $frac{N-1}{3N}$.



          For (b), the denominator is
          $$sum_{k=0}^N left(frac{N-k}{N}right)^2frac{1}{N+1}
          =frac{1}{N(N+1)}sum_{k=0}^{N}(N-k)^2.$$
          Similarly you should arrive in $frac{2N+1}{3N}$.






          share|cite|improve this answer





















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            For (a), you've mixed up the summation index -- and what is $a_n$? The denominator should be
            $$sum_{k=0}^N 2frac{k}{N}frac{N-k}{N}frac{1}{N+1}=
            frac{2}{N^2(N+1)}sum_{k=0}^{N}(kN-k^2)=frac{2}{N^2(N+1)}
            left(Nsum_{k=0}^{N}k-sum_{k=0}^{N}k^2right).
            $$
            The last two sums are all known; finally you should arrive in $frac{N-1}{3N}$.



            For (b), the denominator is
            $$sum_{k=0}^N left(frac{N-k}{N}right)^2frac{1}{N+1}
            =frac{1}{N(N+1)}sum_{k=0}^{N}(N-k)^2.$$
            Similarly you should arrive in $frac{2N+1}{3N}$.






            share|cite|improve this answer


























              0














              For (a), you've mixed up the summation index -- and what is $a_n$? The denominator should be
              $$sum_{k=0}^N 2frac{k}{N}frac{N-k}{N}frac{1}{N+1}=
              frac{2}{N^2(N+1)}sum_{k=0}^{N}(kN-k^2)=frac{2}{N^2(N+1)}
              left(Nsum_{k=0}^{N}k-sum_{k=0}^{N}k^2right).
              $$
              The last two sums are all known; finally you should arrive in $frac{N-1}{3N}$.



              For (b), the denominator is
              $$sum_{k=0}^N left(frac{N-k}{N}right)^2frac{1}{N+1}
              =frac{1}{N(N+1)}sum_{k=0}^{N}(N-k)^2.$$
              Similarly you should arrive in $frac{2N+1}{3N}$.






              share|cite|improve this answer
























                0












                0








                0






                For (a), you've mixed up the summation index -- and what is $a_n$? The denominator should be
                $$sum_{k=0}^N 2frac{k}{N}frac{N-k}{N}frac{1}{N+1}=
                frac{2}{N^2(N+1)}sum_{k=0}^{N}(kN-k^2)=frac{2}{N^2(N+1)}
                left(Nsum_{k=0}^{N}k-sum_{k=0}^{N}k^2right).
                $$
                The last two sums are all known; finally you should arrive in $frac{N-1}{3N}$.



                For (b), the denominator is
                $$sum_{k=0}^N left(frac{N-k}{N}right)^2frac{1}{N+1}
                =frac{1}{N(N+1)}sum_{k=0}^{N}(N-k)^2.$$
                Similarly you should arrive in $frac{2N+1}{3N}$.






                share|cite|improve this answer












                For (a), you've mixed up the summation index -- and what is $a_n$? The denominator should be
                $$sum_{k=0}^N 2frac{k}{N}frac{N-k}{N}frac{1}{N+1}=
                frac{2}{N^2(N+1)}sum_{k=0}^{N}(kN-k^2)=frac{2}{N^2(N+1)}
                left(Nsum_{k=0}^{N}k-sum_{k=0}^{N}k^2right).
                $$
                The last two sums are all known; finally you should arrive in $frac{N-1}{3N}$.



                For (b), the denominator is
                $$sum_{k=0}^N left(frac{N-k}{N}right)^2frac{1}{N+1}
                =frac{1}{N(N+1)}sum_{k=0}^{N}(N-k)^2.$$
                Similarly you should arrive in $frac{2N+1}{3N}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 '18 at 21:40









                Michael Hoppe

                10.8k31834




                10.8k31834






























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