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bluemartini

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In a recent consumer survey, 85% of those surveyed liked at least one of three products: 1, 2 and 3. 50% of those liked product 1. 30% liked 2, and 20% liked 3. if 5 % of the people in the survey liked all three products, what percentage of the survey participants liked more than one of the three products?OA seems to be 10, but I am getting something else.
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Here is the correct formula for set theory

 

n(A U B U C) = n(A) + n(B) + n© - n(A ^ B) - n(A ^ C) - n(B ^ C) + n(A ^ B ^C)

 

I am using the caret symbol to mean intersection.

 

In this case no pf people who liked at least one i.e. 1 or 2 or 3 is 85%

Number of people who liked 1 is 50

Number of people who liked 2 is 30

Number of people who liked 3 is 20

Number of people who like all 3 is 5

 

Substitute in to the formula and you get no of people who like A and B or A and C or B and C = 20

i.e. 85 = 50 + 30 + 20 - say X + 5

X = 25

 

Now 25 is the no of people who like at least 2 brands i.e. 2 or more and the 5 people who like all three brands are included in these 25 people 3 times so no of people who like more than one of the brands is 25 - 3(5) = 25 -15 = 10 %

 

To elaborate further on why I multiplied 5, 3 times suppose 15 people like brand 1 and 2 and 5 people like all three these 5 people who like all three are also a part of the group that likes 1 and 2. => If you like 1, 2 and 3 it obviously means you like 1 and 2.

 

HTH,

--smitha.

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Smith i followed your explaination to a point when u substitute the formula 85=50 N(a) + 30 n(b) + 20 n© - x (A^b etc.) + 5 ( a^b^c). i get x to be 20 and not 25 like y got. can u just point out how X ix equal to 25 and not 20. by thw way your explaination is very good. i wished i used this forum before my first painful shot
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After checking out Scoot's link (and following the links from there), I have found the following general formulas for three-component set problems:

 

If there are three sets A, B, and C, then

P(AuBuC) = P(A) + P(B) + P© – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

 

 

Number of people in exactly one set =

P(A) + P(B) + P© – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC)

 

 

Number of people in exactly two of the sets =

P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

 

 

Number of people in exactly three of the sets =

P(AnBnC)

 

 

Number of people in two or more sets =

P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)

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  • 4 weeks later...

To add to TwinSplitter's forumla list:

 

No of people in atleast 1 set =

P(A) + P(B) + P© - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

 

This formula can also be derived from:

No of people in atleast 1 set =

no of people in exactly 1 set + no of people in exactly 2 set + no of people in exactly 3 set

 

 

 

~~~~~~~~

More you look, more you see!

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  • 5 months later...
To add to TwinSplitter's forumla list:

 

No of people in atleast 1 set =

P(A) + P(B) + P© - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

 

This formula can also be derived from:

No of people in atleast 1 set =

no of people in exactly 1 set + no of people in exactly 2 set + no of people in exactly 3 set

 

 

Shouldnt this be

 

P(A) + P(B) + P© - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

When I add no of people in exactly 1 set with no of people in exactly 2 set I see that 3P(AnBnC) gets cancelled and then I add P(AnBnc) for exactly 3 set.

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