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Foodmart customers regularly buy at least one of the [#permalink]
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27 Jan 2008, 23:20
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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



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Re: Sets with >=3 elements [#permalink]
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28 Jan 2008, 04:53
Yup, that looks good 60+50+35 = 145 14510(2) = 125 125100 = 25 25% of customers purchase 2 of the above products



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Re: Sets with >=3 elements [#permalink]
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05 Feb 2008, 12:11
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?



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Re: Sets with >=3 elements [#permalink]
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05 Feb 2008, 12:24
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 P1+P2+P3 = 100 P1+2*P2+3*P3 = 60+50+35 = 145 P3 = 10 > P2 = (145100)2*10 = 4520 = 25 > D



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Re: Sets with >=3 elements [#permalink]
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05 Feb 2008, 13:22
Hi,
x=3/5 x + ½ x +7/20 x  y (people with 2 purchases) – 2z (people with 3 purchases),
can someone explain how this formula is derived. I am not sure how we get 2z. It says 10% but all 3 , so wont it be 1/10 x.
Thanks Jack



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Re: Sets with >=3 elements [#permalink]
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05 Feb 2008, 19:51
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 Venn Diagram is my favorite here. Slow but true. 10= All 3 CA: A10 MA: C10 CA: B10 M: 60  (A10+10+C10) > 70AC C: 50  (A10+10+B10) > 60AB A: 35  (B10+10+C10) > 45BC Add everything 70AC 60AB 45BC A10 B10 C10 10 > 155ABC=100 > A+B+C=55 Now we want only 2 items so now its just A+B+C30 =25 D



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Re: Sets with >=3 elements [#permalink]
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07 Feb 2008, 01:00
jackychamp wrote: Hi,
x=3/5 x + ½ x +7/20 x  y (people with 2 purchases) – 2z (people with 3 purchases),
can someone explain how this formula is derived. I am not sure how we get 2z. It says 10% but all 3 , so wont it be 1/10 x.
Thanks Jack It is a derivation of the Venn diagram you'll see that we have to get rid of the double results and the triple ones. actually in this case worked...



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Re: Sets with >=3 elements [#permalink]
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07 Feb 2008, 16:00
Can someone provide the standard formula for 3 Venn diagram and 2 Venn Diagram.
Thanks JAck



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Re: Sets with >=3 elements [#permalink]
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11 Oct 2011, 15:12
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 P1+P2+P3 = 100 P1+2*P2+3*P3 = 60+50+35 = 145 P3 = 10 > P2 = (145100)2*10 = 4520 = 25 > D how do we know P1+P2+P3 = 100?



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Re: Sets with >=3 elements [#permalink]
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11 Oct 2011, 21:18
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 3set overlap there, but recall in the beginning you added 3x for the 3set overlap for A,B,C, so net is zero for that 3 set overlap. You can't ignore the 3set 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 3set 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.



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Re: Sets with >=3 elements [#permalink]
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11 Oct 2011, 23:07
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 1519 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
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Re: Sets with >=3 elements [#permalink]
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12 Oct 2011, 23:48
@Karishma:
How did you get:
"45 = x + y + z + 2*10 x+ y + z = 25"
I lost the part here..



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Re: Sets with >=3 elements [#permalink]
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13 Oct 2011, 08:46
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 25 = x+y+z
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