Question 1091470: According to a survey, 16 people liked ham, 19 liked eggs, 18 liked mackerel, 8 liked mackerel and eggs, 5 liked ham and eggs, 7 liked ham and mackerel, and 3 liked all three foods. (is it possible to see a Venn diagram of this?)
- how many liked only mackerel?
- how many liked at least two of three foods?
- how many liked only one of three foods?
- how many liked eggs and mackerel but not ham?
- how many people were surveyed?
I'm trying to read up on this more, but as my username suggests, I'm still new and having a bit of a hard time. Thanks in advance!
Answer by ikleyn(52818)   (Show Source):
You can put this solution on YOUR website! .
According to a survey, 16 people liked ham, 19 liked eggs, 18 liked mackerel,
8 liked mackerel and eggs,
5 liked ham and eggs,
7 liked ham and mackerel,
and 3 liked all three foods.
1) how many liked only mackerel? 18 - 8 - 7 + 3.
2) how many liked at least two of three foods? 8 + 5 + 7 - 2*3.
3) how many liked only one of three foods? 16 + 19 + 18 - 8 - 5 - 7 + 2*3.
4) how many liked eggs and mackerel but not ham? 19 + 18 - 8 + 3.
5) how many people were surveyed? 16 + 19 + 18 - 8 - 5 - 7 + 3.
First, I put the numbers only in hope that they will explain everything to you without my words.
But then I put some minimal explanations to worm your mind.
1) how many liked only mackerel ?
Take those who like mackerel; subtract those who like ME; subtract those who like MH; add MHE what were subtracted twice.
2) how many liked at least two of three foods ?
ME + HE + HM - 2*HEM = 8 + 5 + 7 - 2*3.
(when we add ME + HE + HM, we count the triple intersection thrice; therefore, I subtract HEM twice . . . )
The same or similar logic works in other cases . . .
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To get familiar with the subject, look into my lessons
- Counting elements in sub-sets of a given finite set
- Advanced problems on counting elements in sub-sets of a given finite set
in this site.
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