Mixing
Probabilities-extractions of balls
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A box contains m white balls and n black balls. We extract them one by one, without reintroducing them back. What is the probability to obtain the first ball white at the k-th extraction?
probability-theory discrete-mathematics
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
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add a comment |
$begingroup$
A box contains m white balls and n black balls. We extract them one by one, without reintroducing them back. What is the probability to obtain the first ball white at the k-th extraction?
probability-theory discrete-mathematics
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
$endgroup$
add a comment |
$begingroup$
A box contains m white balls and n black balls. We extract them one by one, without reintroducing them back. What is the probability to obtain the first ball white at the k-th extraction?
probability-theory discrete-mathematics
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
$endgroup$
A box contains m white balls and n black balls. We extract them one by one, without reintroducing them back. What is the probability to obtain the first ball white at the k-th extraction?
probability-theory discrete-mathematics
probability-theory discrete-mathematics
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
asked Feb 1 at 21:14
Livia MitricăLivia Mitrică
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
In order for the first white ball to be sampled in the $k^{th}$ step, the first $k-1$ balls had to be black. The probability of sampling $k-1$ black balls is
$$Prlbrack mbox{first $k-1$ balls black} rbrack =prod_{i=0}^{k-2} frac{n-i}{m+n-i} $$
The probability of choosing a white ball on the $k^{th}$ step, given that the first $k-1$ balls were black is then
$$Prlbrack mbox{$k^{th}$ ball is white | first $k-1$ balls black} rbrack =frac{m}{m+n-(k-1)} $$
The overall probability is the product of these two. We condition here, since the event that we are conditioning on must occur for the $k^{th}$ ball to be the first white ball chosen.
answered Feb 1 at 21:27
mm8511mm8511
538210
$endgroup$
$begingroup$
I understand, thank you for your answer!
$endgroup$
– Livia Mitrică
Feb 1 at 21:33
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
In order for the first white ball to be sampled in the $k^{th}$ step, the first $k-1$ balls had to be black. The probability of sampling $k-1$ black balls is
$$Prlbrack mbox{first $k-1$ balls black} rbrack =prod_{i=0}^{k-2} frac{n-i}{m+n-i} $$
The probability of choosing a white ball on the $k^{th}$ step, given that the first $k-1$ balls were black is then
$$Prlbrack mbox{$k^{th}$ ball is white | first $k-1$ balls black} rbrack =frac{m}{m+n-(k-1)} $$
The overall probability is the product of these two. We condition here, since the event that we are conditioning on must occur for the $k^{th}$ ball to be the first white ball chosen.
answered Feb 1 at 21:27
mm8511mm8511
538210
$endgroup$
$begingroup$
I understand, thank you for your answer!
$endgroup$
– Livia Mitrică
Feb 1 at 21:33
add a comment |
$begingroup$
In order for the first white ball to be sampled in the $k^{th}$ step, the first $k-1$ balls had to be black. The probability of sampling $k-1$ black balls is
$$Prlbrack mbox{first $k-1$ balls black} rbrack =prod_{i=0}^{k-2} frac{n-i}{m+n-i} $$
The probability of choosing a white ball on the $k^{th}$ step, given that the first $k-1$ balls were black is then
$$Prlbrack mbox{$k^{th}$ ball is white | first $k-1$ balls black} rbrack =frac{m}{m+n-(k-1)} $$
The overall probability is the product of these two. We condition here, since the event that we are conditioning on must occur for the $k^{th}$ ball to be the first white ball chosen.
answered Feb 1 at 21:27
mm8511mm8511
538210
$endgroup$
$begingroup$
I understand, thank you for your answer!
$endgroup$
– Livia Mitrică
Feb 1 at 21:33
add a comment |
$begingroup$
In order for the first white ball to be sampled in the $k^{th}$ step, the first $k-1$ balls had to be black. The probability of sampling $k-1$ black balls is
$$Prlbrack mbox{first $k-1$ balls black} rbrack =prod_{i=0}^{k-2} frac{n-i}{m+n-i} $$
The probability of choosing a white ball on the $k^{th}$ step, given that the first $k-1$ balls were black is then
$$Prlbrack mbox{$k^{th}$ ball is white | first $k-1$ balls black} rbrack =frac{m}{m+n-(k-1)} $$
The overall probability is the product of these two. We condition here, since the event that we are conditioning on must occur for the $k^{th}$ ball to be the first white ball chosen.
answered Feb 1 at 21:27
mm8511mm8511
538210
$endgroup$
In order for the first white ball to be sampled in the $k^{th}$ step, the first $k-1$ balls had to be black. The probability of sampling $k-1$ black balls is
$$Prlbrack mbox{first $k-1$ balls black} rbrack =prod_{i=0}^{k-2} frac{n-i}{m+n-i} $$
The probability of choosing a white ball on the $k^{th}$ step, given that the first $k-1$ balls were black is then
$$Prlbrack mbox{$k^{th}$ ball is white | first $k-1$ balls black} rbrack =frac{m}{m+n-(k-1)} $$
The overall probability is the product of these two. We condition here, since the event that we are conditioning on must occur for the $k^{th}$ ball to be the first white ball chosen.
answered Feb 1 at 21:27
mm8511mm8511
538210
answered Feb 1 at 21:27
mm8511mm8511
538210
answered Feb 1 at 21:27
mm8511mm8511
538210
answered Feb 1 at 21:27
mm8511mm8511
538210
538210
$begingroup$
I understand, thank you for your answer!
$endgroup$
– Livia Mitrică
Feb 1 at 21:33
add a comment |
$begingroup$
I understand, thank you for your answer!
$endgroup$
– Livia Mitrică
Feb 1 at 21:33
$begingroup$
I understand, thank you for your answer!
$endgroup$
– Livia Mitrică
Feb 1 at 21:33
$begingroup$
I understand, thank you for your answer!
$endgroup$
– Livia Mitrică
Feb 1 at 21:33
add a comment |
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