Intersecting sets formula - PL HELP

Question

I am confused with the 2 available formulas for 3 intersecting sets:

One way to deduce is this,
For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)

There is another formula available too:
For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) -2P(A n B n C)

Which one is correct pls?

firasath · Answer

First of all P(A u B u C) is the UNION of 3 sets not the intersection. The Union refers to the total of the 3 sets without any overlaps. (In the venn diagram, this is everything inside the blue line without overlaps). The intersection means an overlap of the 3 sets. This is section 4 in the venn diagram. the UNION of 3 sets A, B, C is: P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C) Your second formula only applies in a specific case: You probably got it from here: https://www.gmathacks.com/gmat-math/thre ... -sets.html Of the shoe stores in City X, 30 carry Brand A shoes, 40 carry Brand B shoes, and 25 carry Brand C shoes. If each store carries at least one of the brands, 32 of the stores carry two of the three shoe brands, and none of the stores carry all three, how many shoe stores are there in City X? Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither Your 2nd equation could be used for this example because from the text of the stimulus, the 2-group overlaps do not include 3-group overlaps. We know this because there are no groups(read stores) that have 3-group overlaps so this has to mean that the two group overlaps ONLY count stores where there are two group overlaps. Therefore the 2nd equation is valid here. Visualize the Venn diagram. P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C), the 2-group sets (1,2 & 3 in the diagram) each include the 3-group set (section 4 in the diagram) and we are subtracting each of them, therefore we don't account for any 3-group set, thus we need to add the 3-group set and we get: P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C) now if the 2-group sets (1, 2 & 3 in the venn diagram) did NOT include any 3-group sets as in the example problem and since we are subtracting each of the 2-group sets, this means that the 3-group sets have been counted 3 times so we need to subtract 2 of them and we get: P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) -2P(A n B n C) The "neither" (read none) part is only used for calculating totals. In a union problem, we don't include it since it is not in any of the sets, therefore it cannot be part of the union. Hope this helps, it's a little confusing but if you try to visualize it on the Venn diagram it should be clear.

ExecMBA2010 · Answer

Thanks for such an elaborate explanation and the diagram. And yes, you have guesseed it right about the source of the 2nd formula.

Now, help me to clear another confusion in the below formula:

P(AUBUC) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)

What we are actually trying to do is to add the number of members in each of the sets A, B & C. In case of overlapping sets this would be the sum of each set less the numbers for each overlapping areas. For this reason we add P(A), P(B) and P(C) and then 
subtract P(A n B), P(A n C), P(B n C) which is actually area 1,2 & 3 of your diagram. Understood till this part.

Now while subtracting shaded areas 1,2 & 3 we also have deducted the shaded area 4 but thrice. So, in order to balance that we have to add P(A n B n C), which is actually the area 4 but why only once? Should it not be added twice?

firasath · Answer

No, it should be added only once.
In each of the formulas and in any union, you want the members of all of the sets in the union WITHOUT any overlap.

So, in formula one, section 4 has been accounted for 3 times ( from P(A) + P(B) + P(C)) and then removed 3 times (by the subtraction of the 2-group sections). This essentially leaves a hole in the middle of the venn diagram. That is why you need to add it back to the equation so that it is counted once.

In the example problem (second formula), section 4 has been accounted for 3 times ( from P(A) + P(B) + P(C), and not subtracted via the 2-group sections --because in this problem the 2-group sections do not include section 4) so you need to subtract it twice so that it is counted once.

p.s. if you like my explanation, I would appreciate a kudos

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Intersecting sets formula - PL HELP

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