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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.

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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 $$N:=|F\cap V|+|V\cap E|+|E\cap F|-2|F\cap V\cap E|$$ By the inclusion-exclusion principle $$N=|F|+|V|+|E|-|F\cup V\cup E|-|F\cap V\cap E|\\=10+9+7-21-5=0.$$ This can't be because $N\geq |F\cap V\cap E|=5$.

On the other hand, if there are persons that do not eat vegetables, fish or eggs then $$10=\max(|F|,|V|,|E|)\leq |F\cup V\cup E|\leq 21$$ and the above equality implies $$5=|F\cap V\cap E|\leq N=21-|F\cup V\cup E|\leq 11.$$

P.S. By Sander De Dycker's comments, $|F\cup V\cup E|\not=10$, therefore $|F\cup V\cup E|\geq 11$ and $$5\leq N\le 10.$$

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    $\begingroup$ I suspect there are persons here that do not eat vegetables, fish or eggs. This puts us on "the wrong leg". Deceitful question. $\endgroup$ – drhab Oct 23 '18 at 13:14
  • $\begingroup$ can you please elaborate your last line I am not getting it. $\endgroup$ – Geeklovenerds Oct 23 '18 at 13:31
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    $\begingroup$ Is it clear now? $\endgroup$ – Robert Z Oct 23 '18 at 13:34
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    $\begingroup$ max(|F|,|V|,|E|) would be a good lower bound for |F ∪ V ∪ E| if we weren't given the extra information that |F ∩ V ∩ E| = 5 . That raises the lower bound to 11, since there's no way to have only 10 persons eat at least one dish, when there are just 5 that eat 3 dishes. That would mean that the remaining 5 people need to eat 26-(3*5) dishes, which means at least one of them needs to eat more than 2. $\endgroup$ – Sander De Dycker Oct 23 '18 at 14:35
  • $\begingroup$ @SanderDeDycker You are right, thanks for pointing out. $\endgroup$ – Robert Z Oct 23 '18 at 14:48
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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 use the 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 number 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

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  • $\begingroup$ (1) IMHO, you wasted a lot of space making and then disproving an assumption that was unwarranted in the first place. (For all we know, 21 co-workers got together in a tavern after work. Some drank but didn’t eat; maybe for religious reasons, maybe because they had a big dinner waiting for them at home.) (2) So I disagree with your assertion that the problem statement is misleading. (3) Based on my quick scan through of all the answers, yours is the first one to get the correct result.  And so, you’re right: the answer in the book is wrong.  And this is not about probability. $\endgroup$ – Scott Oct 23 '18 at 18:28
  • $\begingroup$ P.S. But, to be pedantic, you should say ‘‘Any number of people between 5 and 10 had two or more of the three dishes.’’  Also, you have a typo: “we need to usethe knowledge”. $\endgroup$ – Scott Oct 23 '18 at 18:28
  • $\begingroup$ thank for simplifying the answer given by gnasher $\endgroup$ – Geeklovenerds Oct 24 '18 at 5:52
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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.

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  • $\begingroup$ how there are total 26 dishes. $\endgroup$ – Geeklovenerds Oct 23 '18 at 13:57
  • $\begingroup$ @Geeklovenerds: 9+10+7 dishes ordered totals $26$. This is the best answer. $\endgroup$ – Ross Millikan Oct 23 '18 at 14:02
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    $\begingroup$ 5 times 3, plus 6 times 2 is also > 26, so the maximum can't be 11 ... the maximum is 10. $\endgroup$ – Sander De Dycker Oct 23 '18 at 14:06
  • $\begingroup$ (1) Well, it’s somewhat of a given that nobody eats 5 dishes.  And nobody eats 4 dishes, or 42.  I believe that you meant to say “nobody eats three dishes”. (2) You say “We can extract the 5 people who eat all three dishes from the problem.  Now we have 16 people left, …. We see that at most 5 of these eat two dishes, …. / Add the 5 eating three dishes back in” (i.e., adding the 5 people who eat all three dishes to the at most 5 (of the 16) who eat two dishes), “and you get that … at most 11 eat two or more dishes.” … (Cont’d) $\endgroup$ – Scott Oct 23 '18 at 18:47
  • $\begingroup$ (Cont’d) …  I would say that @SanderDeDycker is right, except you seem to have made a simple arithmetic mistake (5+5=11) rather than a logic error. $\endgroup$ – Scott Oct 23 '18 at 18:47
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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

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  • $\begingroup$ 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 $\endgroup$ – Sander De Dycker Oct 23 '18 at 14:16
  • $\begingroup$ i have assumed "Let Y be the no. of persons who eat at least one item." in the solution above. So, according to this MIN. VALUE OF Y= 10 cover all possibilities. $\endgroup$ – Geeklovenerds Oct 23 '18 at 14:26
  • $\begingroup$ if you don't see how it's impossible to have Y = 10, may I suggest you draw a Venn diagram, and then try adding 10 people into it to fit the known values of 9, 10, 7 and 5 from the problem statement ? You'll find you need at least 11 people to achieve that. $\endgroup$ – Sander De Dycker Oct 23 '18 at 14:45
  • $\begingroup$ I think the no. of persons eating at most 2 items =11 $\endgroup$ – Geeklovenerds Oct 23 '18 at 14:49

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