Mixing
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How many persons eat at least two out of the three dishes?
Clash Royale CLAN TAG#URR8PPP
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Out of a group of 21 persons, 9 eat vegetables, 10 eat fish and 7 eat eggs. 5 persons eat all three. How many persons eat at least two out of the three dishes?
My take:
Let $A∩B∩C = x$, then $(A∩B+B∩C+A∩C)$, this already contains $3x$. therefore subtracting $2x$ from this should result into POSITIVE value, but it is zero.
Moreover, they are asking for "at least 2" which means $(A∩B+B∩C+A∩C) - 2x$.
Is something wrong with given data ?
The answer given in book is any number between [5,11].
Please help me understand & solve this question.
Any help is appreciated in advance.
probability combinatorics
edited 10 mins ago
Robert Z
87.3k1056127
asked 2 hours ago
Geeklovenerds
258
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up vote
2
down vote
favorite
Out of a group of 21 persons, 9 eat vegetables, 10 eat fish and 7 eat eggs. 5 persons eat all three. How many persons eat at least two out of the three dishes?
My take:
Let $A∩B∩C = x$, then $(A∩B+B∩C+A∩C)$, this already contains $3x$. therefore subtracting $2x$ from this should result into POSITIVE value, but it is zero.
Moreover, they are asking for "at least 2" which means $(A∩B+B∩C+A∩C) - 2x$.
Is something wrong with given data ?
The answer given in book is any number between [5,11].
Please help me understand & solve this question.
Any help is appreciated in advance.
probability combinatorics
edited 10 mins ago
Robert Z
87.3k1056127
asked 2 hours ago
Geeklovenerds
258
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Out of a group of 21 persons, 9 eat vegetables, 10 eat fish and 7 eat eggs. 5 persons eat all three. How many persons eat at least two out of the three dishes?
My take:
Let $A∩B∩C = x$, then $(A∩B+B∩C+A∩C)$, this already contains $3x$. therefore subtracting $2x$ from this should result into POSITIVE value, but it is zero.
Moreover, they are asking for "at least 2" which means $(A∩B+B∩C+A∩C) - 2x$.
Is something wrong with given data ?
The answer given in book is any number between [5,11].
Please help me understand & solve this question.
Any help is appreciated in advance.
probability combinatorics
edited 10 mins ago
Robert Z
87.3k1056127
asked 2 hours ago
Geeklovenerds
258
Out of a group of 21 persons, 9 eat vegetables, 10 eat fish and 7 eat eggs. 5 persons eat all three. How many persons eat at least two out of the three dishes?
My take:
Let $A∩B∩C = x$, then $(A∩B+B∩C+A∩C)$, this already contains $3x$. therefore subtracting $2x$ from this should result into POSITIVE value, but it is zero.
Moreover, they are asking for "at least 2" which means $(A∩B+B∩C+A∩C) - 2x$.
Is something wrong with given data ?
The answer given in book is any number between [5,11].
Please help me understand & solve this question.
Any help is appreciated in advance.
probability combinatorics
probability combinatorics
edited 10 mins ago
Robert Z
87.3k1056127
asked 2 hours ago
Geeklovenerds
258
edited 10 mins ago
Robert Z
87.3k1056127
asked 2 hours ago
Geeklovenerds
258
edited 10 mins ago
Robert Z
87.3k1056127
edited 10 mins ago
Robert Z
87.3k1056127
edited 10 mins ago
Robert Z
87.3k1056127
87.3k1056127
asked 2 hours ago
Geeklovenerds
258
asked 2 hours ago
Geeklovenerds
258
asked 2 hours ago
Geeklovenerds
258
258
add a comment |Â
add a comment |Â
5 Answers
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up vote
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Something is wrong with the data in this problem, assuming that each person eats at least one dish: vegetables, fish or eggs.
The number of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Fcap Vcap E|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|E|-|Fcup Vcup E|-|Fcap Vcap E|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Fcap Vcap E|=5$.
On the other hand, if there are persons that do not eat vegetables, fish or eggs then
$$10=max(|F|,|V|,|E|)leq |Fcup Vcup E|leq 21$$
and the above equality implies
$$5=|Fcap Vcap E|leq N=21-|Fcup Vcup E|leq 11.$$
answered 1 hour ago
Robert Z
87.3k1056127
2
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
1 hour ago
can you please elaborate your last line I am not getting it.
– Geeklovenerds
54 mins ago
1
Is it clear now?
– Robert Z
50 mins ago
add a comment |Â
up vote
1
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My approach- don't know if it is correct.
N(A∪B∪C)=N(A)+N(B)+N(C)−N(A∩B)−N(A∩C)−N(B∩C)+N(A∩B∩C)
Let Y be the no. of persons who eat at least one item. 21−Y people do not eat anything.
Y=9+10+7−[N(A∩B)+N(A∩C)+N(B∩C)]+5
[N(A∩B)+N(A∩C)+N(B∩C)]=31−Y.
Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as
31−Y−2∗5=21−Y.
The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items =(9−5)+(10−5)+(7−5)=11. And adding 5 we get 16 people who eat at least one item.
So, our required answer is 21−10≥X≥21−16⟹5≤X≤11
answered 1 hour ago
Geeklovenerds
258
The minimum value of Y cannot be 10 - you need to also take into account the fact that 5 people ate 3 items. After those 5 people ate their 3 items, there are 11 items left. You need at least 6 people to eat these items if none can eat more than 2 : ceil(11/2) = 6
– Sander De Dycker
8 mins ago
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up vote
1
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We can extract the 5 people who eat all three dishes from the problem. Now we have 16 people left, 4 eat vegetables, 5 eat fish, 2 eat eggs, nobody eats three dishes. We see that at most 5 of these eat two dishes, because there are only 11 dishes in total. It is also possible that 11 people eat one dish each, 5 eat nothing, and nobody eats 5 dishes.
Add the 5 eating three dishes back in, and you get that at least 5 and at most 11 eat two or more dishes.
At least five people eat two or more dishes, because there are already five eating three dishes. At most eleven eat three or more, because for 12 people you would need 5 times 3 dishes, plus 7 times 2 dishes, that's 29, but there are only 26.
answered 35 mins ago
gnasher729
5,8911028
how there are total 26 dishes.
– Geeklovenerds
27 mins ago
@Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer.
– Ross Millikan
23 mins ago
5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10.
– Sander De Dycker
18 mins ago
add a comment |Â
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1
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The problem as supplied certainly is misleading. It leads us to the following assumption:
Assumption: that every person in the group eats at least one dish.
This however is impossible given the premise. If we just do this quite simply, without all the probability stuff:
The main pertinent information given is that 5 people eat all three, so that leaves us with
21 -5 = 16 people
9 - 5 = 4 vegetables
10 - 5 = 5 fish
7 - 5 = 2 eggs
This leaves us with a total of 11 dishes to split up among 16 people... You can see the issue with our above assumption.
Therefore we have to discard that assumption. Now, we need to usethe knowledge that no more people ate 3 dishes and the new set of data:
P = 16; V = 4; F = 5; E = 2
Really at this point the amount of people is pointless as we can easily give away one dish per person and not run out of people so let's ignore the amount of people. This also gives us our minimum value. At least 5 people ate at least 2 dishes (the 5 that ate 3).
So, we now need to find the maximum value. how can we combine the veggies, fish and eggs left over to have the most possible who eat two dishes.
As we have an odd number of dishes, we know we will have one left over. Now, let's pair them off as easily as possible, let's say, everyone who had vegetables also had fish, and then the last person who had fish also had eggs. So in a list:
VF, VF, VF, VF, FE
leaving us with one eggs left over. This adds 5 dishes to our minimum of 5 and so our maximum is 10. Therefore our answer should be:
"Any amount of people between 5 and 10 had two or more out of the three dishes"
In conclusion:
- The problem statement is misleading
- The answer in the book is wrong
- This problem doesn't really belong to probability, it's more of a logic puzzle
answered 26 mins ago
Aglahir
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I don't believe that the accepted answer is correct. As the OP pointed out in his question (although not at all clearly) adding up $|Acap B| + |Bcap C| + |Ccap A| $ counts anyone in $Acap B cap C$ three times, so we need to subtract $2cdot|Acap B cap C|$ from $10$ to get the number who eat at least two dishes. Unfortunately, this gives $0,$ contradicting the fact that $5$ people eat all three dishes.
The problem is insoluble as given.
answered 1 hour ago
saulspatz
12.3k21326
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Something is wrong with the data in this problem, assuming that each person eats at least one dish: vegetables, fish or eggs.
The number of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Fcap Vcap E|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|E|-|Fcup Vcup E|-|Fcap Vcap E|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Fcap Vcap E|=5$.
On the other hand, if there are persons that do not eat vegetables, fish or eggs then
$$10=max(|F|,|V|,|E|)leq |Fcup Vcup E|leq 21$$
and the above equality implies
$$5=|Fcap Vcap E|leq N=21-|Fcup Vcup E|leq 11.$$
answered 1 hour ago
Robert Z
87.3k1056127
2
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
1 hour ago
can you please elaborate your last line I am not getting it.
– Geeklovenerds
54 mins ago
1
Is it clear now?
– Robert Z
50 mins ago
add a comment |Â
up vote
2
down vote
Something is wrong with the data in this problem, assuming that each person eats at least one dish: vegetables, fish or eggs.
The number of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Fcap Vcap E|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|E|-|Fcup Vcup E|-|Fcap Vcap E|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Fcap Vcap E|=5$.
On the other hand, if there are persons that do not eat vegetables, fish or eggs then
$$10=max(|F|,|V|,|E|)leq |Fcup Vcup E|leq 21$$
and the above equality implies
$$5=|Fcap Vcap E|leq N=21-|Fcup Vcup E|leq 11.$$
answered 1 hour ago
Robert Z
87.3k1056127
2
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
1 hour ago
can you please elaborate your last line I am not getting it.
– Geeklovenerds
54 mins ago
1
Is it clear now?
– Robert Z
50 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Something is wrong with the data in this problem, assuming that each person eats at least one dish: vegetables, fish or eggs.
The number of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Fcap Vcap E|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|E|-|Fcup Vcup E|-|Fcap Vcap E|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Fcap Vcap E|=5$.
On the other hand, if there are persons that do not eat vegetables, fish or eggs then
$$10=max(|F|,|V|,|E|)leq |Fcup Vcup E|leq 21$$
and the above equality implies
$$5=|Fcap Vcap E|leq N=21-|Fcup Vcup E|leq 11.$$
answered 1 hour ago
Robert Z
87.3k1056127
Something is wrong with the data in this problem, assuming that each person eats at least one dish: vegetables, fish or eggs.
The number of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Fcap Vcap E|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|E|-|Fcup Vcup E|-|Fcap Vcap E|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Fcap Vcap E|=5$.
On the other hand, if there are persons that do not eat vegetables, fish or eggs then
$$10=max(|F|,|V|,|E|)leq |Fcup Vcup E|leq 21$$
and the above equality implies
$$5=|Fcap Vcap E|leq N=21-|Fcup Vcup E|leq 11.$$
answered 1 hour ago
Robert Z
87.3k1056127
edited 7 mins ago
answered 1 hour ago
Robert Z
87.3k1056127
answered 1 hour ago
Robert Z
87.3k1056127
answered 1 hour ago
Robert Z
87.3k1056127
87.3k1056127
2
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
1 hour ago
can you please elaborate your last line I am not getting it.
– Geeklovenerds
54 mins ago
1
Is it clear now?
– Robert Z
50 mins ago
add a comment |Â
2
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
1 hour ago
can you please elaborate your last line I am not getting it.
– Geeklovenerds
54 mins ago
1
Is it clear now?
– Robert Z
50 mins ago
2
2
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
1 hour ago
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
1 hour ago
can you please elaborate your last line I am not getting it.
– Geeklovenerds
54 mins ago
can you please elaborate your last line I am not getting it.
– Geeklovenerds
54 mins ago
1
1
Is it clear now?
– Robert Z
50 mins ago
Is it clear now?
– Robert Z
50 mins ago
add a comment |Â
up vote
1
down vote
My approach- don't know if it is correct.
N(A∪B∪C)=N(A)+N(B)+N(C)−N(A∩B)−N(A∩C)−N(B∩C)+N(A∩B∩C)
Let Y be the no. of persons who eat at least one item. 21−Y people do not eat anything.
Y=9+10+7−[N(A∩B)+N(A∩C)+N(B∩C)]+5
[N(A∩B)+N(A∩C)+N(B∩C)]=31−Y.
Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as
31−Y−2∗5=21−Y.
The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items =(9−5)+(10−5)+(7−5)=11. And adding 5 we get 16 people who eat at least one item.
So, our required answer is 21−10≥X≥21−16⟹5≤X≤11
answered 1 hour ago
Geeklovenerds
258
The minimum value of Y cannot be 10 - you need to also take into account the fact that 5 people ate 3 items. After those 5 people ate their 3 items, there are 11 items left. You need at least 6 people to eat these items if none can eat more than 2 : ceil(11/2) = 6
– Sander De Dycker
8 mins ago
add a comment |Â
up vote
1
down vote
My approach- don't know if it is correct.
N(A∪B∪C)=N(A)+N(B)+N(C)−N(A∩B)−N(A∩C)−N(B∩C)+N(A∩B∩C)
Let Y be the no. of persons who eat at least one item. 21−Y people do not eat anything.
Y=9+10+7−[N(A∩B)+N(A∩C)+N(B∩C)]+5
[N(A∩B)+N(A∩C)+N(B∩C)]=31−Y.
Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as
31−Y−2∗5=21−Y.
The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items =(9−5)+(10−5)+(7−5)=11. And adding 5 we get 16 people who eat at least one item.
So, our required answer is 21−10≥X≥21−16⟹5≤X≤11
answered 1 hour ago
Geeklovenerds
258
The minimum value of Y cannot be 10 - you need to also take into account the fact that 5 people ate 3 items. After those 5 people ate their 3 items, there are 11 items left. You need at least 6 people to eat these items if none can eat more than 2 : ceil(11/2) = 6
– Sander De Dycker
8 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
My approach- don't know if it is correct.
N(A∪B∪C)=N(A)+N(B)+N(C)−N(A∩B)−N(A∩C)−N(B∩C)+N(A∩B∩C)
Let Y be the no. of persons who eat at least one item. 21−Y people do not eat anything.
Y=9+10+7−[N(A∩B)+N(A∩C)+N(B∩C)]+5
[N(A∩B)+N(A∩C)+N(B∩C)]=31−Y.
Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as
31−Y−2∗5=21−Y.
The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items =(9−5)+(10−5)+(7−5)=11. And adding 5 we get 16 people who eat at least one item.
So, our required answer is 21−10≥X≥21−16⟹5≤X≤11
answered 1 hour ago
Geeklovenerds
258
My approach- don't know if it is correct.
N(A∪B∪C)=N(A)+N(B)+N(C)−N(A∩B)−N(A∩C)−N(B∩C)+N(A∩B∩C)
Let Y be the no. of persons who eat at least one item. 21−Y people do not eat anything.
Y=9+10+7−[N(A∩B)+N(A∩C)+N(B∩C)]+5
[N(A∩B)+N(A∩C)+N(B∩C)]=31−Y.
Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as
31−Y−2∗5=21−Y.
The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items =(9−5)+(10−5)+(7−5)=11. And adding 5 we get 16 people who eat at least one item.
So, our required answer is 21−10≥X≥21−16⟹5≤X≤11
answered 1 hour ago
Geeklovenerds
258
answered 1 hour ago
Geeklovenerds
258
answered 1 hour ago
Geeklovenerds
258
answered 1 hour ago
Geeklovenerds
258
258
The minimum value of Y cannot be 10 - you need to also take into account the fact that 5 people ate 3 items. After those 5 people ate their 3 items, there are 11 items left. You need at least 6 people to eat these items if none can eat more than 2 : ceil(11/2) = 6
– Sander De Dycker
8 mins ago
add a comment |Â
The minimum value of Y cannot be 10 - you need to also take into account the fact that 5 people ate 3 items. After those 5 people ate their 3 items, there are 11 items left. You need at least 6 people to eat these items if none can eat more than 2 : ceil(11/2) = 6
– Sander De Dycker
8 mins ago
The minimum value of Y cannot be 10 - you need to also take into account the fact that 5 people ate 3 items. After those 5 people ate their 3 items, there are 11 items left. You need at least 6 people to eat these items if none can eat more than 2 : ceil(11/2) = 6
– Sander De Dycker
8 mins ago
The minimum value of Y cannot be 10 - you need to also take into account the fact that 5 people ate 3 items. After those 5 people ate their 3 items, there are 11 items left. You need at least 6 people to eat these items if none can eat more than 2 : ceil(11/2) = 6
– Sander De Dycker
8 mins ago
add a comment |Â
up vote
1
down vote
We can extract the 5 people who eat all three dishes from the problem. Now we have 16 people left, 4 eat vegetables, 5 eat fish, 2 eat eggs, nobody eats three dishes. We see that at most 5 of these eat two dishes, because there are only 11 dishes in total. It is also possible that 11 people eat one dish each, 5 eat nothing, and nobody eats 5 dishes.
Add the 5 eating three dishes back in, and you get that at least 5 and at most 11 eat two or more dishes.
At least five people eat two or more dishes, because there are already five eating three dishes. At most eleven eat three or more, because for 12 people you would need 5 times 3 dishes, plus 7 times 2 dishes, that's 29, but there are only 26.
answered 35 mins ago
gnasher729
5,8911028
how there are total 26 dishes.
– Geeklovenerds
27 mins ago
@Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer.
– Ross Millikan
23 mins ago
5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10.
– Sander De Dycker
18 mins ago
add a comment |Â
up vote
1
down vote
We can extract the 5 people who eat all three dishes from the problem. Now we have 16 people left, 4 eat vegetables, 5 eat fish, 2 eat eggs, nobody eats three dishes. We see that at most 5 of these eat two dishes, because there are only 11 dishes in total. It is also possible that 11 people eat one dish each, 5 eat nothing, and nobody eats 5 dishes.
Add the 5 eating three dishes back in, and you get that at least 5 and at most 11 eat two or more dishes.
At least five people eat two or more dishes, because there are already five eating three dishes. At most eleven eat three or more, because for 12 people you would need 5 times 3 dishes, plus 7 times 2 dishes, that's 29, but there are only 26.
answered 35 mins ago
gnasher729
5,8911028
how there are total 26 dishes.
– Geeklovenerds
27 mins ago
@Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer.
– Ross Millikan
23 mins ago
5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10.
– Sander De Dycker
18 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We can extract the 5 people who eat all three dishes from the problem. Now we have 16 people left, 4 eat vegetables, 5 eat fish, 2 eat eggs, nobody eats three dishes. We see that at most 5 of these eat two dishes, because there are only 11 dishes in total. It is also possible that 11 people eat one dish each, 5 eat nothing, and nobody eats 5 dishes.
Add the 5 eating three dishes back in, and you get that at least 5 and at most 11 eat two or more dishes.
At least five people eat two or more dishes, because there are already five eating three dishes. At most eleven eat three or more, because for 12 people you would need 5 times 3 dishes, plus 7 times 2 dishes, that's 29, but there are only 26.
answered 35 mins ago
gnasher729
5,8911028
We can extract the 5 people who eat all three dishes from the problem. Now we have 16 people left, 4 eat vegetables, 5 eat fish, 2 eat eggs, nobody eats three dishes. We see that at most 5 of these eat two dishes, because there are only 11 dishes in total. It is also possible that 11 people eat one dish each, 5 eat nothing, and nobody eats 5 dishes.
Add the 5 eating three dishes back in, and you get that at least 5 and at most 11 eat two or more dishes.
At least five people eat two or more dishes, because there are already five eating three dishes. At most eleven eat three or more, because for 12 people you would need 5 times 3 dishes, plus 7 times 2 dishes, that's 29, but there are only 26.
answered 35 mins ago
gnasher729
5,8911028
answered 35 mins ago
gnasher729
5,8911028
answered 35 mins ago
gnasher729
5,8911028
answered 35 mins ago
gnasher729
5,8911028
5,8911028
how there are total 26 dishes.
– Geeklovenerds
27 mins ago
@Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer.
– Ross Millikan
23 mins ago
5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10.
– Sander De Dycker
18 mins ago
add a comment |Â
how there are total 26 dishes.
– Geeklovenerds
27 mins ago
@Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer.
– Ross Millikan
23 mins ago
5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10.
– Sander De Dycker
18 mins ago
how there are total 26 dishes.
– Geeklovenerds
27 mins ago
how there are total 26 dishes.
– Geeklovenerds
27 mins ago
@Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer.
– Ross Millikan
23 mins ago
@Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer.
– Ross Millikan
23 mins ago
5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10.
– Sander De Dycker
18 mins ago
5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10.
– Sander De Dycker
18 mins ago
add a comment |Â
up vote
1
down vote
The problem as supplied certainly is misleading. It leads us to the following assumption:
Assumption: that every person in the group eats at least one dish.
This however is impossible given the premise. If we just do this quite simply, without all the probability stuff:
The main pertinent information given is that 5 people eat all three, so that leaves us with
21 -5 = 16 people
9 - 5 = 4 vegetables
10 - 5 = 5 fish
7 - 5 = 2 eggs
This leaves us with a total of 11 dishes to split up among 16 people... You can see the issue with our above assumption.
Therefore we have to discard that assumption. Now, we need to usethe knowledge that no more people ate 3 dishes and the new set of data:
P = 16; V = 4; F = 5; E = 2
Really at this point the amount of people is pointless as we can easily give away one dish per person and not run out of people so let's ignore the amount of people. This also gives us our minimum value. At least 5 people ate at least 2 dishes (the 5 that ate 3).
So, we now need to find the maximum value. how can we combine the veggies, fish and eggs left over to have the most possible who eat two dishes.
As we have an odd number of dishes, we know we will have one left over. Now, let's pair them off as easily as possible, let's say, everyone who had vegetables also had fish, and then the last person who had fish also had eggs. So in a list:
VF, VF, VF, VF, FE
leaving us with one eggs left over. This adds 5 dishes to our minimum of 5 and so our maximum is 10. Therefore our answer should be:
"Any amount of people between 5 and 10 had two or more out of the three dishes"
In conclusion:
- The problem statement is misleading
- The answer in the book is wrong
- This problem doesn't really belong to probability, it's more of a logic puzzle
answered 26 mins ago
Aglahir
113
New contributor
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
The problem as supplied certainly is misleading. It leads us to the following assumption:
Assumption: that every person in the group eats at least one dish.
This however is impossible given the premise. If we just do this quite simply, without all the probability stuff:
The main pertinent information given is that 5 people eat all three, so that leaves us with
21 -5 = 16 people
9 - 5 = 4 vegetables
10 - 5 = 5 fish
7 - 5 = 2 eggs
This leaves us with a total of 11 dishes to split up among 16 people... You can see the issue with our above assumption.
Therefore we have to discard that assumption. Now, we need to usethe knowledge that no more people ate 3 dishes and the new set of data:
P = 16; V = 4; F = 5; E = 2
Really at this point the amount of people is pointless as we can easily give away one dish per person and not run out of people so let's ignore the amount of people. This also gives us our minimum value. At least 5 people ate at least 2 dishes (the 5 that ate 3).
So, we now need to find the maximum value. how can we combine the veggies, fish and eggs left over to have the most possible who eat two dishes.
As we have an odd number of dishes, we know we will have one left over. Now, let's pair them off as easily as possible, let's say, everyone who had vegetables also had fish, and then the last person who had fish also had eggs. So in a list:
VF, VF, VF, VF, FE
leaving us with one eggs left over. This adds 5 dishes to our minimum of 5 and so our maximum is 10. Therefore our answer should be:
"Any amount of people between 5 and 10 had two or more out of the three dishes"
In conclusion:
- The problem statement is misleading
- The answer in the book is wrong
- This problem doesn't really belong to probability, it's more of a logic puzzle
answered 26 mins ago
Aglahir
113
New contributor
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The problem as supplied certainly is misleading. It leads us to the following assumption:
Assumption: that every person in the group eats at least one dish.
This however is impossible given the premise. If we just do this quite simply, without all the probability stuff:
The main pertinent information given is that 5 people eat all three, so that leaves us with
21 -5 = 16 people
9 - 5 = 4 vegetables
10 - 5 = 5 fish
7 - 5 = 2 eggs
This leaves us with a total of 11 dishes to split up among 16 people... You can see the issue with our above assumption.
Therefore we have to discard that assumption. Now, we need to usethe knowledge that no more people ate 3 dishes and the new set of data:
P = 16; V = 4; F = 5; E = 2
Really at this point the amount of people is pointless as we can easily give away one dish per person and not run out of people so let's ignore the amount of people. This also gives us our minimum value. At least 5 people ate at least 2 dishes (the 5 that ate 3).
So, we now need to find the maximum value. how can we combine the veggies, fish and eggs left over to have the most possible who eat two dishes.
As we have an odd number of dishes, we know we will have one left over. Now, let's pair them off as easily as possible, let's say, everyone who had vegetables also had fish, and then the last person who had fish also had eggs. So in a list:
VF, VF, VF, VF, FE
leaving us with one eggs left over. This adds 5 dishes to our minimum of 5 and so our maximum is 10. Therefore our answer should be:
"Any amount of people between 5 and 10 had two or more out of the three dishes"
In conclusion:
- The problem statement is misleading
- The answer in the book is wrong
- This problem doesn't really belong to probability, it's more of a logic puzzle
answered 26 mins ago
Aglahir
113
New contributor
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The problem as supplied certainly is misleading. It leads us to the following assumption:
Assumption: that every person in the group eats at least one dish.
This however is impossible given the premise. If we just do this quite simply, without all the probability stuff:
The main pertinent information given is that 5 people eat all three, so that leaves us with
21 -5 = 16 people
9 - 5 = 4 vegetables
10 - 5 = 5 fish
7 - 5 = 2 eggs
This leaves us with a total of 11 dishes to split up among 16 people... You can see the issue with our above assumption.
Therefore we have to discard that assumption. Now, we need to usethe knowledge that no more people ate 3 dishes and the new set of data:
P = 16; V = 4; F = 5; E = 2
Really at this point the amount of people is pointless as we can easily give away one dish per person and not run out of people so let's ignore the amount of people. This also gives us our minimum value. At least 5 people ate at least 2 dishes (the 5 that ate 3).
So, we now need to find the maximum value. how can we combine the veggies, fish and eggs left over to have the most possible who eat two dishes.
As we have an odd number of dishes, we know we will have one left over. Now, let's pair them off as easily as possible, let's say, everyone who had vegetables also had fish, and then the last person who had fish also had eggs. So in a list:
VF, VF, VF, VF, FE
leaving us with one eggs left over. This adds 5 dishes to our minimum of 5 and so our maximum is 10. Therefore our answer should be:
"Any amount of people between 5 and 10 had two or more out of the three dishes"
In conclusion:
- The problem statement is misleading
- The answer in the book is wrong
- This problem doesn't really belong to probability, it's more of a logic puzzle
answered 26 mins ago
Aglahir
113
New contributor
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 26 mins ago
Aglahir
113
New contributor
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 26 mins ago
Aglahir
113
answered 26 mins ago
Aglahir
113
113
New contributor
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aglahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
0
down vote
I don't believe that the accepted answer is correct. As the OP pointed out in his question (although not at all clearly) adding up $|Acap B| + |Bcap C| + |Ccap A| $ counts anyone in $Acap B cap C$ three times, so we need to subtract $2cdot|Acap B cap C|$ from $10$ to get the number who eat at least two dishes. Unfortunately, this gives $0,$ contradicting the fact that $5$ people eat all three dishes.
The problem is insoluble as given.
answered 1 hour ago
saulspatz
12.3k21326
add a comment |Â
up vote
0
down vote
I don't believe that the accepted answer is correct. As the OP pointed out in his question (although not at all clearly) adding up $|Acap B| + |Bcap C| + |Ccap A| $ counts anyone in $Acap B cap C$ three times, so we need to subtract $2cdot|Acap B cap C|$ from $10$ to get the number who eat at least two dishes. Unfortunately, this gives $0,$ contradicting the fact that $5$ people eat all three dishes.
The problem is insoluble as given.
answered 1 hour ago
saulspatz
12.3k21326
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I don't believe that the accepted answer is correct. As the OP pointed out in his question (although not at all clearly) adding up $|Acap B| + |Bcap C| + |Ccap A| $ counts anyone in $Acap B cap C$ three times, so we need to subtract $2cdot|Acap B cap C|$ from $10$ to get the number who eat at least two dishes. Unfortunately, this gives $0,$ contradicting the fact that $5$ people eat all three dishes.
The problem is insoluble as given.
answered 1 hour ago
saulspatz
12.3k21326
I don't believe that the accepted answer is correct. As the OP pointed out in his question (although not at all clearly) adding up $|Acap B| + |Bcap C| + |Ccap A| $ counts anyone in $Acap B cap C$ three times, so we need to subtract $2cdot|Acap B cap C|$ from $10$ to get the number who eat at least two dishes. Unfortunately, this gives $0,$ contradicting the fact that $5$ people eat all three dishes.
The problem is insoluble as given.
answered 1 hour ago
saulspatz
12.3k21326
answered 1 hour ago
saulspatz
12.3k21326
answered 1 hour ago
saulspatz
12.3k21326
answered 1 hour ago
saulspatz
12.3k21326
12.3k21326
add a comment |Â
add a comment |Â
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