Probability there is at least one item left from each type of item after random choose [closed]
$begingroup$
I'm thinking about the following:
I have 3 types of things, 3 identical pieces of each. Each of 5 people randomly choose 1 item. What is the probability that there still is at least 1 item left from each type?
I was thinking about combination with repetition, which however doesnt consider who gets what.
Thank you in advance.
probability
$endgroup$
closed as off-topic by amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs Dec 6 '18 at 20:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm thinking about the following:
I have 3 types of things, 3 identical pieces of each. Each of 5 people randomly choose 1 item. What is the probability that there still is at least 1 item left from each type?
I was thinking about combination with repetition, which however doesnt consider who gets what.
Thank you in advance.
probability
$endgroup$
closed as off-topic by amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs Dec 6 '18 at 20:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm thinking about the following:
I have 3 types of things, 3 identical pieces of each. Each of 5 people randomly choose 1 item. What is the probability that there still is at least 1 item left from each type?
I was thinking about combination with repetition, which however doesnt consider who gets what.
Thank you in advance.
probability
$endgroup$
I'm thinking about the following:
I have 3 types of things, 3 identical pieces of each. Each of 5 people randomly choose 1 item. What is the probability that there still is at least 1 item left from each type?
I was thinking about combination with repetition, which however doesnt consider who gets what.
Thank you in advance.
probability
probability
asked Dec 6 '18 at 10:23
dwarf_dzdwarf_dz
1
1
closed as off-topic by amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs Dec 6 '18 at 20:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs Dec 6 '18 at 20:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, Brahadeesh, user10354138, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's call your items $a_{11}$ through $a_{33}$, with $a_{11},a_{12},a_{13}$ being of type $t_1$, same for $t_2$ and $t_3$.
There are ${9choose 4}$ equiprobable final configurations after 5 items have been selected. Let's compute the number of configurations where at least one item of each kind is left.
There are 3 cases we want to look at:
$C_1$ : 2 of $t_1$, 1 of $t_2$, 1 of $t_3$
$C_2$ : 1 of $t_1$, 2 of $t_2$, 1 of $t_3$
$C_3$ : 1 of $t_1$, 1 of $t_2$, 2 of $t_3$
They're mutually exclusive and symmetric so we can just compute the first case and the end result is 3 times that.
To get a case $C_1$, we remove one of three from $t_1$, we keep one of three from $t_2$, we keep one of three from $t_3$, so the number of cases $C_1$ is $N(C_1) = 3 * 3 * 3 = 3^3$.
The number of admissible configurations is then $3N(C_1) = 3^4$, which means the probability of getting an admissible configuration is :
$$frac{3^4}{{9choose 4}} = frac{9}{14}$$
$endgroup$
add a comment |
$begingroup$
Number the types with $i=1,2,3$.
Let $E$ denote the event that still $1$ item of each kind is left.
Let $A_i$ denote the event that no items are left from the $i$-th type.
Then: $$E^{complement}=A_1cup A_2cup A_3$$
Moreover the events $A_1,A_2,A_3$ are mutually exclusive and equiprobable so that $$P(E)=1-P(E^{complement})=1-P(A_1)-P(A_2)-P(A_3)=1-3P(A_1)$$
Here: $$P(A_1)=frac{binom33binom62}{binom95}=frac5{42}$$ corresponding with a selection of $5$ items out of $9$ where exactly $3$ items of type $1$ are selected.
So we end up with: $$P(E)=1-3cdotfrac5{42}=frac{27}{42}=frac9{14}$$
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's call your items $a_{11}$ through $a_{33}$, with $a_{11},a_{12},a_{13}$ being of type $t_1$, same for $t_2$ and $t_3$.
There are ${9choose 4}$ equiprobable final configurations after 5 items have been selected. Let's compute the number of configurations where at least one item of each kind is left.
There are 3 cases we want to look at:
$C_1$ : 2 of $t_1$, 1 of $t_2$, 1 of $t_3$
$C_2$ : 1 of $t_1$, 2 of $t_2$, 1 of $t_3$
$C_3$ : 1 of $t_1$, 1 of $t_2$, 2 of $t_3$
They're mutually exclusive and symmetric so we can just compute the first case and the end result is 3 times that.
To get a case $C_1$, we remove one of three from $t_1$, we keep one of three from $t_2$, we keep one of three from $t_3$, so the number of cases $C_1$ is $N(C_1) = 3 * 3 * 3 = 3^3$.
The number of admissible configurations is then $3N(C_1) = 3^4$, which means the probability of getting an admissible configuration is :
$$frac{3^4}{{9choose 4}} = frac{9}{14}$$
$endgroup$
add a comment |
$begingroup$
Let's call your items $a_{11}$ through $a_{33}$, with $a_{11},a_{12},a_{13}$ being of type $t_1$, same for $t_2$ and $t_3$.
There are ${9choose 4}$ equiprobable final configurations after 5 items have been selected. Let's compute the number of configurations where at least one item of each kind is left.
There are 3 cases we want to look at:
$C_1$ : 2 of $t_1$, 1 of $t_2$, 1 of $t_3$
$C_2$ : 1 of $t_1$, 2 of $t_2$, 1 of $t_3$
$C_3$ : 1 of $t_1$, 1 of $t_2$, 2 of $t_3$
They're mutually exclusive and symmetric so we can just compute the first case and the end result is 3 times that.
To get a case $C_1$, we remove one of three from $t_1$, we keep one of three from $t_2$, we keep one of three from $t_3$, so the number of cases $C_1$ is $N(C_1) = 3 * 3 * 3 = 3^3$.
The number of admissible configurations is then $3N(C_1) = 3^4$, which means the probability of getting an admissible configuration is :
$$frac{3^4}{{9choose 4}} = frac{9}{14}$$
$endgroup$
add a comment |
$begingroup$
Let's call your items $a_{11}$ through $a_{33}$, with $a_{11},a_{12},a_{13}$ being of type $t_1$, same for $t_2$ and $t_3$.
There are ${9choose 4}$ equiprobable final configurations after 5 items have been selected. Let's compute the number of configurations where at least one item of each kind is left.
There are 3 cases we want to look at:
$C_1$ : 2 of $t_1$, 1 of $t_2$, 1 of $t_3$
$C_2$ : 1 of $t_1$, 2 of $t_2$, 1 of $t_3$
$C_3$ : 1 of $t_1$, 1 of $t_2$, 2 of $t_3$
They're mutually exclusive and symmetric so we can just compute the first case and the end result is 3 times that.
To get a case $C_1$, we remove one of three from $t_1$, we keep one of three from $t_2$, we keep one of three from $t_3$, so the number of cases $C_1$ is $N(C_1) = 3 * 3 * 3 = 3^3$.
The number of admissible configurations is then $3N(C_1) = 3^4$, which means the probability of getting an admissible configuration is :
$$frac{3^4}{{9choose 4}} = frac{9}{14}$$
$endgroup$
Let's call your items $a_{11}$ through $a_{33}$, with $a_{11},a_{12},a_{13}$ being of type $t_1$, same for $t_2$ and $t_3$.
There are ${9choose 4}$ equiprobable final configurations after 5 items have been selected. Let's compute the number of configurations where at least one item of each kind is left.
There are 3 cases we want to look at:
$C_1$ : 2 of $t_1$, 1 of $t_2$, 1 of $t_3$
$C_2$ : 1 of $t_1$, 2 of $t_2$, 1 of $t_3$
$C_3$ : 1 of $t_1$, 1 of $t_2$, 2 of $t_3$
They're mutually exclusive and symmetric so we can just compute the first case and the end result is 3 times that.
To get a case $C_1$, we remove one of three from $t_1$, we keep one of three from $t_2$, we keep one of three from $t_3$, so the number of cases $C_1$ is $N(C_1) = 3 * 3 * 3 = 3^3$.
The number of admissible configurations is then $3N(C_1) = 3^4$, which means the probability of getting an admissible configuration is :
$$frac{3^4}{{9choose 4}} = frac{9}{14}$$
edited Dec 6 '18 at 11:03
answered Dec 6 '18 at 10:49
RcnScRcnSc
1015
1015
add a comment |
add a comment |
$begingroup$
Number the types with $i=1,2,3$.
Let $E$ denote the event that still $1$ item of each kind is left.
Let $A_i$ denote the event that no items are left from the $i$-th type.
Then: $$E^{complement}=A_1cup A_2cup A_3$$
Moreover the events $A_1,A_2,A_3$ are mutually exclusive and equiprobable so that $$P(E)=1-P(E^{complement})=1-P(A_1)-P(A_2)-P(A_3)=1-3P(A_1)$$
Here: $$P(A_1)=frac{binom33binom62}{binom95}=frac5{42}$$ corresponding with a selection of $5$ items out of $9$ where exactly $3$ items of type $1$ are selected.
So we end up with: $$P(E)=1-3cdotfrac5{42}=frac{27}{42}=frac9{14}$$
$endgroup$
add a comment |
$begingroup$
Number the types with $i=1,2,3$.
Let $E$ denote the event that still $1$ item of each kind is left.
Let $A_i$ denote the event that no items are left from the $i$-th type.
Then: $$E^{complement}=A_1cup A_2cup A_3$$
Moreover the events $A_1,A_2,A_3$ are mutually exclusive and equiprobable so that $$P(E)=1-P(E^{complement})=1-P(A_1)-P(A_2)-P(A_3)=1-3P(A_1)$$
Here: $$P(A_1)=frac{binom33binom62}{binom95}=frac5{42}$$ corresponding with a selection of $5$ items out of $9$ where exactly $3$ items of type $1$ are selected.
So we end up with: $$P(E)=1-3cdotfrac5{42}=frac{27}{42}=frac9{14}$$
$endgroup$
add a comment |
$begingroup$
Number the types with $i=1,2,3$.
Let $E$ denote the event that still $1$ item of each kind is left.
Let $A_i$ denote the event that no items are left from the $i$-th type.
Then: $$E^{complement}=A_1cup A_2cup A_3$$
Moreover the events $A_1,A_2,A_3$ are mutually exclusive and equiprobable so that $$P(E)=1-P(E^{complement})=1-P(A_1)-P(A_2)-P(A_3)=1-3P(A_1)$$
Here: $$P(A_1)=frac{binom33binom62}{binom95}=frac5{42}$$ corresponding with a selection of $5$ items out of $9$ where exactly $3$ items of type $1$ are selected.
So we end up with: $$P(E)=1-3cdotfrac5{42}=frac{27}{42}=frac9{14}$$
$endgroup$
Number the types with $i=1,2,3$.
Let $E$ denote the event that still $1$ item of each kind is left.
Let $A_i$ denote the event that no items are left from the $i$-th type.
Then: $$E^{complement}=A_1cup A_2cup A_3$$
Moreover the events $A_1,A_2,A_3$ are mutually exclusive and equiprobable so that $$P(E)=1-P(E^{complement})=1-P(A_1)-P(A_2)-P(A_3)=1-3P(A_1)$$
Here: $$P(A_1)=frac{binom33binom62}{binom95}=frac5{42}$$ corresponding with a selection of $5$ items out of $9$ where exactly $3$ items of type $1$ are selected.
So we end up with: $$P(E)=1-3cdotfrac5{42}=frac{27}{42}=frac9{14}$$
edited Dec 6 '18 at 11:46
answered Dec 6 '18 at 11:35
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |