Mixing
Why is the multinomial coefficient best suited in this case?
$begingroup$
Let there be $m$ boxes and $k$ balls where the probability that a ball is laid in box $i$ ($1leq i leq m)$ is $p_{i}$
Define a probability space $(Omega, mathcal{F}, P)$:
Idea:
Define $Omega:={(omega_{1},...,omega_{m})in mathbb N_{0}^{m}:sum_{i=1}^{m}omega_{i}=k}$
Since $m in mathbb N$, we're in a discrete case and therefore $mathcal{F}=mathcal{P}(Omega)$
Now onto the distribution:
Let $omega:=(omega_{1},...,omega_{m})$
$P({w})=prod_{i=1}^{m}(cdot)p_{i}^{omega_{i}}$
where $(cdot)$ is meant to represent a particular combination. I would have thought that the binomial coefficient would have been a good idea, but I have been told it is rather the multinomial coefficient, but I do not understand why.
probability combinatorics probability-theory probability-distributions
asked Jan 14 at 16:34
SABOYSABOY
694311
$endgroup$
add a comment |
$begingroup$
Let there be $m$ boxes and $k$ balls where the probability that a ball is laid in box $i$ ($1leq i leq m)$ is $p_{i}$
Define a probability space $(Omega, mathcal{F}, P)$:
Idea:
Define $Omega:={(omega_{1},...,omega_{m})in mathbb N_{0}^{m}:sum_{i=1}^{m}omega_{i}=k}$
Since $m in mathbb N$, we're in a discrete case and therefore $mathcal{F}=mathcal{P}(Omega)$
Now onto the distribution:
Let $omega:=(omega_{1},...,omega_{m})$
$P({w})=prod_{i=1}^{m}(cdot)p_{i}^{omega_{i}}$
where $(cdot)$ is meant to represent a particular combination. I would have thought that the binomial coefficient would have been a good idea, but I have been told it is rather the multinomial coefficient, but I do not understand why.
probability combinatorics probability-theory probability-distributions
asked Jan 14 at 16:34
SABOYSABOY
694311
$endgroup$
add a comment |
$begingroup$
Let there be $m$ boxes and $k$ balls where the probability that a ball is laid in box $i$ ($1leq i leq m)$ is $p_{i}$
Define a probability space $(Omega, mathcal{F}, P)$:
Idea:
Define $Omega:={(omega_{1},...,omega_{m})in mathbb N_{0}^{m}:sum_{i=1}^{m}omega_{i}=k}$
Since $m in mathbb N$, we're in a discrete case and therefore $mathcal{F}=mathcal{P}(Omega)$
Now onto the distribution:
Let $omega:=(omega_{1},...,omega_{m})$
$P({w})=prod_{i=1}^{m}(cdot)p_{i}^{omega_{i}}$
where $(cdot)$ is meant to represent a particular combination. I would have thought that the binomial coefficient would have been a good idea, but I have been told it is rather the multinomial coefficient, but I do not understand why.
probability combinatorics probability-theory probability-distributions
asked Jan 14 at 16:34
SABOYSABOY
694311
$endgroup$
Let there be $m$ boxes and $k$ balls where the probability that a ball is laid in box $i$ ($1leq i leq m)$ is $p_{i}$
Define a probability space $(Omega, mathcal{F}, P)$:
Idea:
Define $Omega:={(omega_{1},...,omega_{m})in mathbb N_{0}^{m}:sum_{i=1}^{m}omega_{i}=k}$
Since $m in mathbb N$, we're in a discrete case and therefore $mathcal{F}=mathcal{P}(Omega)$
Now onto the distribution:
Let $omega:=(omega_{1},...,omega_{m})$
$P({w})=prod_{i=1}^{m}(cdot)p_{i}^{omega_{i}}$
where $(cdot)$ is meant to represent a particular combination. I would have thought that the binomial coefficient would have been a good idea, but I have been told it is rather the multinomial coefficient, but I do not understand why.
probability combinatorics probability-theory probability-distributions
probability combinatorics probability-theory probability-distributions
asked Jan 14 at 16:34
SABOYSABOY
694311
asked Jan 14 at 16:34
SABOYSABOY
694311
edited Jan 14 at 16:53
SABOY
asked Jan 14 at 16:34
SABOYSABOY
694311
asked Jan 14 at 16:34
SABOYSABOY
694311
asked Jan 14 at 16:34
SABOYSABOY
694311
694311
add a comment |
add a comment |
1 Answer
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$begingroup$
A binomial distribution arises from the sum of independent and identically distributed Bernoulli trials. The outcome of a Bernoulli trial is either $0$ or $1$; or more generally, a binary set of outcomes.
A multinomial distribution arises from the sum of independent and identically distributed categorical trials. The outcome of a categorical trial is some number in ${0, 1, 2, ldots, m-1}$, where $m ge 2$; or more generally, some discrete and finite set of outcomes.
I chose $m$ in such a way that it matches the $m$ in your problem. If you have only $2$ boxes, then the resulting distribution of balls is binomial, because for any given ball, there are only two choices for which box to place it. If you have more than $2$ boxes, then it is not possible to characterize the distribution of balls using only a single binomial variable. you'd need to describe whether a given ball is put into the first, second, or third box, for example.
answered Jan 14 at 16:43
heropupheropup
63.8k762102
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add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
A binomial distribution arises from the sum of independent and identically distributed Bernoulli trials. The outcome of a Bernoulli trial is either $0$ or $1$; or more generally, a binary set of outcomes.
A multinomial distribution arises from the sum of independent and identically distributed categorical trials. The outcome of a categorical trial is some number in ${0, 1, 2, ldots, m-1}$, where $m ge 2$; or more generally, some discrete and finite set of outcomes.
I chose $m$ in such a way that it matches the $m$ in your problem. If you have only $2$ boxes, then the resulting distribution of balls is binomial, because for any given ball, there are only two choices for which box to place it. If you have more than $2$ boxes, then it is not possible to characterize the distribution of balls using only a single binomial variable. you'd need to describe whether a given ball is put into the first, second, or third box, for example.
answered Jan 14 at 16:43
heropupheropup
63.8k762102
$endgroup$
add a comment |
$begingroup$
A binomial distribution arises from the sum of independent and identically distributed Bernoulli trials. The outcome of a Bernoulli trial is either $0$ or $1$; or more generally, a binary set of outcomes.
A multinomial distribution arises from the sum of independent and identically distributed categorical trials. The outcome of a categorical trial is some number in ${0, 1, 2, ldots, m-1}$, where $m ge 2$; or more generally, some discrete and finite set of outcomes.
I chose $m$ in such a way that it matches the $m$ in your problem. If you have only $2$ boxes, then the resulting distribution of balls is binomial, because for any given ball, there are only two choices for which box to place it. If you have more than $2$ boxes, then it is not possible to characterize the distribution of balls using only a single binomial variable. you'd need to describe whether a given ball is put into the first, second, or third box, for example.
answered Jan 14 at 16:43
heropupheropup
63.8k762102
$endgroup$
add a comment |
$begingroup$
A binomial distribution arises from the sum of independent and identically distributed Bernoulli trials. The outcome of a Bernoulli trial is either $0$ or $1$; or more generally, a binary set of outcomes.
A multinomial distribution arises from the sum of independent and identically distributed categorical trials. The outcome of a categorical trial is some number in ${0, 1, 2, ldots, m-1}$, where $m ge 2$; or more generally, some discrete and finite set of outcomes.
I chose $m$ in such a way that it matches the $m$ in your problem. If you have only $2$ boxes, then the resulting distribution of balls is binomial, because for any given ball, there are only two choices for which box to place it. If you have more than $2$ boxes, then it is not possible to characterize the distribution of balls using only a single binomial variable. you'd need to describe whether a given ball is put into the first, second, or third box, for example.
answered Jan 14 at 16:43
heropupheropup
63.8k762102
$endgroup$
A binomial distribution arises from the sum of independent and identically distributed Bernoulli trials. The outcome of a Bernoulli trial is either $0$ or $1$; or more generally, a binary set of outcomes.
A multinomial distribution arises from the sum of independent and identically distributed categorical trials. The outcome of a categorical trial is some number in ${0, 1, 2, ldots, m-1}$, where $m ge 2$; or more generally, some discrete and finite set of outcomes.
I chose $m$ in such a way that it matches the $m$ in your problem. If you have only $2$ boxes, then the resulting distribution of balls is binomial, because for any given ball, there are only two choices for which box to place it. If you have more than $2$ boxes, then it is not possible to characterize the distribution of balls using only a single binomial variable. you'd need to describe whether a given ball is put into the first, second, or third box, for example.
answered Jan 14 at 16:43
heropupheropup
63.8k762102
answered Jan 14 at 16:43
heropupheropup
63.8k762102
answered Jan 14 at 16:43
heropupheropup
63.8k762102
answered Jan 14 at 16:43
heropupheropup
63.8k762102
63.8k762102
add a comment |
add a comment |
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