Mixing
An urn contains $4$ red balls, $6$ green balls and $8$ blue balls. Probability with versus without putting...
$begingroup$
If we had to determine the probability of getting $3$ red balls $2$ green balls and $4$ blue balls after picking a ball, noting its color and putting it back it would be ${9choose 3,2,4}cdot({4over18})^3cdot({6over18})^2cdot({8over18})^4$
But what if we didn't put each the ball back after picking it out?
probability balls-in-bins multinomial-coefficients
asked Jan 29 at 17:17
H. WalterH. Walter
1047
$endgroup$
add a comment |
$begingroup$
If we had to determine the probability of getting $3$ red balls $2$ green balls and $4$ blue balls after picking a ball, noting its color and putting it back it would be ${9choose 3,2,4}cdot({4over18})^3cdot({6over18})^2cdot({8over18})^4$
But what if we didn't put each the ball back after picking it out?
probability balls-in-bins multinomial-coefficients
asked Jan 29 at 17:17
H. WalterH. Walter
1047
$endgroup$
1
$begingroup$
You can use a modification of the hypergeometric distribution. In your case the probability is $$frac{binom{4}{3}cdot binom{6}{2}cdot binom{8}{4}}{binom{18}{9}}$$
$endgroup$
– callculus
Jan 29 at 17:23
add a comment |
$begingroup$
If we had to determine the probability of getting $3$ red balls $2$ green balls and $4$ blue balls after picking a ball, noting its color and putting it back it would be ${9choose 3,2,4}cdot({4over18})^3cdot({6over18})^2cdot({8over18})^4$
But what if we didn't put each the ball back after picking it out?
probability balls-in-bins multinomial-coefficients
asked Jan 29 at 17:17
H. WalterH. Walter
1047
$endgroup$
If we had to determine the probability of getting $3$ red balls $2$ green balls and $4$ blue balls after picking a ball, noting its color and putting it back it would be ${9choose 3,2,4}cdot({4over18})^3cdot({6over18})^2cdot({8over18})^4$
But what if we didn't put each the ball back after picking it out?
probability balls-in-bins multinomial-coefficients
probability balls-in-bins multinomial-coefficients
asked Jan 29 at 17:17
H. WalterH. Walter
1047
asked Jan 29 at 17:17
H. WalterH. Walter
1047
asked Jan 29 at 17:17
H. WalterH. Walter
1047
asked Jan 29 at 17:17
H. WalterH. Walter
1047
asked Jan 29 at 17:17
H. WalterH. Walter
1047
1047
1
$begingroup$
You can use a modification of the hypergeometric distribution. In your case the probability is $$frac{binom{4}{3}cdot binom{6}{2}cdot binom{8}{4}}{binom{18}{9}}$$
$endgroup$
– callculus
Jan 29 at 17:23
add a comment |
1
$begingroup$
You can use a modification of the hypergeometric distribution. In your case the probability is $$frac{binom{4}{3}cdot binom{6}{2}cdot binom{8}{4}}{binom{18}{9}}$$
$endgroup$
– callculus
Jan 29 at 17:23
1
1
$begingroup$
You can use a modification of the hypergeometric distribution. In your case the probability is $$frac{binom{4}{3}cdot binom{6}{2}cdot binom{8}{4}}{binom{18}{9}}$$
$endgroup$
– callculus
Jan 29 at 17:23
$begingroup$
You can use a modification of the hypergeometric distribution. In your case the probability is $$frac{binom{4}{3}cdot binom{6}{2}cdot binom{8}{4}}{binom{18}{9}}$$
$endgroup$
– callculus
Jan 29 at 17:23
add a comment |
1 Answer
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$begingroup$
This is an example of the multivariate hypergeometric distribution.
With $N$ balls total, $r$ colors of balls, $c_i$ of each type of color of ball available, $n_i$ of each type of color of ball desired, and $n_1+n_2+dots+n_r = n$ balls desired in total, the probability of selecting our desired count of balls when pulling without replacement will be:
$$dfrac{binom{c_1}{n_1}binom{c_2}{n_2}cdotsbinom{c_r}{n_r}}{binom{N}{n}}$$
In your specific case
$$dfrac{binom{4}{3}binom{6}{2}binom{8}{4}}{binom{18}{9}}$$
It can be seen by imagining each ball is distinctly labeled (even if they aren't originally) since in doing so each selection of $n$ balls are then equally likely to occur. The desired ways of taking the amounts of each colored ball can be counted via multiplication principle by first counting the number of ways of selecting three out of the four available red balls, multiplying this by the number of ways of selecting two out of the six available green balls, etc... Performing the division completes the calculation and gives the probability.
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
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$begingroup$
This is an example of the multivariate hypergeometric distribution.
With $N$ balls total, $r$ colors of balls, $c_i$ of each type of color of ball available, $n_i$ of each type of color of ball desired, and $n_1+n_2+dots+n_r = n$ balls desired in total, the probability of selecting our desired count of balls when pulling without replacement will be:
$$dfrac{binom{c_1}{n_1}binom{c_2}{n_2}cdotsbinom{c_r}{n_r}}{binom{N}{n}}$$
In your specific case
$$dfrac{binom{4}{3}binom{6}{2}binom{8}{4}}{binom{18}{9}}$$
It can be seen by imagining each ball is distinctly labeled (even if they aren't originally) since in doing so each selection of $n$ balls are then equally likely to occur. The desired ways of taking the amounts of each colored ball can be counted via multiplication principle by first counting the number of ways of selecting three out of the four available red balls, multiplying this by the number of ways of selecting two out of the six available green balls, etc... Performing the division completes the calculation and gives the probability.
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
This is an example of the multivariate hypergeometric distribution.
With $N$ balls total, $r$ colors of balls, $c_i$ of each type of color of ball available, $n_i$ of each type of color of ball desired, and $n_1+n_2+dots+n_r = n$ balls desired in total, the probability of selecting our desired count of balls when pulling without replacement will be:
$$dfrac{binom{c_1}{n_1}binom{c_2}{n_2}cdotsbinom{c_r}{n_r}}{binom{N}{n}}$$
In your specific case
$$dfrac{binom{4}{3}binom{6}{2}binom{8}{4}}{binom{18}{9}}$$
It can be seen by imagining each ball is distinctly labeled (even if they aren't originally) since in doing so each selection of $n$ balls are then equally likely to occur. The desired ways of taking the amounts of each colored ball can be counted via multiplication principle by first counting the number of ways of selecting three out of the four available red balls, multiplying this by the number of ways of selecting two out of the six available green balls, etc... Performing the division completes the calculation and gives the probability.
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
This is an example of the multivariate hypergeometric distribution.
With $N$ balls total, $r$ colors of balls, $c_i$ of each type of color of ball available, $n_i$ of each type of color of ball desired, and $n_1+n_2+dots+n_r = n$ balls desired in total, the probability of selecting our desired count of balls when pulling without replacement will be:
$$dfrac{binom{c_1}{n_1}binom{c_2}{n_2}cdotsbinom{c_r}{n_r}}{binom{N}{n}}$$
In your specific case
$$dfrac{binom{4}{3}binom{6}{2}binom{8}{4}}{binom{18}{9}}$$
It can be seen by imagining each ball is distinctly labeled (even if they aren't originally) since in doing so each selection of $n$ balls are then equally likely to occur. The desired ways of taking the amounts of each colored ball can be counted via multiplication principle by first counting the number of ways of selecting three out of the four available red balls, multiplying this by the number of ways of selecting two out of the six available green balls, etc... Performing the division completes the calculation and gives the probability.
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
$endgroup$
This is an example of the multivariate hypergeometric distribution.
With $N$ balls total, $r$ colors of balls, $c_i$ of each type of color of ball available, $n_i$ of each type of color of ball desired, and $n_1+n_2+dots+n_r = n$ balls desired in total, the probability of selecting our desired count of balls when pulling without replacement will be:
$$dfrac{binom{c_1}{n_1}binom{c_2}{n_2}cdotsbinom{c_r}{n_r}}{binom{N}{n}}$$
In your specific case
$$dfrac{binom{4}{3}binom{6}{2}binom{8}{4}}{binom{18}{9}}$$
It can be seen by imagining each ball is distinctly labeled (even if they aren't originally) since in doing so each selection of $n$ balls are then equally likely to occur. The desired ways of taking the amounts of each colored ball can be counted via multiplication principle by first counting the number of ways of selecting three out of the four available red balls, multiplying this by the number of ways of selecting two out of the six available green balls, etc... Performing the division completes the calculation and gives the probability.
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
answered Jan 29 at 17:54
JMoravitzJMoravitz
48.7k43988
48.7k43988
add a comment |
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$begingroup$
You can use a modification of the hypergeometric distribution. In your case the probability is $$frac{binom{4}{3}cdot binom{6}{2}cdot binom{8}{4}}{binom{18}{9}}$$
$endgroup$
– callculus
Jan 29 at 17:23