ExecMBA2010 wrote:
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?
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.htmlOf 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.
Attachments
Union_3sets.gif [ 11.63 KiB | Viewed 14964 times ]