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The New Marketing Journal conducted a survey of wealthy

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The New Marketing Journal conducted a survey of wealthy [#permalink] New post 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] New post 14 Aug 2010, 13:10
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 :shock:
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Re: Overlapping Sets [#permalink] New post 14 Aug 2010, 15:16
Not sure if wording of my reply is correct.

Total= Set1+Set2-Both+Neither( in our case it's 0)

X=45+38+27-(8+15+12+5)=70
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Re: Overlapping Sets [#permalink] New post 14 Aug 2010, 15:44
vasili wrote:
Not sure if wording of my reply is correct.

Total= Set1+Set2-Both+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] New post 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] New post 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] New post 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] New post 16 Aug 2010, 10:53
Image

I'm getting 65 with this diagram...

6=38-15-12-5
2=27-12-8-5
17=45-8-5-15

17+8+5+15+2+12+6=65

What is wrong with it?
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Re: Overlapping Sets [#permalink] New post 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] New post 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+P-BM-MP-BP+BMP=45+38+27-15-12-8+8=80

Answer is (c)
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Re: Overlapping Sets [#permalink] New post 10 Oct 2010, 22:45
C

start from the end of the q and work up...

all 3 - 5
B+M = 15-5
M+P = 12-5
B+P = 8-5
B = 45-10-5-3 = 27
M = 38-10-5-7 = 16
P = 27-3-5-7 = 12

5 + 10 + 7 + 3 + 27 + 16 + 12 = 80
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Re: Overlapping Sets [#permalink] New post 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+y-xy = 20+15-5 = 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.

15-5=10
12-5=7
08-5=3

At this point you can find the number of unique drivers of each model.

BMW=45-10-5-3=27
Mercedes=38-10-5-7=16
Porsche=12-7-5-3=12

27+16+12+10+3+7+5=80

Alternatively, we can use a formula.

45+38+27-15-8-7+5=80

Another way to look at it is to to subtract the overlap from the overlapping sets:

45+38+27 - (15+8+12-5) = 80

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Re: Overlapping Sets [#permalink] New post 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] New post 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] New post 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] New post 12 Oct 2010, 02:36
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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


Two formulas of 3 overlapping sets: formulae-for-3-overlapping-sets-69014.html#p729340

Hope it helps.
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Re: Overlapping Sets [#permalink] New post 12 Dec 2010, 19:17
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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


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:
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Ques1.jpg [ 17.34 KiB | Viewed 1959 times ]

16+10+5+7+12+3+27 = 80
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Re: Overlapping Sets [#permalink] New post 14 Dec 2010, 21:54
Use Venn Diagram
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Re: Overlapping Sets [#permalink] New post 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] New post 10 Apr 2012, 12:30
Bunuel wrote:
Two formulas of 3 overlapping sets: formulae-for-3-overlapping-sets-69014.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.
Re: Overlapping Sets   [#permalink] 10 Apr 2012, 12:30
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