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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?
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## 2 Answers

+7 votes
Best answer
Answer: 10

P(AuBuC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

Let $Y$ be the no. of persons who eat at least one item.

Y = 9 + 10 + 7 - [P(AnB) + P(AnC) + P(BnC)] + 5

[P(AnB) + P(AnC) + P(BnC)] = 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 as

31 - Y - 2*5 = 21 - Y.

Minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.

Maximum value of Y is 21. Is this possible? No. Because 5 person eat all three item. 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 $21-10 \geq X \geq 21 - 16 \implies 5 \leq X \leq 11.$
answered by Veteran (35k points)
edited by
atleast 2 dishes means  2dishes and all 3 dishes   should be (10+5)????

"at least two out of the three " both  means 2 and 3

red shaded gives both 2 and 3

so no need of again finding for 3

so ans should be only 10

@Arjun Sir ,I think the data in this question is wrong...

Let A$\cap$B$\cap$C = x

A$\cap$B+B$\cap$C+A$\cap$C contains x 3 times...

And atleast 2 means ( A$\cap$B+B$\cap$C+A$\cap$C ) - 2x

which evaluates to 0 ...which is contradict to the given data i.e.5 persons eat all three items.

@Rajesh Pradhan
I got the same doubt.
Let A∩B∩C = x
then ( A∩B+B∩C+A∩C ), this already contains $3$x. therefore subtracting $2$x from this should result into POSITIVE value, but it is zero.
Moreover, u r right. They are asking for "atleast 2" which means ( A∩B+B∩C+A∩C ) - 2x.

Something wrong with given Data ?

@Arjun sir, plz see this.

Sir, i can't understand how this is possible?

Minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
Maximum value of Y is 21. Is this possible? No. Because 5 person eat all three item.
–3 votes
15 (10+5)
answered by Junior (727 points)

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