Selecting n<m elements from a set of m elements containing repeated elements, What's the SIZE of the...
$begingroup$
For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?
I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.
combinatorics
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
asked Dec 3 '18 at 2:17
KrioKrio
306
$endgroup$
add a comment |
$begingroup$
For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?
I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.
combinatorics
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
asked Dec 3 '18 at 2:17
KrioKrio
306
$endgroup$
add a comment |
$begingroup$
For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?
I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.
combinatorics
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
asked Dec 3 '18 at 2:17
KrioKrio
306
$endgroup$
For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?
I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.
combinatorics
combinatorics
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
asked Dec 3 '18 at 2:17
KrioKrio
306
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
asked Dec 3 '18 at 2:17
KrioKrio
306
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
edited Dec 3 '18 at 9:27
Tianlalu
3,08621038
3,08621038
asked Dec 3 '18 at 2:17
KrioKrio
306
asked Dec 3 '18 at 2:17
KrioKrio
306
asked Dec 3 '18 at 2:17
KrioKrio
306
306
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".
The first is just $C(21,5)$, but as you noted not all of them are distinguishable.
To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.
This is $C(5+2,2)$ with the following stars and bars argument:
- Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"
- Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow
- Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5
- The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
$endgroup$
$begingroup$
This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
$endgroup$
– Krio
Dec 3 '18 at 2:39
1
$begingroup$
If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
$endgroup$
– obscurans
Dec 3 '18 at 2:43
$begingroup$
thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
$endgroup$
– Krio
Dec 3 '18 at 2:52
1
$begingroup$
When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
$endgroup$
– obscurans
Dec 3 '18 at 2:56
add a comment |
$begingroup$
It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
$endgroup$
$begingroup$
for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
$endgroup$
– Krio
Dec 3 '18 at 2:35
$begingroup$
Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
$endgroup$
– Boshu
Dec 3 '18 at 2:37
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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$begingroup$
This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".
The first is just $C(21,5)$, but as you noted not all of them are distinguishable.
To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.
This is $C(5+2,2)$ with the following stars and bars argument:
- Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"
- Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow
- Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5
- The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
$endgroup$
$begingroup$
This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
$endgroup$
– Krio
Dec 3 '18 at 2:39
1
$begingroup$
If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
$endgroup$
– obscurans
Dec 3 '18 at 2:43
$begingroup$
thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
$endgroup$
– Krio
Dec 3 '18 at 2:52
1
$begingroup$
When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
$endgroup$
– obscurans
Dec 3 '18 at 2:56
add a comment |
$begingroup$
This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".
The first is just $C(21,5)$, but as you noted not all of them are distinguishable.
To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.
This is $C(5+2,2)$ with the following stars and bars argument:
- Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"
- Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow
- Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5
- The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
$endgroup$
$begingroup$
This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
$endgroup$
– Krio
Dec 3 '18 at 2:39
1
$begingroup$
If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
$endgroup$
– obscurans
Dec 3 '18 at 2:43
$begingroup$
thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
$endgroup$
– Krio
Dec 3 '18 at 2:52
1
$begingroup$
When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
$endgroup$
– obscurans
Dec 3 '18 at 2:56
add a comment |
$begingroup$
This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".
The first is just $C(21,5)$, but as you noted not all of them are distinguishable.
To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.
This is $C(5+2,2)$ with the following stars and bars argument:
- Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"
- Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow
- Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5
- The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
$endgroup$
This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".
The first is just $C(21,5)$, but as you noted not all of them are distinguishable.
To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.
This is $C(5+2,2)$ with the following stars and bars argument:
- Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"
- Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow
- Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5
- The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
answered Dec 3 '18 at 2:33
obscuransobscurans
1,152311
1,152311
$begingroup$
This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
$endgroup$
– Krio
Dec 3 '18 at 2:39
1
$begingroup$
If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
$endgroup$
– obscurans
Dec 3 '18 at 2:43
$begingroup$
thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
$endgroup$
– Krio
Dec 3 '18 at 2:52
1
$begingroup$
When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
$endgroup$
– obscurans
Dec 3 '18 at 2:56
add a comment |
$begingroup$
This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
$endgroup$
– Krio
Dec 3 '18 at 2:39
1
$begingroup$
If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
$endgroup$
– obscurans
Dec 3 '18 at 2:43
$begingroup$
thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
$endgroup$
– Krio
Dec 3 '18 at 2:52
1
$begingroup$
When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
$endgroup$
– obscurans
Dec 3 '18 at 2:56
$begingroup$
This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
$endgroup$
– Krio
Dec 3 '18 at 2:39
$begingroup$
This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
$endgroup$
– Krio
Dec 3 '18 at 2:39
1
1
$begingroup$
If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
$endgroup$
– obscurans
Dec 3 '18 at 2:43
$begingroup$
If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
$endgroup$
– obscurans
Dec 3 '18 at 2:43
$begingroup$
thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
$endgroup$
– Krio
Dec 3 '18 at 2:52
$begingroup$
thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
$endgroup$
– Krio
Dec 3 '18 at 2:52
1
1
$begingroup$
When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
$endgroup$
– obscurans
Dec 3 '18 at 2:56
$begingroup$
When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
$endgroup$
– obscurans
Dec 3 '18 at 2:56
add a comment |
$begingroup$
It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
$endgroup$
$begingroup$
for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
$endgroup$
– Krio
Dec 3 '18 at 2:35
$begingroup$
Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
$endgroup$
– Boshu
Dec 3 '18 at 2:37
add a comment |
$begingroup$
It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
$endgroup$
$begingroup$
for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
$endgroup$
– Krio
Dec 3 '18 at 2:35
$begingroup$
Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
$endgroup$
– Boshu
Dec 3 '18 at 2:37
add a comment |
$begingroup$
It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
$endgroup$
It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
answered Dec 3 '18 at 2:29
BoshuBoshu
703315
703315
$begingroup$
for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
$endgroup$
– Krio
Dec 3 '18 at 2:35
$begingroup$
Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
$endgroup$
– Boshu
Dec 3 '18 at 2:37
add a comment |
$begingroup$
for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
$endgroup$
– Krio
Dec 3 '18 at 2:35
$begingroup$
Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
$endgroup$
– Boshu
Dec 3 '18 at 2:37
$begingroup$
for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
$endgroup$
– Krio
Dec 3 '18 at 2:35
$begingroup$
for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
$endgroup$
– Krio
Dec 3 '18 at 2:35
$begingroup$
Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
$endgroup$
– Boshu
Dec 3 '18 at 2:37
$begingroup$
Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
$endgroup$
– Boshu
Dec 3 '18 at 2:37
add a comment |
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