P(A), P(B), P(C); Range of Intersection of A, B and C

Question

I came across this question today in an exam held locally. Answers haven't been given out yet.

A, B and C are subsets of S. P(X) indicates the number of elements in X upon the the total number of elements in S. 
P(A) = .7
P(B) = .8
P(C) = .9

What is the range of P(A and B and C)?

Can someone help me out with this question please?

mikemcgarry · Answer

I'm happy to help with this. :-)

First of all, I'll say --- of course, this is not in GMAT form (either PS or DS), and it strikes me as not quite something the GMAT would ask. Nevertheless, I'll talk about how to answer it. To make this concrete, let's suppose the sample set consists of a hundred individuals, to whom I will refer as the digits 1-100, and A & B & C are some qualities or grouping that we ascribe to these individuals. Let's start with the maximum value of P(A and B and C) ---- Let' s say the individuals in A are members 1 - 70, which constitutes 0.7 of the population. If all 70 of those individuals are in B & C as well as A, then P(A and B and C) = 0.7. Furthermore, P(A and B and C) can't possibly be any higher than 0.7, because only 0.7 of the population is a member of A. If we tried to go any higher than 0.7, we would run out of A's. The max value of P(A and B and C) is 0.7 In general, when you have three groups, the maximum value of P(A and B and C) would be the case of maximum overlap, and would be equal to the minimum value of the set {P(A), P(B), P(C)}. The minimum value is a little trickier. We want minimum overlap. Let's push set A & B as far apart as we can. Well say Set A consists of folks at the beginning, members 1 - 70, and Set B consists of folks at the end, member 21-100. That means the overlap of P(A and B) consists of 50 individuals, from 21-70. Now, we will hand-place the membership of Set C, so that it has the absolute minimum possible overlap with P(A and B). We know there are 50 individuals in P(A and B), so that means there must be another 50 outside of it (the 1-20 folks and the 71-100 folks). Put the first 50 C folks there, so we use up all the "non A-and-B" space first. That leaves 40 more members of C that have to go somewhere inside the P(A and B) space, which means the absolute minimum for P(A and B and C) is 0.4. 0.4 <= P(A and B and C) <= 0.7 Does all this make sense? Mike :-)

m0h1t · Answer

Thanks a lot mike! I got .7 as the max but was struggling with the minimum.
It's also good to know that this isn't something the GMAT would ask.

PiyushK · Answer

P(A n B n C)/P(S)

Yes, 0.4 to 0.7.
P(A)/P(S) = 7/10
P(B)/P(S) = 8/10
P(C)/P(S) = 9/10

P(A) can give max overlap of 7.
7-(2+1) = 4 will give least possible overlap.

Attachment:

Solution2i.jpg [ 12.54 KiB | Viewed 7775 times ]

thus, 4/10 to 7/10 is the range of P(A n B n C)

P(A), P(B), P(C); Range of Intersection of A, B and C

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P(A), P(B), P(C); Range of Intersection of A, B and C

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