Question about probability theory.

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A forest has a population of $b$ animals. A uniformly random sample of $a$ of them are picked, marked and then released back into the forest. After that, we keep on capturing the animals in a uniformly random order until we have got $m$ marked animals. What is the probability that we need to capture exactly $n$ animals in this experiment?



My attempt at the solution -



Each outcome of the experiment is a sequence of animals.



Let A be the event that exactly $m$ marked animals exist among the first $n$ animals of the sequence and $n^th$ animal of the sequence is a marked animal.
Each sequence is equally likely.



Total number of outcomes = $b!$



probability,
$P(A) = cfracn - 1 choose m - 1b - n choose a - ma!(b - a)!b!$



$Rightarrow P(A) = cfracn - 1 choose m - 1b - n choose a - mb choose a$



Is this answer correct and can it be simplified further?







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  • 1




    Are there exactly $a$ animals marked? If yes (as I believe it is), it is irrelevant and possibly confusing that they are picked as a uniformly random sample. It's just any $a$ animals.
    – Arnaud Mortier
    Aug 19 at 8:03











  • Each outcome is a sequence of animals. Not a sequence of size $n$. The question is "how likely is it that the sequence is of size $n$?"
    – Arnaud Mortier
    Aug 19 at 8:04











  • After that, we keep on capturing the animals : are they replaced or not?
    – Arnaud Mortier
    Aug 19 at 8:20










  • Animals are not replaced.
    – Shashank Kumar
    Aug 20 at 8:07










  • @ArnaudMortier That's correct. Thanks for pointing it out.
    – Shashank Kumar
    Aug 20 at 8:08














up vote
-1
down vote

favorite












A forest has a population of $b$ animals. A uniformly random sample of $a$ of them are picked, marked and then released back into the forest. After that, we keep on capturing the animals in a uniformly random order until we have got $m$ marked animals. What is the probability that we need to capture exactly $n$ animals in this experiment?



My attempt at the solution -



Each outcome of the experiment is a sequence of animals.



Let A be the event that exactly $m$ marked animals exist among the first $n$ animals of the sequence and $n^th$ animal of the sequence is a marked animal.
Each sequence is equally likely.



Total number of outcomes = $b!$



probability,
$P(A) = cfracn - 1 choose m - 1b - n choose a - ma!(b - a)!b!$



$Rightarrow P(A) = cfracn - 1 choose m - 1b - n choose a - mb choose a$



Is this answer correct and can it be simplified further?







share|cite|improve this question


















  • 1




    Are there exactly $a$ animals marked? If yes (as I believe it is), it is irrelevant and possibly confusing that they are picked as a uniformly random sample. It's just any $a$ animals.
    – Arnaud Mortier
    Aug 19 at 8:03











  • Each outcome is a sequence of animals. Not a sequence of size $n$. The question is "how likely is it that the sequence is of size $n$?"
    – Arnaud Mortier
    Aug 19 at 8:04











  • After that, we keep on capturing the animals : are they replaced or not?
    – Arnaud Mortier
    Aug 19 at 8:20










  • Animals are not replaced.
    – Shashank Kumar
    Aug 20 at 8:07










  • @ArnaudMortier That's correct. Thanks for pointing it out.
    – Shashank Kumar
    Aug 20 at 8:08












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











A forest has a population of $b$ animals. A uniformly random sample of $a$ of them are picked, marked and then released back into the forest. After that, we keep on capturing the animals in a uniformly random order until we have got $m$ marked animals. What is the probability that we need to capture exactly $n$ animals in this experiment?



My attempt at the solution -



Each outcome of the experiment is a sequence of animals.



Let A be the event that exactly $m$ marked animals exist among the first $n$ animals of the sequence and $n^th$ animal of the sequence is a marked animal.
Each sequence is equally likely.



Total number of outcomes = $b!$



probability,
$P(A) = cfracn - 1 choose m - 1b - n choose a - ma!(b - a)!b!$



$Rightarrow P(A) = cfracn - 1 choose m - 1b - n choose a - mb choose a$



Is this answer correct and can it be simplified further?







share|cite|improve this question














A forest has a population of $b$ animals. A uniformly random sample of $a$ of them are picked, marked and then released back into the forest. After that, we keep on capturing the animals in a uniformly random order until we have got $m$ marked animals. What is the probability that we need to capture exactly $n$ animals in this experiment?



My attempt at the solution -



Each outcome of the experiment is a sequence of animals.



Let A be the event that exactly $m$ marked animals exist among the first $n$ animals of the sequence and $n^th$ animal of the sequence is a marked animal.
Each sequence is equally likely.



Total number of outcomes = $b!$



probability,
$P(A) = cfracn - 1 choose m - 1b - n choose a - ma!(b - a)!b!$



$Rightarrow P(A) = cfracn - 1 choose m - 1b - n choose a - mb choose a$



Is this answer correct and can it be simplified further?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 20 at 8:08

























asked Aug 19 at 7:43









Shashank Kumar

76




76







  • 1




    Are there exactly $a$ animals marked? If yes (as I believe it is), it is irrelevant and possibly confusing that they are picked as a uniformly random sample. It's just any $a$ animals.
    – Arnaud Mortier
    Aug 19 at 8:03











  • Each outcome is a sequence of animals. Not a sequence of size $n$. The question is "how likely is it that the sequence is of size $n$?"
    – Arnaud Mortier
    Aug 19 at 8:04











  • After that, we keep on capturing the animals : are they replaced or not?
    – Arnaud Mortier
    Aug 19 at 8:20










  • Animals are not replaced.
    – Shashank Kumar
    Aug 20 at 8:07










  • @ArnaudMortier That's correct. Thanks for pointing it out.
    – Shashank Kumar
    Aug 20 at 8:08












  • 1




    Are there exactly $a$ animals marked? If yes (as I believe it is), it is irrelevant and possibly confusing that they are picked as a uniformly random sample. It's just any $a$ animals.
    – Arnaud Mortier
    Aug 19 at 8:03











  • Each outcome is a sequence of animals. Not a sequence of size $n$. The question is "how likely is it that the sequence is of size $n$?"
    – Arnaud Mortier
    Aug 19 at 8:04











  • After that, we keep on capturing the animals : are they replaced or not?
    – Arnaud Mortier
    Aug 19 at 8:20










  • Animals are not replaced.
    – Shashank Kumar
    Aug 20 at 8:07










  • @ArnaudMortier That's correct. Thanks for pointing it out.
    – Shashank Kumar
    Aug 20 at 8:08







1




1




Are there exactly $a$ animals marked? If yes (as I believe it is), it is irrelevant and possibly confusing that they are picked as a uniformly random sample. It's just any $a$ animals.
– Arnaud Mortier
Aug 19 at 8:03





Are there exactly $a$ animals marked? If yes (as I believe it is), it is irrelevant and possibly confusing that they are picked as a uniformly random sample. It's just any $a$ animals.
– Arnaud Mortier
Aug 19 at 8:03













Each outcome is a sequence of animals. Not a sequence of size $n$. The question is "how likely is it that the sequence is of size $n$?"
– Arnaud Mortier
Aug 19 at 8:04





Each outcome is a sequence of animals. Not a sequence of size $n$. The question is "how likely is it that the sequence is of size $n$?"
– Arnaud Mortier
Aug 19 at 8:04













After that, we keep on capturing the animals : are they replaced or not?
– Arnaud Mortier
Aug 19 at 8:20




After that, we keep on capturing the animals : are they replaced or not?
– Arnaud Mortier
Aug 19 at 8:20












Animals are not replaced.
– Shashank Kumar
Aug 20 at 8:07




Animals are not replaced.
– Shashank Kumar
Aug 20 at 8:07












@ArnaudMortier That's correct. Thanks for pointing it out.
– Shashank Kumar
Aug 20 at 8:08




@ArnaudMortier That's correct. Thanks for pointing it out.
– Shashank Kumar
Aug 20 at 8:08















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