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safirshahbaaz90
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5 lb. Probability, Combinatorics and Sets Question

by safirshahbaaz90 Sun Sep 07, 2014 12:40 am

Hi Tommy,

In the 5 lb. Probability, Combinatorics, and Overlapping sets section, I have a question regarding problem 26. Could you please explain why we have to multiply the (all three) by 2? I understand that we subtract them for since we would have already counted them in the single subject parts but I did not understand the reason for the multiplication. Also, is there a general formula for computing this?

Thanks and Regards,
Safir

P.S.: Explanation for question 19 in the same section seems to have a mistake. It has 12*11=121 instead of 132. Please correct me if I might have missed something.
tommywallach
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Re: 5 lb. Probability, Combinatorics and Sets Question

by tommywallach Mon Sep 08, 2014 8:41 pm

Hey Safir,

Because if someone is in all three groups, they have actually been counted 3 separate times. You only want each person to be counted twice, so you have to remove two of those three "counts."

As for equations, you can use:

Total = Group 1 + Group 2 + Both - Neither

That works for two group overlaps. For three groups, there is an equation, but it's quite complicated (I should add that I've never seen a three group question on the GRE).

Total = Group 1 + Group 2 + Group 3 - (# of people in ONLY two groups) - 2 * (# of people in ALL three groups) + None

-t
safirshahbaaz90
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Re: 5 lb. Probability, Combinatorics and Sets Question

by safirshahbaaz90 Wed Sep 10, 2014 11:24 am

Thanks Tommy!
tommywallach
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Re: 5 lb. Probability, Combinatorics and Sets Question

by tommywallach Fri Sep 12, 2014 5:50 pm

Glad to help!

-t