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Foodmart customers regularly buy at least one of the [#permalink]
27 Jan 2008, 22:20

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Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase 2 of the above products? 5% 10% 15% 25% 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg. Thus, substituting, we obtain that y=1/4 x

Re: Sets with >=3 elements [#permalink]
05 Feb 2008, 11:11

The explanation on the practice test to the problem goes...

Let bi denote the percentage of buyers who regularly buy i products, and bx the percentage of buyers who regularly purchase product x.

We can construct these equations: b1 + 2b2 + 3b3 = bm + bc + ba = 60% + 50% + 35% = 145% b1 + b2 + b3 = 100% b3 = 10%

Subtract the second one from the first one: b2 + 2b3 = 45% b2 = 25%

How does one explain the multipliers for terms in b1 + 2b2 + 3b3 above? Does that mean that if there are 4 products that the equation would read b1 + 2b2 + 3b3 + 4b4?

Re: Sets with >=3 elements [#permalink]
05 Feb 2008, 11:24

marcodonzelli wrote:

Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase 2 of the above products? 5% 10% 15% 25% 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg. Thus, substituting, we obtain that y=1/4 x

P1 - one product, P2 - two poducts, P3 - three products

Re: Sets with >=3 elements [#permalink]
05 Feb 2008, 18:51

marcodonzelli wrote:

Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase 2 of the above products? 5% 10% 15% 25% 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg. Thus, substituting, we obtain that y=1/4 x

Re: Sets with >=3 elements [#permalink]
11 Oct 2011, 14:12

maratikus wrote:

marcodonzelli wrote:

Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase 2 of the above products? 5% 10% 15% 25% 30%

What is the right formula to apply? I used : x=3/5 x + ½ x +7/20 x - y (people with 2 purchases) – 2z (people with 3 purchases), considering that everybody bought smtg. Thus, substituting, we obtain that y=1/4 x

P1 - one product, P2 - two poducts, P3 - three products

Re: Sets with >=3 elements [#permalink]
11 Oct 2011, 20:18

i'm also a bit confused by everyone else's work, this is the way I solved it

the equation is

P(A or B or C ) = P (A) + P (B) + P (C) - P(A & B) - P(A & C) - P(B & C) + P (A & B & C) also, the probability of A or B or C is 100% since A,B,C are the only sets in this space.

when you add up P(A), P(B), P(C), you are also over adding in case they have intersections of 2 or 3 *draw a venn diagram with 3 intersection circles and it can help make sense . Therefore you subtract out the places where two of them overlap. However, as you were subtracting the 2 set overlaps, each time you're subtracting any 3 set overlap as well. So you end up subtracting 3x for the 3-set overlap there, but recall in the beginning you added 3x for the 3-set overlap for A,B,C, so net is zero for that 3 set overlap. You can't ignore the 3-set overlap so you add it back in as the last component.

Anyway, the problem goes

60+50+35 - (2 set overlaps) + (3 set overlap) = 100 145 - (2 set overlaps) + 10 = 100 155 - (2 set overlaps) = 100 (2 set overlaps) = 55

That means that 2 set overlaps add up to 55%. Meaning P(A&B) + P(A&C) + P(C&B) = 55

What's important here is you're back to the same problem of over adding because there are probabilities where all 3 sets overlap. so if you notice you're actually adding the 3 set overlap 3x, once in each term obviously. Now normally you would think ok, I have to subtract the 3-set overlap out 2x, leaving just net 1. And that would be 100% correct if you are looking for Probability of people buying 2things AND 3 things. But the question asks for only people that buy both (in other words the two set overlaps).

Therefore you would subtract 55 - 3(3 set overlaps) = 25% getting rid of the 3 set overlaps entirely.

Re: Sets with >=3 elements [#permalink]
11 Oct 2011, 22:07

Expert's post

Some good methods have been discussed above. Let me add a small discussion on Venn diagrams

You will not have any confusion if you visualize it. The total number of people is 100 (assume it since numbers are in %). These 100 people are spread around in the 3 circles. One person can be in only one area.

Attachment:

Ques6.jpg [ 17.57 KiB | Viewed 1093 times ]

60+50+35 (= 145) is more than 100 because 60 = the entire left top circle = the left top red part + x + z + 10. 50 = the entire right top circle = the right top red part + x + y + 10 35 = the bottom circle = the bottom red part + y + z + 10 so x, y and z are counted twice and 10 is counted thrice

45 = x + y + z + 2*10 x+ y + z = 25 _________________

Re: Sets with >=3 elements [#permalink]
13 Oct 2011, 07:46

Expert's post

OptimusPrimea1 wrote:

@Karishma:

How did you get:

"45 = x + y + z + 2*10 x+ y + z = 25"

I lost the part here..

145 is 45 more than 100. Why is it 45 extra? Because x, y and z were counted twice (so they appear once extra) and 10 was counted thrice ( so it appears twice extra). check out the diagram. These extras make up the 45. In 100, there is no double/triple counting. It is equal to the actual number of people. That is why 45 = x + y + z + 2*10

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