Mixing
Probability- 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 "atleast 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
asked 1 hour ago
Geeklovenerds
208
add a comment |Â
up vote
1
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 "atleast 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
asked 1 hour ago
Geeklovenerds
208
add a comment |Â
up vote
1
down vote
favorite
up vote
1
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 "atleast 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
asked 1 hour ago
Geeklovenerds
208
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 "atleast 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
probability
asked 1 hour ago
Geeklovenerds
208
asked 1 hour ago
Geeklovenerds
208
asked 1 hour ago
Geeklovenerds
208
asked 1 hour ago
Geeklovenerds
208
asked 1 hour ago
Geeklovenerds
208
208
add a comment |Â
add a comment |Â
3 Answers
3
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up vote
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Something is wrong with the data in this problem (assuming that there are no persons which do not eat vegetables, fish or eggs).
The number $N$ of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Ecap Fcap V|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|C|-|Fcup Vcup E|-|Ecap Fcap V|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Ecap Fcap V|=5$.
P.S. If there are persons that do not eat vegetables, fish or eggs then $|Fcup Vcup E|leq 21$ and the above equality implies $Ngeq 0$.
answered 28 mins ago
Robert Z
87.3k1056127
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
14 mins ago
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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 22 mins ago
Geeklovenerds
208
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 39 mins ago
saulspatz
12.3k21326
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
Something is wrong with the data in this problem (assuming that there are no persons which do not eat vegetables, fish or eggs).
The number $N$ of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Ecap Fcap V|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|C|-|Fcup Vcup E|-|Ecap Fcap V|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Ecap Fcap V|=5$.
P.S. If there are persons that do not eat vegetables, fish or eggs then $|Fcup Vcup E|leq 21$ and the above equality implies $Ngeq 0$.
answered 28 mins ago
Robert Z
87.3k1056127
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
14 mins ago
add a comment |Â
up vote
2
down vote
Something is wrong with the data in this problem (assuming that there are no persons which do not eat vegetables, fish or eggs).
The number $N$ of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Ecap Fcap V|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|C|-|Fcup Vcup E|-|Ecap Fcap V|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Ecap Fcap V|=5$.
P.S. If there are persons that do not eat vegetables, fish or eggs then $|Fcup Vcup E|leq 21$ and the above equality implies $Ngeq 0$.
answered 28 mins ago
Robert Z
87.3k1056127
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
14 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 there are no persons which do not eat vegetables, fish or eggs).
The number $N$ of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Ecap Fcap V|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|C|-|Fcup Vcup E|-|Ecap Fcap V|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Ecap Fcap V|=5$.
P.S. If there are persons that do not eat vegetables, fish or eggs then $|Fcup Vcup E|leq 21$ and the above equality implies $Ngeq 0$.
answered 28 mins ago
Robert Z
87.3k1056127
Something is wrong with the data in this problem (assuming that there are no persons which do not eat vegetables, fish or eggs).
The number $N$ of persons which eat at least two out of the three dishes is
$$N:=|Fcap V|+|Vcap E|+|Ecap F|-2|Ecap Fcap V|$$
By the inclusion-exclusion principle
$$N=|F|+|V|+|C|-|Fcup Vcup E|-|Ecap Fcap V|\=10+9+7-21-5=0.$$
This can't be because $Ngeq |Ecap Fcap V|=5$.
P.S. If there are persons that do not eat vegetables, fish or eggs then $|Fcup Vcup E|leq 21$ and the above equality implies $Ngeq 0$.
answered 28 mins ago
Robert Z
87.3k1056127
edited 27 secs ago
answered 28 mins ago
Robert Z
87.3k1056127
answered 28 mins ago
Robert Z
87.3k1056127
answered 28 mins ago
Robert Z
87.3k1056127
87.3k1056127
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
14 mins ago
add a comment |Â
I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question.
– drhab
14 mins 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
14 mins 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
14 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 22 mins ago
Geeklovenerds
208
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 22 mins ago
Geeklovenerds
208
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 22 mins ago
Geeklovenerds
208
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 22 mins ago
Geeklovenerds
208
answered 22 mins ago
Geeklovenerds
208
answered 22 mins ago
Geeklovenerds
208
answered 22 mins ago
Geeklovenerds
208
208
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 39 mins 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 39 mins 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 39 mins 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 39 mins ago
saulspatz
12.3k21326
answered 39 mins ago
saulspatz
12.3k21326
answered 39 mins ago
saulspatz
12.3k21326
answered 39 mins ago
saulspatz
12.3k21326
12.3k21326
add a comment |Â
add a comment |Â
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