Mixing
Drawing n balls with replacement and pick at least one ball of each colour
$begingroup$
I know this problem sounds familiar, but it has been a while since I stopped doing probabilities, and I did not find the answer easily by myself or on this site. At the origin, the problem is related to datamasking over a string that contains many alphabets, and we want to be sure that all the alphabets we are using are represented. There is in fact an easy mapping between alphabets and colour of a ball, and so here is the problem:
Given $a_1, dots, a_m$ balls of colour $1, dots m$, we want to pick $n$ balls with replacement, with $n geq m$. What is the probability that all the colours are represented in the sample?
Thanks in advance for your answers.
probability
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
$endgroup$
add a comment |
$begingroup$
I know this problem sounds familiar, but it has been a while since I stopped doing probabilities, and I did not find the answer easily by myself or on this site. At the origin, the problem is related to datamasking over a string that contains many alphabets, and we want to be sure that all the alphabets we are using are represented. There is in fact an easy mapping between alphabets and colour of a ball, and so here is the problem:
Given $a_1, dots, a_m$ balls of colour $1, dots m$, we want to pick $n$ balls with replacement, with $n geq m$. What is the probability that all the colours are represented in the sample?
Thanks in advance for your answers.
probability
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
$endgroup$
$begingroup$
Thanks @JoséCarlosSantos I edited the question, to put some context, but I'm not sure if it's sufficent or not
$endgroup$
– Damien Prot
Jan 24 at 7:58
$begingroup$
Is the order important ? For example is picking a white ball then a black ball different from picking black then white ?
$endgroup$
– P. Quinton
Jan 24 at 8:13
$begingroup$
no, the order is not important
$endgroup$
– Damien Prot
Jan 24 at 8:15
$begingroup$
are the $a_i$'s all different? that makes an important difference ;-)
$endgroup$
– user52227
Jan 24 at 8:54
add a comment |
$begingroup$
I know this problem sounds familiar, but it has been a while since I stopped doing probabilities, and I did not find the answer easily by myself or on this site. At the origin, the problem is related to datamasking over a string that contains many alphabets, and we want to be sure that all the alphabets we are using are represented. There is in fact an easy mapping between alphabets and colour of a ball, and so here is the problem:
Given $a_1, dots, a_m$ balls of colour $1, dots m$, we want to pick $n$ balls with replacement, with $n geq m$. What is the probability that all the colours are represented in the sample?
Thanks in advance for your answers.
probability
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
$endgroup$
I know this problem sounds familiar, but it has been a while since I stopped doing probabilities, and I did not find the answer easily by myself or on this site. At the origin, the problem is related to datamasking over a string that contains many alphabets, and we want to be sure that all the alphabets we are using are represented. There is in fact an easy mapping between alphabets and colour of a ball, and so here is the problem:
Given $a_1, dots, a_m$ balls of colour $1, dots m$, we want to pick $n$ balls with replacement, with $n geq m$. What is the probability that all the colours are represented in the sample?
Thanks in advance for your answers.
probability
probability
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
edited Jan 24 at 7:58
Damien Prot
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
asked Jan 24 at 7:49
Damien ProtDamien Prot
1032
1032
$begingroup$
Thanks @JoséCarlosSantos I edited the question, to put some context, but I'm not sure if it's sufficent or not
$endgroup$
– Damien Prot
Jan 24 at 7:58
$begingroup$
Is the order important ? For example is picking a white ball then a black ball different from picking black then white ?
$endgroup$
– P. Quinton
Jan 24 at 8:13
$begingroup$
no, the order is not important
$endgroup$
– Damien Prot
Jan 24 at 8:15
$begingroup$
are the $a_i$'s all different? that makes an important difference ;-)
$endgroup$
– user52227
Jan 24 at 8:54
add a comment |
$begingroup$
Thanks @JoséCarlosSantos I edited the question, to put some context, but I'm not sure if it's sufficent or not
$endgroup$
– Damien Prot
Jan 24 at 7:58
$begingroup$
Is the order important ? For example is picking a white ball then a black ball different from picking black then white ?
$endgroup$
– P. Quinton
Jan 24 at 8:13
$begingroup$
no, the order is not important
$endgroup$
– Damien Prot
Jan 24 at 8:15
$begingroup$
are the $a_i$'s all different? that makes an important difference ;-)
$endgroup$
– user52227
Jan 24 at 8:54
$begingroup$
Thanks @JoséCarlosSantos I edited the question, to put some context, but I'm not sure if it's sufficent or not
$endgroup$
– Damien Prot
Jan 24 at 7:58
$begingroup$
Thanks @JoséCarlosSantos I edited the question, to put some context, but I'm not sure if it's sufficent or not
$endgroup$
– Damien Prot
Jan 24 at 7:58
$begingroup$
Is the order important ? For example is picking a white ball then a black ball different from picking black then white ?
$endgroup$
– P. Quinton
Jan 24 at 8:13
$begingroup$
Is the order important ? For example is picking a white ball then a black ball different from picking black then white ?
$endgroup$
– P. Quinton
Jan 24 at 8:13
$begingroup$
no, the order is not important
$endgroup$
– Damien Prot
Jan 24 at 8:15
$begingroup$
no, the order is not important
$endgroup$
– Damien Prot
Jan 24 at 8:15
$begingroup$
are the $a_i$'s all different? that makes an important difference ;-)
$endgroup$
– user52227
Jan 24 at 8:54
$begingroup$
are the $a_i$'s all different? that makes an important difference ;-)
$endgroup$
– user52227
Jan 24 at 8:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $a:=a_1+cdots+a_m$ and $p_i=frac{a_i}{a}$.
Let $X_i$ denote the number of balls picked from color $i$.
To be found is $P(X_1>0,dots,X_m>0)$.
Here $(X_1,dots,X_m)$ has multinomial distribution with parameters $n$ and $(p_1,dots,p_m)$ so that in general:$$P(X_1=r_1,dots,X_m=r_m)=frac{n!}{r_1!cdots r_m!}p_1^{r_1}cdots p_m^{r_m}$$whenever the $r_i$ are nonnegative integers that satisfy $r_1+cdots+r_m=n$.
It is handsome here to go for $$P(X_1>0,dots,X_m>0)=1-Pleft(bigcup_{i=1}^m{X_i=0}right)$$and to find an expression of the RHS by means of the principle of inclusion/exclusion.
In this expression we meet terms like $P(X_2=0, X_4=0, X_5=0)=left(1-p_2-p_4-p_5right)^n$.
answered Jan 24 at 8:57
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thanks for your answer. Is there any way to avoid the ugly inclusion/exclusion principle, in order to be able to compute this probability efficiently ? Or to have an estimation of this probability very quickly ?
$endgroup$
– Damien Prot
Jan 24 at 9:26
$begingroup$
Not that I know of. Ugly?...It is beautiful ;-) and IMV much better then the ugly summation of the $P(X_1=r_1,dots,X_m=r_m)$ where the $r_i$ are positive.
$endgroup$
– drhab
Jan 24 at 9:29
$begingroup$
Why $ P(X_1>0, dots, X_m > 0 ) = 1 - P( cup X_i = 0) $ and not $ = 1- P(cap X_i = 0) $
$endgroup$
– bm1125
Jan 24 at 10:33
$begingroup$
@bm1125 The negation of "$X_1>0text{ and }cdotstext{ and }X_m>0$" is "$X_1=0text{ or }cdotstext{ or }X_m=0$". It is not "$X_1=0text{ and }cdotstext{ and }X_m=0$" as you seem to think.
$endgroup$
– drhab
Jan 24 at 10:37
$begingroup$
Ohh.. that's like $ P(A_1 cap A_2 cap dots cap A_m) = 1- P(A_1^C cup A_2^C cup dots cup A_m^C) $ ??
$endgroup$
– bm1125
Jan 24 at 10:43
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
Let $a:=a_1+cdots+a_m$ and $p_i=frac{a_i}{a}$.
Let $X_i$ denote the number of balls picked from color $i$.
To be found is $P(X_1>0,dots,X_m>0)$.
Here $(X_1,dots,X_m)$ has multinomial distribution with parameters $n$ and $(p_1,dots,p_m)$ so that in general:$$P(X_1=r_1,dots,X_m=r_m)=frac{n!}{r_1!cdots r_m!}p_1^{r_1}cdots p_m^{r_m}$$whenever the $r_i$ are nonnegative integers that satisfy $r_1+cdots+r_m=n$.
It is handsome here to go for $$P(X_1>0,dots,X_m>0)=1-Pleft(bigcup_{i=1}^m{X_i=0}right)$$and to find an expression of the RHS by means of the principle of inclusion/exclusion.
In this expression we meet terms like $P(X_2=0, X_4=0, X_5=0)=left(1-p_2-p_4-p_5right)^n$.
answered Jan 24 at 8:57
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thanks for your answer. Is there any way to avoid the ugly inclusion/exclusion principle, in order to be able to compute this probability efficiently ? Or to have an estimation of this probability very quickly ?
$endgroup$
– Damien Prot
Jan 24 at 9:26
$begingroup$
Not that I know of. Ugly?...It is beautiful ;-) and IMV much better then the ugly summation of the $P(X_1=r_1,dots,X_m=r_m)$ where the $r_i$ are positive.
$endgroup$
– drhab
Jan 24 at 9:29
$begingroup$
Why $ P(X_1>0, dots, X_m > 0 ) = 1 - P( cup X_i = 0) $ and not $ = 1- P(cap X_i = 0) $
$endgroup$
– bm1125
Jan 24 at 10:33
$begingroup$
@bm1125 The negation of "$X_1>0text{ and }cdotstext{ and }X_m>0$" is "$X_1=0text{ or }cdotstext{ or }X_m=0$". It is not "$X_1=0text{ and }cdotstext{ and }X_m=0$" as you seem to think.
$endgroup$
– drhab
Jan 24 at 10:37
$begingroup$
Ohh.. that's like $ P(A_1 cap A_2 cap dots cap A_m) = 1- P(A_1^C cup A_2^C cup dots cup A_m^C) $ ??
$endgroup$
– bm1125
Jan 24 at 10:43
|
show 2 more comments
$begingroup$
Let $a:=a_1+cdots+a_m$ and $p_i=frac{a_i}{a}$.
Let $X_i$ denote the number of balls picked from color $i$.
To be found is $P(X_1>0,dots,X_m>0)$.
Here $(X_1,dots,X_m)$ has multinomial distribution with parameters $n$ and $(p_1,dots,p_m)$ so that in general:$$P(X_1=r_1,dots,X_m=r_m)=frac{n!}{r_1!cdots r_m!}p_1^{r_1}cdots p_m^{r_m}$$whenever the $r_i$ are nonnegative integers that satisfy $r_1+cdots+r_m=n$.
It is handsome here to go for $$P(X_1>0,dots,X_m>0)=1-Pleft(bigcup_{i=1}^m{X_i=0}right)$$and to find an expression of the RHS by means of the principle of inclusion/exclusion.
In this expression we meet terms like $P(X_2=0, X_4=0, X_5=0)=left(1-p_2-p_4-p_5right)^n$.
answered Jan 24 at 8:57
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thanks for your answer. Is there any way to avoid the ugly inclusion/exclusion principle, in order to be able to compute this probability efficiently ? Or to have an estimation of this probability very quickly ?
$endgroup$
– Damien Prot
Jan 24 at 9:26
$begingroup$
Not that I know of. Ugly?...It is beautiful ;-) and IMV much better then the ugly summation of the $P(X_1=r_1,dots,X_m=r_m)$ where the $r_i$ are positive.
$endgroup$
– drhab
Jan 24 at 9:29
$begingroup$
Why $ P(X_1>0, dots, X_m > 0 ) = 1 - P( cup X_i = 0) $ and not $ = 1- P(cap X_i = 0) $
$endgroup$
– bm1125
Jan 24 at 10:33
$begingroup$
@bm1125 The negation of "$X_1>0text{ and }cdotstext{ and }X_m>0$" is "$X_1=0text{ or }cdotstext{ or }X_m=0$". It is not "$X_1=0text{ and }cdotstext{ and }X_m=0$" as you seem to think.
$endgroup$
– drhab
Jan 24 at 10:37
$begingroup$
Ohh.. that's like $ P(A_1 cap A_2 cap dots cap A_m) = 1- P(A_1^C cup A_2^C cup dots cup A_m^C) $ ??
$endgroup$
– bm1125
Jan 24 at 10:43
|
show 2 more comments
$begingroup$
Let $a:=a_1+cdots+a_m$ and $p_i=frac{a_i}{a}$.
Let $X_i$ denote the number of balls picked from color $i$.
To be found is $P(X_1>0,dots,X_m>0)$.
Here $(X_1,dots,X_m)$ has multinomial distribution with parameters $n$ and $(p_1,dots,p_m)$ so that in general:$$P(X_1=r_1,dots,X_m=r_m)=frac{n!}{r_1!cdots r_m!}p_1^{r_1}cdots p_m^{r_m}$$whenever the $r_i$ are nonnegative integers that satisfy $r_1+cdots+r_m=n$.
It is handsome here to go for $$P(X_1>0,dots,X_m>0)=1-Pleft(bigcup_{i=1}^m{X_i=0}right)$$and to find an expression of the RHS by means of the principle of inclusion/exclusion.
In this expression we meet terms like $P(X_2=0, X_4=0, X_5=0)=left(1-p_2-p_4-p_5right)^n$.
answered Jan 24 at 8:57
drhabdrhab
103k545136
$endgroup$
Let $a:=a_1+cdots+a_m$ and $p_i=frac{a_i}{a}$.
Let $X_i$ denote the number of balls picked from color $i$.
To be found is $P(X_1>0,dots,X_m>0)$.
Here $(X_1,dots,X_m)$ has multinomial distribution with parameters $n$ and $(p_1,dots,p_m)$ so that in general:$$P(X_1=r_1,dots,X_m=r_m)=frac{n!}{r_1!cdots r_m!}p_1^{r_1}cdots p_m^{r_m}$$whenever the $r_i$ are nonnegative integers that satisfy $r_1+cdots+r_m=n$.
It is handsome here to go for $$P(X_1>0,dots,X_m>0)=1-Pleft(bigcup_{i=1}^m{X_i=0}right)$$and to find an expression of the RHS by means of the principle of inclusion/exclusion.
In this expression we meet terms like $P(X_2=0, X_4=0, X_5=0)=left(1-p_2-p_4-p_5right)^n$.
answered Jan 24 at 8:57
drhabdrhab
103k545136
answered Jan 24 at 8:57
drhabdrhab
103k545136
answered Jan 24 at 8:57
drhabdrhab
103k545136
answered Jan 24 at 8:57
drhabdrhab
103k545136
103k545136
$begingroup$
Thanks for your answer. Is there any way to avoid the ugly inclusion/exclusion principle, in order to be able to compute this probability efficiently ? Or to have an estimation of this probability very quickly ?
$endgroup$
– Damien Prot
Jan 24 at 9:26
$begingroup$
Not that I know of. Ugly?...It is beautiful ;-) and IMV much better then the ugly summation of the $P(X_1=r_1,dots,X_m=r_m)$ where the $r_i$ are positive.
$endgroup$
– drhab
Jan 24 at 9:29
$begingroup$
Why $ P(X_1>0, dots, X_m > 0 ) = 1 - P( cup X_i = 0) $ and not $ = 1- P(cap X_i = 0) $
$endgroup$
– bm1125
Jan 24 at 10:33
$begingroup$
@bm1125 The negation of "$X_1>0text{ and }cdotstext{ and }X_m>0$" is "$X_1=0text{ or }cdotstext{ or }X_m=0$". It is not "$X_1=0text{ and }cdotstext{ and }X_m=0$" as you seem to think.
$endgroup$
– drhab
Jan 24 at 10:37
$begingroup$
Ohh.. that's like $ P(A_1 cap A_2 cap dots cap A_m) = 1- P(A_1^C cup A_2^C cup dots cup A_m^C) $ ??
$endgroup$
– bm1125
Jan 24 at 10:43
|
show 2 more comments
$begingroup$
Thanks for your answer. Is there any way to avoid the ugly inclusion/exclusion principle, in order to be able to compute this probability efficiently ? Or to have an estimation of this probability very quickly ?
$endgroup$
– Damien Prot
Jan 24 at 9:26
$begingroup$
Not that I know of. Ugly?...It is beautiful ;-) and IMV much better then the ugly summation of the $P(X_1=r_1,dots,X_m=r_m)$ where the $r_i$ are positive.
$endgroup$
– drhab
Jan 24 at 9:29
$begingroup$
Why $ P(X_1>0, dots, X_m > 0 ) = 1 - P( cup X_i = 0) $ and not $ = 1- P(cap X_i = 0) $
$endgroup$
– bm1125
Jan 24 at 10:33
$begingroup$
@bm1125 The negation of "$X_1>0text{ and }cdotstext{ and }X_m>0$" is "$X_1=0text{ or }cdotstext{ or }X_m=0$". It is not "$X_1=0text{ and }cdotstext{ and }X_m=0$" as you seem to think.
$endgroup$
– drhab
Jan 24 at 10:37
$begingroup$
Ohh.. that's like $ P(A_1 cap A_2 cap dots cap A_m) = 1- P(A_1^C cup A_2^C cup dots cup A_m^C) $ ??
$endgroup$
– bm1125
Jan 24 at 10:43
$begingroup$
Thanks for your answer. Is there any way to avoid the ugly inclusion/exclusion principle, in order to be able to compute this probability efficiently ? Or to have an estimation of this probability very quickly ?
$endgroup$
– Damien Prot
Jan 24 at 9:26
$begingroup$
Thanks for your answer. Is there any way to avoid the ugly inclusion/exclusion principle, in order to be able to compute this probability efficiently ? Or to have an estimation of this probability very quickly ?
$endgroup$
– Damien Prot
Jan 24 at 9:26
$begingroup$
Not that I know of. Ugly?...It is beautiful ;-) and IMV much better then the ugly summation of the $P(X_1=r_1,dots,X_m=r_m)$ where the $r_i$ are positive.
$endgroup$
– drhab
Jan 24 at 9:29
$begingroup$
Not that I know of. Ugly?...It is beautiful ;-) and IMV much better then the ugly summation of the $P(X_1=r_1,dots,X_m=r_m)$ where the $r_i$ are positive.
$endgroup$
– drhab
Jan 24 at 9:29
$begingroup$
Why $ P(X_1>0, dots, X_m > 0 ) = 1 - P( cup X_i = 0) $ and not $ = 1- P(cap X_i = 0) $
$endgroup$
– bm1125
Jan 24 at 10:33
$begingroup$
Why $ P(X_1>0, dots, X_m > 0 ) = 1 - P( cup X_i = 0) $ and not $ = 1- P(cap X_i = 0) $
$endgroup$
– bm1125
Jan 24 at 10:33
$begingroup$
@bm1125 The negation of "$X_1>0text{ and }cdotstext{ and }X_m>0$" is "$X_1=0text{ or }cdotstext{ or }X_m=0$". It is not "$X_1=0text{ and }cdotstext{ and }X_m=0$" as you seem to think.
$endgroup$
– drhab
Jan 24 at 10:37
$begingroup$
@bm1125 The negation of "$X_1>0text{ and }cdotstext{ and }X_m>0$" is "$X_1=0text{ or }cdotstext{ or }X_m=0$". It is not "$X_1=0text{ and }cdotstext{ and }X_m=0$" as you seem to think.
$endgroup$
– drhab
Jan 24 at 10:37
$begingroup$
Ohh.. that's like $ P(A_1 cap A_2 cap dots cap A_m) = 1- P(A_1^C cup A_2^C cup dots cup A_m^C) $ ??
$endgroup$
– bm1125
Jan 24 at 10:43
$begingroup$
Ohh.. that's like $ P(A_1 cap A_2 cap dots cap A_m) = 1- P(A_1^C cup A_2^C cup dots cup A_m^C) $ ??
$endgroup$
– bm1125
Jan 24 at 10:43
|
show 2 more comments
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$begingroup$
Thanks @JoséCarlosSantos I edited the question, to put some context, but I'm not sure if it's sufficent or not
$endgroup$
– Damien Prot
Jan 24 at 7:58
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Is the order important ? For example is picking a white ball then a black ball different from picking black then white ?
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– P. Quinton
Jan 24 at 8:13
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no, the order is not important
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– Damien Prot
Jan 24 at 8:15
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are the $a_i$'s all different? that makes an important difference ;-)
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– user52227
Jan 24 at 8:54