Mixing
Set Theory Problem: Survey of $200$ people asks “Do like Apples (A), Bananas (B), and Cherries (C) , …†[closed]
Clash Royale CLAN TAG#URR8PPP
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I have the following problem:
In a survey of $200$ people asked about whether they like
apples (A), bananas (B), and cherries (C), the following data was obtained:
$|A| = 112$, $|B| = 89$, $|C| = 71$, $|A cap B| = 32$, $|A cap C| = 26$, $|B cap C| = 43$, $|A cap B cap C| = 20$.
a) How many people like apples or bananas?
b) How many people like exactly one of these fruit?
c) How many people like none of these fruit?
My work is as follows:
a) Assuming this problem is using inclusive or, we get
$|A| + |B| - |A cap B| = 112 + 89 - 32$
Reasoning:
If we take all of group A and all of group B, then we are double-counting the individuals that are simultaneously in group A and B, so we subtract this group of individuals from the total amount once, leaving us with the set of all people in group A or group B.
b) $|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C| = 112 + 89 +71 - 64 - 52 - 86 - 60 = 10$
Reasoning:
Since we're trying to find the number of people who like exactly one of these fruit, we must be sure to exclude all of the people who like more than one. When we take the cardinality of any two sets, we are, as I said in the last problem, double-counting the elements that are simultaneously elements in both sets. However, unlike the last problem, we want to totally exclude the people who are in both groups, and so we must subtract twice the intersection of those 2 sets. The special-case to this is the case where individuals are simultaneously present in all 3 groups, since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets.
c) $|A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| - 2|A cap B cap C| = 112 + 89 + 71 - 32 - 26 - 43 - 40 = 131$ people like at least one of A, B, or C.
Therefore, 200 - 131 = 69 people like none
Reasoning:
I've attempted to eliminate all double-counting and triple-counting of people, in order to get the total number of people who like at least one of the fruits. I then subtract this number from total number of people surveyed (200).
I must have made an error here, because I have that 169/200 people like apples or bananas, but 69/200 like none of the fruits, which makes no sense.
I would greatly appreciate it if people could please take the time to provide feedback.
elementary-set-theory
edited Jul 22 at 17:18
amWhy
189k25219431
asked Jul 22 at 17:11
The Pointer
2,4392829
closed as too broad by Shaun, Shailesh, Xander Henderson, Leucippus, Parcly Taxel Jul 23 at 1:41
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |Â
up vote
-2
down vote
favorite
I have the following problem:
In a survey of $200$ people asked about whether they like
apples (A), bananas (B), and cherries (C), the following data was obtained:
$|A| = 112$, $|B| = 89$, $|C| = 71$, $|A cap B| = 32$, $|A cap C| = 26$, $|B cap C| = 43$, $|A cap B cap C| = 20$.
a) How many people like apples or bananas?
b) How many people like exactly one of these fruit?
c) How many people like none of these fruit?
My work is as follows:
a) Assuming this problem is using inclusive or, we get
$|A| + |B| - |A cap B| = 112 + 89 - 32$
Reasoning:
If we take all of group A and all of group B, then we are double-counting the individuals that are simultaneously in group A and B, so we subtract this group of individuals from the total amount once, leaving us with the set of all people in group A or group B.
b) $|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C| = 112 + 89 +71 - 64 - 52 - 86 - 60 = 10$
Reasoning:
Since we're trying to find the number of people who like exactly one of these fruit, we must be sure to exclude all of the people who like more than one. When we take the cardinality of any two sets, we are, as I said in the last problem, double-counting the elements that are simultaneously elements in both sets. However, unlike the last problem, we want to totally exclude the people who are in both groups, and so we must subtract twice the intersection of those 2 sets. The special-case to this is the case where individuals are simultaneously present in all 3 groups, since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets.
c) $|A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| - 2|A cap B cap C| = 112 + 89 + 71 - 32 - 26 - 43 - 40 = 131$ people like at least one of A, B, or C.
Therefore, 200 - 131 = 69 people like none
Reasoning:
I've attempted to eliminate all double-counting and triple-counting of people, in order to get the total number of people who like at least one of the fruits. I then subtract this number from total number of people surveyed (200).
I must have made an error here, because I have that 169/200 people like apples or bananas, but 69/200 like none of the fruits, which makes no sense.
I would greatly appreciate it if people could please take the time to provide feedback.
elementary-set-theory
edited Jul 22 at 17:18
amWhy
189k25219431
asked Jul 22 at 17:11
The Pointer
2,4392829
closed as too broad by Shaun, Shailesh, Xander Henderson, Leucippus, Parcly Taxel Jul 23 at 1:41
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
What's with the downvotes?
– The Pointer
Jul 22 at 18:36
1
" since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets." You must subtract them from the ones you were subtracting! So you must add them!
– fleablood
Jul 22 at 18:44
@fleablood Yes, your answer clarified this for me. Thanks for the help.
– The Pointer
Jul 22 at 19:07
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I have the following problem:
In a survey of $200$ people asked about whether they like
apples (A), bananas (B), and cherries (C), the following data was obtained:
$|A| = 112$, $|B| = 89$, $|C| = 71$, $|A cap B| = 32$, $|A cap C| = 26$, $|B cap C| = 43$, $|A cap B cap C| = 20$.
a) How many people like apples or bananas?
b) How many people like exactly one of these fruit?
c) How many people like none of these fruit?
My work is as follows:
a) Assuming this problem is using inclusive or, we get
$|A| + |B| - |A cap B| = 112 + 89 - 32$
Reasoning:
If we take all of group A and all of group B, then we are double-counting the individuals that are simultaneously in group A and B, so we subtract this group of individuals from the total amount once, leaving us with the set of all people in group A or group B.
b) $|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C| = 112 + 89 +71 - 64 - 52 - 86 - 60 = 10$
Reasoning:
Since we're trying to find the number of people who like exactly one of these fruit, we must be sure to exclude all of the people who like more than one. When we take the cardinality of any two sets, we are, as I said in the last problem, double-counting the elements that are simultaneously elements in both sets. However, unlike the last problem, we want to totally exclude the people who are in both groups, and so we must subtract twice the intersection of those 2 sets. The special-case to this is the case where individuals are simultaneously present in all 3 groups, since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets.
c) $|A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| - 2|A cap B cap C| = 112 + 89 + 71 - 32 - 26 - 43 - 40 = 131$ people like at least one of A, B, or C.
Therefore, 200 - 131 = 69 people like none
Reasoning:
I've attempted to eliminate all double-counting and triple-counting of people, in order to get the total number of people who like at least one of the fruits. I then subtract this number from total number of people surveyed (200).
I must have made an error here, because I have that 169/200 people like apples or bananas, but 69/200 like none of the fruits, which makes no sense.
I would greatly appreciate it if people could please take the time to provide feedback.
elementary-set-theory
edited Jul 22 at 17:18
amWhy
189k25219431
asked Jul 22 at 17:11
The Pointer
2,4392829
I have the following problem:
In a survey of $200$ people asked about whether they like
apples (A), bananas (B), and cherries (C), the following data was obtained:
$|A| = 112$, $|B| = 89$, $|C| = 71$, $|A cap B| = 32$, $|A cap C| = 26$, $|B cap C| = 43$, $|A cap B cap C| = 20$.
a) How many people like apples or bananas?
b) How many people like exactly one of these fruit?
c) How many people like none of these fruit?
My work is as follows:
a) Assuming this problem is using inclusive or, we get
$|A| + |B| - |A cap B| = 112 + 89 - 32$
Reasoning:
If we take all of group A and all of group B, then we are double-counting the individuals that are simultaneously in group A and B, so we subtract this group of individuals from the total amount once, leaving us with the set of all people in group A or group B.
b) $|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C| = 112 + 89 +71 - 64 - 52 - 86 - 60 = 10$
Reasoning:
Since we're trying to find the number of people who like exactly one of these fruit, we must be sure to exclude all of the people who like more than one. When we take the cardinality of any two sets, we are, as I said in the last problem, double-counting the elements that are simultaneously elements in both sets. However, unlike the last problem, we want to totally exclude the people who are in both groups, and so we must subtract twice the intersection of those 2 sets. The special-case to this is the case where individuals are simultaneously present in all 3 groups, since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets.
c) $|A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| - 2|A cap B cap C| = 112 + 89 + 71 - 32 - 26 - 43 - 40 = 131$ people like at least one of A, B, or C.
Therefore, 200 - 131 = 69 people like none
Reasoning:
I've attempted to eliminate all double-counting and triple-counting of people, in order to get the total number of people who like at least one of the fruits. I then subtract this number from total number of people surveyed (200).
I must have made an error here, because I have that 169/200 people like apples or bananas, but 69/200 like none of the fruits, which makes no sense.
I would greatly appreciate it if people could please take the time to provide feedback.
elementary-set-theory
edited Jul 22 at 17:18
amWhy
189k25219431
asked Jul 22 at 17:11
The Pointer
2,4392829
edited Jul 22 at 17:18
amWhy
189k25219431
edited Jul 22 at 17:18
amWhy
189k25219431
edited Jul 22 at 17:18
amWhy
189k25219431
189k25219431
asked Jul 22 at 17:11
The Pointer
2,4392829
asked Jul 22 at 17:11
The Pointer
2,4392829
asked Jul 22 at 17:11
The Pointer
2,4392829
2,4392829
closed as too broad by Shaun, Shailesh, Xander Henderson, Leucippus, Parcly Taxel Jul 23 at 1:41
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by Shaun, Shailesh, Xander Henderson, Leucippus, Parcly Taxel Jul 23 at 1:41
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
What's with the downvotes?
– The Pointer
Jul 22 at 18:36
1
" since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets." You must subtract them from the ones you were subtracting! So you must add them!
– fleablood
Jul 22 at 18:44
@fleablood Yes, your answer clarified this for me. Thanks for the help.
– The Pointer
Jul 22 at 19:07
add a comment |Â
What's with the downvotes?
– The Pointer
Jul 22 at 18:36
1
" since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets." You must subtract them from the ones you were subtracting! So you must add them!
– fleablood
Jul 22 at 18:44
@fleablood Yes, your answer clarified this for me. Thanks for the help.
– The Pointer
Jul 22 at 19:07
What's with the downvotes?
– The Pointer
Jul 22 at 18:36
What's with the downvotes?
– The Pointer
Jul 22 at 18:36
1
1
" since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets." You must subtract them from the ones you were subtracting! So you must add them!
– fleablood
Jul 22 at 18:44
" since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets." You must subtract them from the ones you were subtracting! So you must add them!
– fleablood
Jul 22 at 18:44
@fleablood Yes, your answer clarified this for me. Thanks for the help.
– The Pointer
Jul 22 at 19:07
@fleablood Yes, your answer clarified this for me. Thanks for the help.
– The Pointer
Jul 22 at 19:07
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
A Venn diagram is always helpful.....
answered Jul 22 at 18:40
Phil H
1,8232311
Thanks for taking the time to do the Venn-Diagram. How did you calculate (c)?
– The Pointer
Jul 22 at 19:01
1
200 minus all the numbers inside the Venn $200 - (74+12+34+6+20+23+22) = 200 - 191$
– Phil H
Jul 22 at 19:07
add a comment |Â
up vote
1
down vote
The quantity you want for number of people who like at least one fruit is
$$|Acup Bcup C| = |A| + |B| + |C| - |Acap B| - |Acap C| - |Bcap C| + |A cap Bcap C|$$ You subtracted two times the number of people who like all fruits, but you should actually add this quantity back in because note that any person counted in $|Acap Bcap C|$ is also counted in $|Acap B|$, $|Acap C|$, and $|Bcap C|$ because they like fruits A, B, and C. Thus, you have counted them three times (in each individual $|A|$, $|B|$, $|C|$), and also subtracted them three times, so you have not counted these people who like all fruits at all. Thus, you must add them back in after doing the first six additions and subtractions.
answered Jul 22 at 17:47
CptQ
662
Yes, I think you're right. Thanks for the clarification.
– The Pointer
Jul 22 at 19:33
add a comment |Â
up vote
0
down vote
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C|$
should be
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| +3|A cap B cap C| $
and $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|−2|A∩B∩C|$ should be $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$
....
Let's do b).
The people who only like apples is $|A| - |$people who like apples and other fruits other fruit$|$
And... well you understand the exclusion inclusion principal well enough to get that $|$people who like apples and other fruits other fruit$|= |Acap B| + |Acap C| - |Acap Bcap C|$
So the people who only like apples is $|A| - (|Acap B| + |Acap C| - |Acap Bcap C|) = |A| - |Acap B| - |Acap C| + |Acap Bcap C|$.
And
So on...
upshot: Alternate the signs the "deeper" you get into the inclusion/exclusion.
answered Jul 22 at 18:41
fleablood
60.4k22575
Thanks for the response. But doesn't $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$ = 211? We only surveyed 200 people.
– The Pointer
Jul 22 at 19:00
See CptQ's answer. I think he is correct in that it should be $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|$. $|A|+|B|+|C|$ triple-counts the individuals who like all 3 fruits. $−|A∩B|−|A∩C|−|B∩C|$ then subtracts them 3 times, meaning that we have now not counted any of the individuals who like all 3 fruits; and so we add |A∩B∩C| once to account for these individuals.
– The Pointer
Jul 22 at 19:37
I didn't go over your answer c) with a fine tooth comb. The who like any fruit at all will be $|A|+|B|+|C| - |Acap B| - |Acap C|-|Acap B cap C|$ I'm not sure why you doubled $|Acap Bcap C|$. So those who don't like any will be $200$ minus that.
– fleablood
Jul 22 at 20:21
I still don't think that's correct. See CptQ's answer.
– The Pointer
Jul 22 at 20:36
That's a #@#%# typo. That is the exact same as CptQ's answer. $|A|+|B|+|C| - |Acap B| - |Bcap C|- |Ccap A| + |Acap Bcap C|$.
– fleablood
Jul 22 at 20:40
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A Venn diagram is always helpful.....
answered Jul 22 at 18:40
Phil H
1,8232311
Thanks for taking the time to do the Venn-Diagram. How did you calculate (c)?
– The Pointer
Jul 22 at 19:01
1
200 minus all the numbers inside the Venn $200 - (74+12+34+6+20+23+22) = 200 - 191$
– Phil H
Jul 22 at 19:07
add a comment |Â
up vote
2
down vote
accepted
A Venn diagram is always helpful.....
answered Jul 22 at 18:40
Phil H
1,8232311
Thanks for taking the time to do the Venn-Diagram. How did you calculate (c)?
– The Pointer
Jul 22 at 19:01
1
200 minus all the numbers inside the Venn $200 - (74+12+34+6+20+23+22) = 200 - 191$
– Phil H
Jul 22 at 19:07
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A Venn diagram is always helpful.....
answered Jul 22 at 18:40
Phil H
1,8232311
A Venn diagram is always helpful.....
answered Jul 22 at 18:40
Phil H
1,8232311
answered Jul 22 at 18:40
Phil H
1,8232311
answered Jul 22 at 18:40
Phil H
1,8232311
answered Jul 22 at 18:40
Phil H
1,8232311
1,8232311
Thanks for taking the time to do the Venn-Diagram. How did you calculate (c)?
– The Pointer
Jul 22 at 19:01
1
200 minus all the numbers inside the Venn $200 - (74+12+34+6+20+23+22) = 200 - 191$
– Phil H
Jul 22 at 19:07
add a comment |Â
Thanks for taking the time to do the Venn-Diagram. How did you calculate (c)?
– The Pointer
Jul 22 at 19:01
1
200 minus all the numbers inside the Venn $200 - (74+12+34+6+20+23+22) = 200 - 191$
– Phil H
Jul 22 at 19:07
Thanks for taking the time to do the Venn-Diagram. How did you calculate (c)?
– The Pointer
Jul 22 at 19:01
Thanks for taking the time to do the Venn-Diagram. How did you calculate (c)?
– The Pointer
Jul 22 at 19:01
1
1
200 minus all the numbers inside the Venn $200 - (74+12+34+6+20+23+22) = 200 - 191$
– Phil H
Jul 22 at 19:07
200 minus all the numbers inside the Venn $200 - (74+12+34+6+20+23+22) = 200 - 191$
– Phil H
Jul 22 at 19:07
add a comment |Â
up vote
1
down vote
The quantity you want for number of people who like at least one fruit is
$$|Acup Bcup C| = |A| + |B| + |C| - |Acap B| - |Acap C| - |Bcap C| + |A cap Bcap C|$$ You subtracted two times the number of people who like all fruits, but you should actually add this quantity back in because note that any person counted in $|Acap Bcap C|$ is also counted in $|Acap B|$, $|Acap C|$, and $|Bcap C|$ because they like fruits A, B, and C. Thus, you have counted them three times (in each individual $|A|$, $|B|$, $|C|$), and also subtracted them three times, so you have not counted these people who like all fruits at all. Thus, you must add them back in after doing the first six additions and subtractions.
answered Jul 22 at 17:47
CptQ
662
Yes, I think you're right. Thanks for the clarification.
– The Pointer
Jul 22 at 19:33
add a comment |Â
up vote
1
down vote
The quantity you want for number of people who like at least one fruit is
$$|Acup Bcup C| = |A| + |B| + |C| - |Acap B| - |Acap C| - |Bcap C| + |A cap Bcap C|$$ You subtracted two times the number of people who like all fruits, but you should actually add this quantity back in because note that any person counted in $|Acap Bcap C|$ is also counted in $|Acap B|$, $|Acap C|$, and $|Bcap C|$ because they like fruits A, B, and C. Thus, you have counted them three times (in each individual $|A|$, $|B|$, $|C|$), and also subtracted them three times, so you have not counted these people who like all fruits at all. Thus, you must add them back in after doing the first six additions and subtractions.
answered Jul 22 at 17:47
CptQ
662
Yes, I think you're right. Thanks for the clarification.
– The Pointer
Jul 22 at 19:33
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The quantity you want for number of people who like at least one fruit is
$$|Acup Bcup C| = |A| + |B| + |C| - |Acap B| - |Acap C| - |Bcap C| + |A cap Bcap C|$$ You subtracted two times the number of people who like all fruits, but you should actually add this quantity back in because note that any person counted in $|Acap Bcap C|$ is also counted in $|Acap B|$, $|Acap C|$, and $|Bcap C|$ because they like fruits A, B, and C. Thus, you have counted them three times (in each individual $|A|$, $|B|$, $|C|$), and also subtracted them three times, so you have not counted these people who like all fruits at all. Thus, you must add them back in after doing the first six additions and subtractions.
answered Jul 22 at 17:47
CptQ
662
The quantity you want for number of people who like at least one fruit is
$$|Acup Bcup C| = |A| + |B| + |C| - |Acap B| - |Acap C| - |Bcap C| + |A cap Bcap C|$$ You subtracted two times the number of people who like all fruits, but you should actually add this quantity back in because note that any person counted in $|Acap Bcap C|$ is also counted in $|Acap B|$, $|Acap C|$, and $|Bcap C|$ because they like fruits A, B, and C. Thus, you have counted them three times (in each individual $|A|$, $|B|$, $|C|$), and also subtracted them three times, so you have not counted these people who like all fruits at all. Thus, you must add them back in after doing the first six additions and subtractions.
answered Jul 22 at 17:47
CptQ
662
answered Jul 22 at 17:47
CptQ
662
answered Jul 22 at 17:47
CptQ
662
answered Jul 22 at 17:47
CptQ
662
662
Yes, I think you're right. Thanks for the clarification.
– The Pointer
Jul 22 at 19:33
add a comment |Â
Yes, I think you're right. Thanks for the clarification.
– The Pointer
Jul 22 at 19:33
Yes, I think you're right. Thanks for the clarification.
– The Pointer
Jul 22 at 19:33
Yes, I think you're right. Thanks for the clarification.
– The Pointer
Jul 22 at 19:33
add a comment |Â
up vote
0
down vote
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C|$
should be
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| +3|A cap B cap C| $
and $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|−2|A∩B∩C|$ should be $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$
....
Let's do b).
The people who only like apples is $|A| - |$people who like apples and other fruits other fruit$|$
And... well you understand the exclusion inclusion principal well enough to get that $|$people who like apples and other fruits other fruit$|= |Acap B| + |Acap C| - |Acap Bcap C|$
So the people who only like apples is $|A| - (|Acap B| + |Acap C| - |Acap Bcap C|) = |A| - |Acap B| - |Acap C| + |Acap Bcap C|$.
And
So on...
upshot: Alternate the signs the "deeper" you get into the inclusion/exclusion.
answered Jul 22 at 18:41
fleablood
60.4k22575
Thanks for the response. But doesn't $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$ = 211? We only surveyed 200 people.
– The Pointer
Jul 22 at 19:00
See CptQ's answer. I think he is correct in that it should be $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|$. $|A|+|B|+|C|$ triple-counts the individuals who like all 3 fruits. $−|A∩B|−|A∩C|−|B∩C|$ then subtracts them 3 times, meaning that we have now not counted any of the individuals who like all 3 fruits; and so we add |A∩B∩C| once to account for these individuals.
– The Pointer
Jul 22 at 19:37
I didn't go over your answer c) with a fine tooth comb. The who like any fruit at all will be $|A|+|B|+|C| - |Acap B| - |Acap C|-|Acap B cap C|$ I'm not sure why you doubled $|Acap Bcap C|$. So those who don't like any will be $200$ minus that.
– fleablood
Jul 22 at 20:21
I still don't think that's correct. See CptQ's answer.
– The Pointer
Jul 22 at 20:36
That's a #@#%# typo. That is the exact same as CptQ's answer. $|A|+|B|+|C| - |Acap B| - |Bcap C|- |Ccap A| + |Acap Bcap C|$.
– fleablood
Jul 22 at 20:40
add a comment |Â
up vote
0
down vote
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C|$
should be
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| +3|A cap B cap C| $
and $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|−2|A∩B∩C|$ should be $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$
....
Let's do b).
The people who only like apples is $|A| - |$people who like apples and other fruits other fruit$|$
And... well you understand the exclusion inclusion principal well enough to get that $|$people who like apples and other fruits other fruit$|= |Acap B| + |Acap C| - |Acap Bcap C|$
So the people who only like apples is $|A| - (|Acap B| + |Acap C| - |Acap Bcap C|) = |A| - |Acap B| - |Acap C| + |Acap Bcap C|$.
And
So on...
upshot: Alternate the signs the "deeper" you get into the inclusion/exclusion.
answered Jul 22 at 18:41
fleablood
60.4k22575
Thanks for the response. But doesn't $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$ = 211? We only surveyed 200 people.
– The Pointer
Jul 22 at 19:00
See CptQ's answer. I think he is correct in that it should be $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|$. $|A|+|B|+|C|$ triple-counts the individuals who like all 3 fruits. $−|A∩B|−|A∩C|−|B∩C|$ then subtracts them 3 times, meaning that we have now not counted any of the individuals who like all 3 fruits; and so we add |A∩B∩C| once to account for these individuals.
– The Pointer
Jul 22 at 19:37
I didn't go over your answer c) with a fine tooth comb. The who like any fruit at all will be $|A|+|B|+|C| - |Acap B| - |Acap C|-|Acap B cap C|$ I'm not sure why you doubled $|Acap Bcap C|$. So those who don't like any will be $200$ minus that.
– fleablood
Jul 22 at 20:21
I still don't think that's correct. See CptQ's answer.
– The Pointer
Jul 22 at 20:36
That's a #@#%# typo. That is the exact same as CptQ's answer. $|A|+|B|+|C| - |Acap B| - |Bcap C|- |Ccap A| + |Acap Bcap C|$.
– fleablood
Jul 22 at 20:40
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C|$
should be
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| +3|A cap B cap C| $
and $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|−2|A∩B∩C|$ should be $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$
....
Let's do b).
The people who only like apples is $|A| - |$people who like apples and other fruits other fruit$|$
And... well you understand the exclusion inclusion principal well enough to get that $|$people who like apples and other fruits other fruit$|= |Acap B| + |Acap C| - |Acap Bcap C|$
So the people who only like apples is $|A| - (|Acap B| + |Acap C| - |Acap Bcap C|) = |A| - |Acap B| - |Acap C| + |Acap Bcap C|$.
And
So on...
upshot: Alternate the signs the "deeper" you get into the inclusion/exclusion.
answered Jul 22 at 18:41
fleablood
60.4k22575
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| - 3|A cap B cap C|$
should be
$|A| + |B| + |C| - 2|A cap B| - 2|A cap C| - 2|B cap C| +3|A cap B cap C| $
and $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|−2|A∩B∩C|$ should be $ |A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$
....
Let's do b).
The people who only like apples is $|A| - |$people who like apples and other fruits other fruit$|$
And... well you understand the exclusion inclusion principal well enough to get that $|$people who like apples and other fruits other fruit$|= |Acap B| + |Acap C| - |Acap Bcap C|$
So the people who only like apples is $|A| - (|Acap B| + |Acap C| - |Acap Bcap C|) = |A| - |Acap B| - |Acap C| + |Acap Bcap C|$.
And
So on...
upshot: Alternate the signs the "deeper" you get into the inclusion/exclusion.
answered Jul 22 at 18:41
fleablood
60.4k22575
answered Jul 22 at 18:41
fleablood
60.4k22575
answered Jul 22 at 18:41
fleablood
60.4k22575
answered Jul 22 at 18:41
fleablood
60.4k22575
60.4k22575
Thanks for the response. But doesn't $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$ = 211? We only surveyed 200 people.
– The Pointer
Jul 22 at 19:00
See CptQ's answer. I think he is correct in that it should be $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|$. $|A|+|B|+|C|$ triple-counts the individuals who like all 3 fruits. $−|A∩B|−|A∩C|−|B∩C|$ then subtracts them 3 times, meaning that we have now not counted any of the individuals who like all 3 fruits; and so we add |A∩B∩C| once to account for these individuals.
– The Pointer
Jul 22 at 19:37
I didn't go over your answer c) with a fine tooth comb. The who like any fruit at all will be $|A|+|B|+|C| - |Acap B| - |Acap C|-|Acap B cap C|$ I'm not sure why you doubled $|Acap Bcap C|$. So those who don't like any will be $200$ minus that.
– fleablood
Jul 22 at 20:21
I still don't think that's correct. See CptQ's answer.
– The Pointer
Jul 22 at 20:36
That's a #@#%# typo. That is the exact same as CptQ's answer. $|A|+|B|+|C| - |Acap B| - |Bcap C|- |Ccap A| + |Acap Bcap C|$.
– fleablood
Jul 22 at 20:40
add a comment |Â
Thanks for the response. But doesn't $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$ = 211? We only surveyed 200 people.
– The Pointer
Jul 22 at 19:00
See CptQ's answer. I think he is correct in that it should be $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|$. $|A|+|B|+|C|$ triple-counts the individuals who like all 3 fruits. $−|A∩B|−|A∩C|−|B∩C|$ then subtracts them 3 times, meaning that we have now not counted any of the individuals who like all 3 fruits; and so we add |A∩B∩C| once to account for these individuals.
– The Pointer
Jul 22 at 19:37
I didn't go over your answer c) with a fine tooth comb. The who like any fruit at all will be $|A|+|B|+|C| - |Acap B| - |Acap C|-|Acap B cap C|$ I'm not sure why you doubled $|Acap Bcap C|$. So those who don't like any will be $200$ minus that.
– fleablood
Jul 22 at 20:21
I still don't think that's correct. See CptQ's answer.
– The Pointer
Jul 22 at 20:36
That's a #@#%# typo. That is the exact same as CptQ's answer. $|A|+|B|+|C| - |Acap B| - |Bcap C|- |Ccap A| + |Acap Bcap C|$.
– fleablood
Jul 22 at 20:40
Thanks for the response. But doesn't $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$ = 211? We only surveyed 200 people.
– The Pointer
Jul 22 at 19:00
Thanks for the response. But doesn't $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+2|A∩B∩C|$ = 211? We only surveyed 200 people.
– The Pointer
Jul 22 at 19:00
See CptQ's answer. I think he is correct in that it should be $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|$. $|A|+|B|+|C|$ triple-counts the individuals who like all 3 fruits. $−|A∩B|−|A∩C|−|B∩C|$ then subtracts them 3 times, meaning that we have now not counted any of the individuals who like all 3 fruits; and so we add |A∩B∩C| once to account for these individuals.
– The Pointer
Jul 22 at 19:37
See CptQ's answer. I think he is correct in that it should be $|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|$. $|A|+|B|+|C|$ triple-counts the individuals who like all 3 fruits. $−|A∩B|−|A∩C|−|B∩C|$ then subtracts them 3 times, meaning that we have now not counted any of the individuals who like all 3 fruits; and so we add |A∩B∩C| once to account for these individuals.
– The Pointer
Jul 22 at 19:37
I didn't go over your answer c) with a fine tooth comb. The who like any fruit at all will be $|A|+|B|+|C| - |Acap B| - |Acap C|-|Acap B cap C|$ I'm not sure why you doubled $|Acap Bcap C|$. So those who don't like any will be $200$ minus that.
– fleablood
Jul 22 at 20:21
I didn't go over your answer c) with a fine tooth comb. The who like any fruit at all will be $|A|+|B|+|C| - |Acap B| - |Acap C|-|Acap B cap C|$ I'm not sure why you doubled $|Acap Bcap C|$. So those who don't like any will be $200$ minus that.
– fleablood
Jul 22 at 20:21
I still don't think that's correct. See CptQ's answer.
– The Pointer
Jul 22 at 20:36
I still don't think that's correct. See CptQ's answer.
– The Pointer
Jul 22 at 20:36
That's a #@#%# typo. That is the exact same as CptQ's answer. $|A|+|B|+|C| - |Acap B| - |Bcap C|- |Ccap A| + |Acap Bcap C|$.
– fleablood
Jul 22 at 20:40
That's a #@#%# typo. That is the exact same as CptQ's answer. $|A|+|B|+|C| - |Acap B| - |Bcap C|- |Ccap A| + |Acap Bcap C|$.
– fleablood
Jul 22 at 20:40
add a comment |Â
What's with the downvotes?
– The Pointer
Jul 22 at 18:36
1
" since we are then triple-counting them, and we must therefore subtract three times the intersection of these 3 sets." You must subtract them from the ones you were subtracting! So you must add them!
– fleablood
Jul 22 at 18:44
@fleablood Yes, your answer clarified this for me. Thanks for the help.
– The Pointer
Jul 22 at 19:07