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?
probability combinatorics
 |Â
show 1 more comment
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?
probability combinatorics
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.
â 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 $n$?"
â Arnaud Mortier
Aug 19 at 8:04
After that, we keep on capturing the animals : are they replaced or not?
â Arnaud Mortier
Aug 19 at 8:20
Animals are not replaced.
â Shashank Kumar
Aug 20 at 8:07
@ArnaudMortier That's correct. Thanks for pointing it out.
â Shashank Kumar
Aug 20 at 8:08
 |Â
show 1 more comment
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?
probability combinatorics
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?
probability combinatorics
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.
â 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 $n$?"
â Arnaud Mortier
Aug 19 at 8:04
After that, we keep on capturing the animals : are they replaced or not?
â Arnaud Mortier
Aug 19 at 8:20
Animals are not replaced.
â Shashank Kumar
Aug 20 at 8:07
@ArnaudMortier That's correct. Thanks for pointing it out.
â Shashank Kumar
Aug 20 at 8:08
 |Â
show 1 more comment
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.
â 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 $n$?"
â Arnaud Mortier
Aug 19 at 8:04
After that, we keep on capturing the animals : are they replaced or not?
â Arnaud Mortier
Aug 19 at 8:20
Animals are not replaced.
â Shashank Kumar
Aug 20 at 8:07
@ArnaudMortier That's correct. Thanks for pointing it out.
â 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.
â 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.
â 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 $n$?"
â 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 $n$?"
â Arnaud Mortier
Aug 19 at 8:04
After that, we keep on capturing the animals : are they replaced or not?
â Arnaud Mortier
Aug 19 at 8:20
After that, we keep on capturing the animals : are they replaced or not?
â Arnaud Mortier
Aug 19 at 8:20
Animals are not replaced.
â Shashank Kumar
Aug 20 at 8:07
Animals are not replaced.
â Shashank Kumar
Aug 20 at 8:07
@ArnaudMortier That's correct. Thanks for pointing it out.
â Shashank Kumar
Aug 20 at 8:08
@ArnaudMortier That's correct. Thanks for pointing it out.
â Shashank Kumar
Aug 20 at 8:08
 |Â
show 1 more comment
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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.
â 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 $n$?"
â Arnaud Mortier
Aug 19 at 8:04
After that, we keep on capturing the animals : are they replaced or not?
â Arnaud Mortier
Aug 19 at 8:20
Animals are not replaced.
â Shashank Kumar
Aug 20 at 8:07
@ArnaudMortier That's correct. Thanks for pointing it out.
â Shashank Kumar
Aug 20 at 8:08