Problem of $N+1$ identical urns

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Suppose there are $(N +1)$ identical urns marked $0,1,2, . . . ,N$ which contains $N$ balls. The $k^th$ urn contains $ k $ black and $N−k$ white balls, $k =0,1,2, . . . ,N.$ An urn is chosen at random,
and $n$ random drawings are made from it, the ball drawn being always replaced. If
all the $n$ draws result in black balls, find the probability that the $(n+1)^th$ draw will
also produce a black ball. How does this probability behave as $Nto infty$?



What I attempted:- Since all the $n$ balls drawn are found black, we can consider only those urns which contain at least $n$ black balls. There are $N-n+1$ urns which contain at least $n$ balls. The probability of choosing any one of them is $frac1N-n+1$



Suppose, $A$ =The event of getting black balls in $n$ consecutive draws of a randomly chosen urn

$A_i$ = The $i^th$ urn is chosen. Here $nle i le N$



Now,beginequation
beginaligned
P(A)&=P(AA_n)+P(AA_n+1)+......+P(AA_N) \
&=sum_i=n^N P(A_i) P(A|A_i)\
&=frac1N-n+1 sum_i=n^N left(fraciNright)^n
endaligned
endequation



Similarly, we define

$B$= The event of getting $n+1$ black balls in $n+1$ consecutive draws of a randomly chosen urn.
We obtain that $P(B)=frac1N-n sum_i=n+1^N left(fraciNright)^n+1$



We are required to find, $P(B|A)$ which is given as
beginequation
beginaligned
P(B|A)=&fracP(AB)P(A)\
&=fracfrac1N-n sum_i=n+1^N left(fraciNright)^n+1frac1N-n+1 sum_i=n^N left(fraciNright)^n\
endaligned
endequation



Am I correct?




probability




share|cite|improve this question















  • 1




    Since the balls are being replaced after each draw, all you need is $1$ black ball in the urn.
    – lulu
    Jul 28 at 10:08










  • "Since all the n balls drawn are found black, we can consider only those urns which contain at least n black balls. " No, since you are assuming replacement.
    – Lord Shark the Unknown
    Jul 28 at 10:24










  • Thanks for pointing the mistake. Probably, the correct answer should be $fracfrac1N+1 sum_i=1^N left(fraciNright)^n+1frac1N+1 sum_i=1^N left(fraciNright)^n=frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n$
    – Bhargob
    Jul 28 at 12:41











  • If I am not wrong, the limit of this probability as $N to infty$ can be evaluated as follows:- beginequation lim_N to infty frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n=lim_N to infty N lim_N to infty frac1^n+1+2^n+1+.....+N^n+11^n+2^n+......+N^n=lim_N to inftyN fracint_0^N t^n+1 dtint_0^N t^n+1 dt=fracn+1n+2 endequation I am using Euler-Maclaurin's Summation formula.
    – Bhargob
    Jul 28 at 13:08















up vote
1
down vote

favorite












Suppose there are $(N +1)$ identical urns marked $0,1,2, . . . ,N$ which contains $N$ balls. The $k^th$ urn contains $ k $ black and $N−k$ white balls, $k =0,1,2, . . . ,N.$ An urn is chosen at random,
and $n$ random drawings are made from it, the ball drawn being always replaced. If
all the $n$ draws result in black balls, find the probability that the $(n+1)^th$ draw will
also produce a black ball. How does this probability behave as $Nto infty$?



What I attempted:- Since all the $n$ balls drawn are found black, we can consider only those urns which contain at least $n$ black balls. There are $N-n+1$ urns which contain at least $n$ balls. The probability of choosing any one of them is $frac1N-n+1$



Suppose, $A$ =The event of getting black balls in $n$ consecutive draws of a randomly chosen urn

$A_i$ = The $i^th$ urn is chosen. Here $nle i le N$



Now,beginequation
beginaligned
P(A)&=P(AA_n)+P(AA_n+1)+......+P(AA_N) \
&=sum_i=n^N P(A_i) P(A|A_i)\
&=frac1N-n+1 sum_i=n^N left(fraciNright)^n
endaligned
endequation



Similarly, we define

$B$= The event of getting $n+1$ black balls in $n+1$ consecutive draws of a randomly chosen urn.
We obtain that $P(B)=frac1N-n sum_i=n+1^N left(fraciNright)^n+1$



We are required to find, $P(B|A)$ which is given as
beginequation
beginaligned
P(B|A)=&fracP(AB)P(A)\
&=fracfrac1N-n sum_i=n+1^N left(fraciNright)^n+1frac1N-n+1 sum_i=n^N left(fraciNright)^n\
endaligned
endequation



Am I correct?




probability




share|cite|improve this question















  • 1




    Since the balls are being replaced after each draw, all you need is $1$ black ball in the urn.
    – lulu
    Jul 28 at 10:08










  • "Since all the n balls drawn are found black, we can consider only those urns which contain at least n black balls. " No, since you are assuming replacement.
    – Lord Shark the Unknown
    Jul 28 at 10:24










  • Thanks for pointing the mistake. Probably, the correct answer should be $fracfrac1N+1 sum_i=1^N left(fraciNright)^n+1frac1N+1 sum_i=1^N left(fraciNright)^n=frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n$
    – Bhargob
    Jul 28 at 12:41











  • If I am not wrong, the limit of this probability as $N to infty$ can be evaluated as follows:- beginequation lim_N to infty frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n=lim_N to infty N lim_N to infty frac1^n+1+2^n+1+.....+N^n+11^n+2^n+......+N^n=lim_N to inftyN fracint_0^N t^n+1 dtint_0^N t^n+1 dt=fracn+1n+2 endequation I am using Euler-Maclaurin's Summation formula.
    – Bhargob
    Jul 28 at 13:08













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Suppose there are $(N +1)$ identical urns marked $0,1,2, . . . ,N$ which contains $N$ balls. The $k^th$ urn contains $ k $ black and $N−k$ white balls, $k =0,1,2, . . . ,N.$ An urn is chosen at random,
and $n$ random drawings are made from it, the ball drawn being always replaced. If
all the $n$ draws result in black balls, find the probability that the $(n+1)^th$ draw will
also produce a black ball. How does this probability behave as $Nto infty$?



What I attempted:- Since all the $n$ balls drawn are found black, we can consider only those urns which contain at least $n$ black balls. There are $N-n+1$ urns which contain at least $n$ balls. The probability of choosing any one of them is $frac1N-n+1$



Suppose, $A$ =The event of getting black balls in $n$ consecutive draws of a randomly chosen urn

$A_i$ = The $i^th$ urn is chosen. Here $nle i le N$



Now,beginequation
beginaligned
P(A)&=P(AA_n)+P(AA_n+1)+......+P(AA_N) \
&=sum_i=n^N P(A_i) P(A|A_i)\
&=frac1N-n+1 sum_i=n^N left(fraciNright)^n
endaligned
endequation



Similarly, we define

$B$= The event of getting $n+1$ black balls in $n+1$ consecutive draws of a randomly chosen urn.
We obtain that $P(B)=frac1N-n sum_i=n+1^N left(fraciNright)^n+1$



We are required to find, $P(B|A)$ which is given as
beginequation
beginaligned
P(B|A)=&fracP(AB)P(A)\
&=fracfrac1N-n sum_i=n+1^N left(fraciNright)^n+1frac1N-n+1 sum_i=n^N left(fraciNright)^n\
endaligned
endequation



Am I correct?




probability




share|cite|improve this question











Suppose there are $(N +1)$ identical urns marked $0,1,2, . . . ,N$ which contains $N$ balls. The $k^th$ urn contains $ k $ black and $N−k$ white balls, $k =0,1,2, . . . ,N.$ An urn is chosen at random,
and $n$ random drawings are made from it, the ball drawn being always replaced. If
all the $n$ draws result in black balls, find the probability that the $(n+1)^th$ draw will
also produce a black ball. How does this probability behave as $Nto infty$?



What I attempted:- Since all the $n$ balls drawn are found black, we can consider only those urns which contain at least $n$ black balls. There are $N-n+1$ urns which contain at least $n$ balls. The probability of choosing any one of them is $frac1N-n+1$



Suppose, $A$ =The event of getting black balls in $n$ consecutive draws of a randomly chosen urn

$A_i$ = The $i^th$ urn is chosen. Here $nle i le N$



Now,beginequation
beginaligned
P(A)&=P(AA_n)+P(AA_n+1)+......+P(AA_N) \
&=sum_i=n^N P(A_i) P(A|A_i)\
&=frac1N-n+1 sum_i=n^N left(fraciNright)^n
endaligned
endequation



Similarly, we define

$B$= The event of getting $n+1$ black balls in $n+1$ consecutive draws of a randomly chosen urn.
We obtain that $P(B)=frac1N-n sum_i=n+1^N left(fraciNright)^n+1$



We are required to find, $P(B|A)$ which is given as
beginequation
beginaligned
P(B|A)=&fracP(AB)P(A)\
&=fracfrac1N-n sum_i=n+1^N left(fraciNright)^n+1frac1N-n+1 sum_i=n^N left(fraciNright)^n\
endaligned
endequation



Am I correct?





probability





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 28 at 8:30









Bhargob

434214




434214







  • 1




    Since the balls are being replaced after each draw, all you need is $1$ black ball in the urn.
    – lulu
    Jul 28 at 10:08










  • "Since all the n balls drawn are found black, we can consider only those urns which contain at least n black balls. " No, since you are assuming replacement.
    – Lord Shark the Unknown
    Jul 28 at 10:24










  • Thanks for pointing the mistake. Probably, the correct answer should be $fracfrac1N+1 sum_i=1^N left(fraciNright)^n+1frac1N+1 sum_i=1^N left(fraciNright)^n=frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n$
    – Bhargob
    Jul 28 at 12:41











  • If I am not wrong, the limit of this probability as $N to infty$ can be evaluated as follows:- beginequation lim_N to infty frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n=lim_N to infty N lim_N to infty frac1^n+1+2^n+1+.....+N^n+11^n+2^n+......+N^n=lim_N to inftyN fracint_0^N t^n+1 dtint_0^N t^n+1 dt=fracn+1n+2 endequation I am using Euler-Maclaurin's Summation formula.
    – Bhargob
    Jul 28 at 13:08













  • 1




    Since the balls are being replaced after each draw, all you need is $1$ black ball in the urn.
    – lulu
    Jul 28 at 10:08










  • "Since all the n balls drawn are found black, we can consider only those urns which contain at least n black balls. " No, since you are assuming replacement.
    – Lord Shark the Unknown
    Jul 28 at 10:24










  • Thanks for pointing the mistake. Probably, the correct answer should be $fracfrac1N+1 sum_i=1^N left(fraciNright)^n+1frac1N+1 sum_i=1^N left(fraciNright)^n=frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n$
    – Bhargob
    Jul 28 at 12:41











  • If I am not wrong, the limit of this probability as $N to infty$ can be evaluated as follows:- beginequation lim_N to infty frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n=lim_N to infty N lim_N to infty frac1^n+1+2^n+1+.....+N^n+11^n+2^n+......+N^n=lim_N to inftyN fracint_0^N t^n+1 dtint_0^N t^n+1 dt=fracn+1n+2 endequation I am using Euler-Maclaurin's Summation formula.
    – Bhargob
    Jul 28 at 13:08








1




1




Since the balls are being replaced after each draw, all you need is $1$ black ball in the urn.
– lulu
Jul 28 at 10:08




Since the balls are being replaced after each draw, all you need is $1$ black ball in the urn.
– lulu
Jul 28 at 10:08












"Since all the n balls drawn are found black, we can consider only those urns which contain at least n black balls. " No, since you are assuming replacement.
– Lord Shark the Unknown
Jul 28 at 10:24




"Since all the n balls drawn are found black, we can consider only those urns which contain at least n black balls. " No, since you are assuming replacement.
– Lord Shark the Unknown
Jul 28 at 10:24












Thanks for pointing the mistake. Probably, the correct answer should be $fracfrac1N+1 sum_i=1^N left(fraciNright)^n+1frac1N+1 sum_i=1^N left(fraciNright)^n=frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n$
– Bhargob
Jul 28 at 12:41





Thanks for pointing the mistake. Probably, the correct answer should be $fracfrac1N+1 sum_i=1^N left(fraciNright)^n+1frac1N+1 sum_i=1^N left(fraciNright)^n=frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n$
– Bhargob
Jul 28 at 12:41













If I am not wrong, the limit of this probability as $N to infty$ can be evaluated as follows:- beginequation lim_N to infty frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n=lim_N to infty N lim_N to infty frac1^n+1+2^n+1+.....+N^n+11^n+2^n+......+N^n=lim_N to inftyN fracint_0^N t^n+1 dtint_0^N t^n+1 dt=fracn+1n+2 endequation I am using Euler-Maclaurin's Summation formula.
– Bhargob
Jul 28 at 13:08





If I am not wrong, the limit of this probability as $N to infty$ can be evaluated as follows:- beginequation lim_N to infty frac sum_i=1^N left(fraciNright)^n+1sum_i=1^N left(fraciNright)^n=lim_N to infty N lim_N to infty frac1^n+1+2^n+1+.....+N^n+11^n+2^n+......+N^n=lim_N to inftyN fracint_0^N t^n+1 dtint_0^N t^n+1 dt=fracn+1n+2 endequation I am using Euler-Maclaurin's Summation formula.
– Bhargob
Jul 28 at 13:08
















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