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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?

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)

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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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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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 ?

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.

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

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

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

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.

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

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.

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