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Sets Question

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Intern
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Sets Question [#permalink]

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New post 04 Sep 2008, 21:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone explain to me what the difference between the below formulae is?

I actually don't understand how the first formula could be correct: If you've subtracted AnB, AnC, BnC, you've taken out AnBnC three times. So you need to add it back in twice, right?

For 3 sets A, B, and C: P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

Any ideas?

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Re: Sets Question [#permalink]

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New post 05 Sep 2008, 00:17
I am not sure I understant your expressions, so I'll go with mine.

We have 3 sets: A, B, and C. These sets overlap, so that there are three subsets of the type AB and one subset ABC (say, students who study all three languages). The total number (of people), elimitnating overlaps, looks like this:

A + B + C -AB -AC -BC -2*ABC
or:
Aonly + Bonly + Conly +AB + AC + BC + ABC

Math wizards, please correct me if the above is wrong.
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Re: Sets Question [#permalink]

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New post 05 Sep 2008, 04:47
P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC) is correct.

You can understand it by drawing 3 circles intersecting each other.

when you add P(A) + P(B) + P(C) , you have included each of P(AnB) , P(AnC) and P(BnC) twice.

for example P(AnB) is a part of P(A) as well as P(B) ; so you deduct once.

Now when you deduct -P(AnB) – P(AnC) – P(BnC) , you are deducting P(AnBnC) three times and also P(A) + P(B) + P(C) has added P(AnBnC) 3 times. So you need to add 1 P(AnBnC) back to the expression.

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Re: Sets Question   [#permalink] 05 Sep 2008, 04:47
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