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The New Marketing Journal conducted a survey of wealthy [#permalink]
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14 Aug 2010, 12:59
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The New Marketing Journal conducted a survey of wealthy German car owners. According to the survey, all wealthy car owners owned one or more of the following three brands: BMW, Mercedes, or Porsche. Respondents' answers were grouped as follows: 45 owned BMW cars 38 owned Mercedes cars, and 27 owned Porsche cars. Of these, 15 owned both BMW and Mercedes cars, 12 owned both Mercedes and Porsche cars, 8 owned both BMW and Porsche cars, and 5 persons owned all three types of cars. How many different individuals were surveyed? A. 70 B. 75 C. 80 D. 110 E. 130
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Re: Overlapping Sets [#permalink]
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14 Aug 2010, 13:10
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zmaster85 wrote: The New Marketing Journal conducted a survey of wealthy German car owners. According to the survey, all wealthy car owners owned one or more of the following three brands: BMW, Mercedes, or Porsche. Respondents' answers were grouped as follows: 45 owned BMW cars 38 owned Mercedes cars, and 27 owned Porsche cars. Of these, 15 owned both BMW and Mercedes cars, 12 owned both Mercedes and Porsche cars, 8 owned both BMW and Porsche cars, and 5 persons owned all three types of cars. How many different individuals were surveyed?
A) 70 B) 75 C) 80 D) 110 E) 130 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) = 48 + 38+ 27  (15 + 12 + 8) + 5 = 80
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Re: Overlapping Sets [#permalink]
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14 Aug 2010, 15:16
Not sure if wording of my reply is correct.
Total= Set1+Set2Both+Neither( in our case it's 0)
X=45+38+27(8+15+12+5)=70



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Re: Overlapping Sets [#permalink]
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14 Aug 2010, 15:44
vasili wrote: Not sure if wording of my reply is correct.
Total= Set1+Set2Both+Neither( in our case it's 0)
X=45+38+27(8+15+12+5)=70 I think your formula (Bunuel has this in one of his other posts) is for the case when "Both" applies to the population such that they only own those 2 cars, that does not seem to be the case here. So we shoudl go with the usual formula above and get 80. I don't see 70. X = A+B+C(only2)2AandBandC  I think this is the formula you were attempting...
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Re: Overlapping Sets [#permalink]
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14 Aug 2010, 20:41
I got 65 which is not one of the answers.
I believe the formula is Total = A + B + C  (A&B)  (B&C)  (A&C)  2(A&B&C) Total = 45 + 38 + 27  15  12  8  2(5) Total = 65
What am I doing wrong?



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Re: Overlapping Sets [#permalink]
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14 Aug 2010, 21:10
swdatta wrote: I got 65 which is not one of the answers.
I believe the formula is Total = A + B + C  (A&B)  (B&C)  (A&C)  2(A&B&C) Total = 45 + 38 + 27  15  12  8  2(5) Total = 65
What am I doing wrong? A&B.. etc. in the formula is for A&B only... i think the first post answer is correct
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Re: Overlapping Sets [#permalink]
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14 Aug 2010, 21:18
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)
= 48 + 38+ 27  (15 + 12 + 8) + 5 = 80



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Re: Overlapping Sets [#permalink]
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16 Aug 2010, 10:53
I'm getting 65 with this diagram... 6=3815125 2=271285 17=458515 17+8+5+15+2+12+6=65 What is wrong with it?



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Re: Overlapping Sets [#permalink]
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19 Aug 2010, 12:33
I agree with swdatta and Financier but I don't understand why it's wrong... You should deduct the people who own all three types of cars twice otherwise you would count them several times.
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Re: Overlapping Sets [#permalink]
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07 Oct 2010, 11:28
zmaster85 wrote: The New Marketing Journal conducted a survey of wealthy German car owners. According to the survey, all wealthy car owners owned one or more of the following three brands: BMW, Mercedes, or Porsche. Respondents' answers were grouped as follows: 45 owned BMW cars 38 owned Mercedes cars, and 27 owned Porsche cars. Of these, 15 owned both BMW and Mercedes cars, 12 owned both Mercedes and Porsche cars, 8 owned both BMW and Porsche cars, and 5 persons owned all three types of cars. How many different individuals were surveyed?
A) 70 B) 75 C) 80 D) 110 E) 130 Let the set of owners be B,M,P B=45 M=38 P=27 \(B \cap M=15\) \(P \cap M=12\) \(B \cap P=8\) \(B \cap M \cap P=5\) Q : How many were questioned ? Or in other words size of the universe ? Universe = B+M+PBMMPBP+BMP=45+38+2715128+8=80 Answer is (c)
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Re: Overlapping Sets [#permalink]
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10 Oct 2010, 22:45
C start from the end of the q and work up... all 3  5 B+M = 155 M+P = 125 B+P = 85 B = 451053 = 27 M = 381057 = 16 P = 27357 = 12 5 + 10 + 7 + 3 + 27 + 16 + 12 = 80
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Re: Overlapping Sets [#permalink]
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11 Oct 2010, 06:52
To help clarify some of the confusion:
Conisder a situation where there are only two things in an overlapping set (X and Y). There are 20 people in X; 15 people in Y; 5 people in X and Y. In this case, the size of the population would be x+yxy = 20+155 = 30.
Apply this same principle to the overlapping sets of the three cars:
15 people own BMW and Mercedes. 5 people own a BMW, Mercedes and Porsche. However, these two data points are counting owners of BMW and Mercedes twice. The same applies for the 12 owners of a Mercedes and Porsche and the 8 owners of BMW and Porsche.
If you solved this using a Venn Diagram, you must first account for the overlap in the overlapping sets.
155=10 125=7 085=3
At this point you can find the number of unique drivers of each model.
BMW=451053=27 Mercedes=381057=16 Porsche=12753=12
27+16+12+10+3+7+5=80
Alternatively, we can use a formula.
45+38+271587+5=80
Another way to look at it is to to subtract the overlap from the overlapping sets:
45+38+27  (15+8+125) = 80
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Re: Overlapping Sets [#permalink]
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12 Oct 2010, 01:54
i believe formula should be : P(A) + P(B) + P(C)  P(AnB)  P(AnC)  P(BnC) + 2 P(AnBnC), using this the ans would be 65, but 65 is no where in the option, but using P(A) + P(B) + P(C)  P(AnB)  P(AnC)  P(BnC) + P(AnBnC) we get 80. which is one of the options(ans seems to be the correct option). which formula is correct. Any one? i cant figure out what am i doing wrong here



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Re: Overlapping Sets [#permalink]
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12 Oct 2010, 01:56
prab wrote: i believe formula should be : P(A) + P(B) + P(C)  P(AnB)  P(AnC)  P(BnC) + 2 P(AnBnC), using this the ans would be 65, but 65 is no where in the option, but using P(A) + P(B) + P(C)  P(AnB)  P(AnC)  P(BnC) + P(AnBnC) we get 80. which is one of the options(ans seems to be the correct option). which formula is correct. Any one? i cant figure out what am i doing wrong here P(A) + P(B) + P(C)  P(AnB)  P(AnC)  P(BnC) + P(AnBnC) this is the correct formula
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Re: Overlapping Sets [#permalink]
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12 Oct 2010, 02:01
the formula to be used is P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC). Thanx



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Re: Overlapping Sets [#permalink]
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Re: Overlapping Sets [#permalink]
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12 Dec 2010, 19:17
zmaster85 wrote: The New Marketing Journal conducted a survey of wealthy German car owners. According to the survey, all wealthy car owners owned one or more of the following three brands: BMW, Mercedes, or Porsche. Respondents' answers were grouped as follows: 45 owned BMW cars 38 owned Mercedes cars, and 27 owned Porsche cars. Of these, 15 owned both BMW and Mercedes cars, 12 owned both Mercedes and Porsche cars, 8 owned both BMW and Porsche cars, and 5 persons owned all three types of cars. How many different individuals were surveyed?
A) 70 B) 75 C) 80 D) 110 E) 130 It is a straight forward question that can be solved using the formula discussed above but if you forget it, you can use a Venn diagram. Start with the region where all 3 sets overlap. That is 5. Next work on each of the three regions where 2 sets overlap. Next work on the 3 regions where people own a single car. Add the number of people in all the regions and you get the total number of people. Attachment:
Ques1.jpg [ 17.34 KiB  Viewed 4036 times ]
16+10+5+7+12+3+27 = 80
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Re: Overlapping Sets [#permalink]
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14 Dec 2010, 21:54
Use Venn Diagram
Attachments
File comment: Venn
Venn.PNG [ 16.13 KiB  Viewed 3304 times ]



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Re: Overlapping Sets [#permalink]
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12 Apr 2011, 18:48
Venn Diagram is very helpful, and I performed some long calculattions to illustrate the point : 45 owned BMW cars 38 owned Mercedes cars 27 owned Porsche cars 15 owned both BMW and Mercedes cars 12 owned both Mercedes and Porsche cars 8 owned both BMW and Porsche cars 5 persons owned all three types of cars BMW OWners = Only owned BMW cars + owned both BMW and Mercedes cars + owned both BMW and Porsche cars  owned all three types of cars Mercedes Owners = Only owned Mercedes cars + owned both BMW and Mercedes cars + owned both Mercedes and Porsche cars  owned all three types of cars Porsche Owners = Only owned Porsche cars + owned both BMW and Porsche cars + owned both Mercedes and Porsche cars  owned all three types of cars 45 = Only owned BMW cars + 15 + 8  5 = > Only owned BMW cars = 45  18 = 27 38 = Only owned Mercedes cars + 15 + 12  5 => Only owned Mercedes cars = 38  22 = 16 27 = Only owned Porsche cars + 8 + 12  5 => Only owned Porsche cars = 27  15 = 12 Total = Only owned BMW cars + Only owned Mercedes cars + Only owned Porsche cars + Only owned both BMW and Mercedes cars + Only owned both Mercedes and Porsche cars + ONly owned both BMW and Porsche cars + owned all three types of cars => Total = 27 + 16 + 12 + (15  5) + (12  5) + (8  5) + 5 = 43 + 12 + 10 + 7 + 3 + 5 = 43 + 22 + 15 = 80
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Re: Overlapping Sets [#permalink]
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10 Apr 2012, 12:30
Bunuel wrote: Two formulas of 3 overlapping sets: formulaefor3overlappingsets69014.html#p729340 Bunuel's link explains it all. I too did the Venn Diagram and came up with 65. However, I overlooked a fact; with 3 groups, the intersection of 2 groups includes the intersection of 3 groups. So everytime you subtract MnB, MnP, BnP, you also subtract MnBnP. In the end, you subtract MnBnP one too many times, therefore, must add it back in once.




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