bluemartini Posted April 29, 2005 Share Posted April 29, 2005 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. Quote Link to comment Share on other sites More sharing options...
scoot Posted April 29, 2005 Share Posted April 29, 2005 x= %participants liked 1 and 2 y= %participants liked 1 and 3 z= %participants liked 2 and 3 50+30+20-5-5-x-y-z= 85 x+y+z= 5. the survey participants liked more than one of the three products = x+y+z+5 =10. Quote Link to comment Share on other sites More sharing options...
ish Posted April 29, 2005 Share Posted April 29, 2005 why are you subtracting 5 twice? could you explain? Quote Link to comment Share on other sites More sharing options...
bluemartini Posted April 29, 2005 Author Share Posted April 29, 2005 I thought there is a catch here , it says 85% liked either, and 50% of THOSE liked 1 ...etc Scoot>> your eqn assumes that 50% of all those surveyed liked 1 etc. But I think the Q says 50% of the 85% liked 1.... Quote Link to comment Share on other sites More sharing options...
scoot Posted April 29, 2005 Share Posted April 29, 2005 why are you subtracting 5 twice? could you explain? formula is like this . A+B+C - (A and B)-(B and C) - (C and A)- (A and B and C) -(A and B and C). Quote Link to comment Share on other sites More sharing options...
securitygeek Posted April 29, 2005 Share Posted April 29, 2005 You are right, there is a catch and thats what you equate the LHS to 85 and not 100. Quote Link to comment Share on other sites More sharing options...
upchucks Posted April 29, 2005 Share Posted April 29, 2005 i am still a bit confused. i understand everything except the 5-5. cam someone please explain this to me. i am mathematically challenged Quote Link to comment Share on other sites More sharing options...
smithavishwanath Posted April 29, 2005 Share Posted April 29, 2005 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. Quote Link to comment Share on other sites More sharing options...
upchucks Posted April 30, 2005 Share Posted April 30, 2005 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 Quote Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 30, 2005 Share Posted April 30, 2005 Here is a very closely related thread: http://www.www.urch.com/forums/showthread.php?t=23076 Read Lego's explanation at the bottom, I believe it is the best way to approach these problems. Quote Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 30, 2005 Share Posted April 30, 2005 realized I screwed up, ignore this post Quote Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 30, 2005 Share Posted April 30, 2005 realized I screwed up, ignore this post:blush: Quote Link to comment Share on other sites More sharing options...
scoot Posted April 30, 2005 Share Posted April 30, 2005 I was looking for a way to draw a venn diagram. I do not know how draw that here. I got this link which has venn diagram. Please check this link. http://www.www.urch.com/forums/showthread.php?t=17900&highlight=85%25+surveyed Quote Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 30, 2005 Share Posted April 30, 2005 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) Quote Link to comment Share on other sites More sharing options...
ish Posted April 30, 2005 Share Posted April 30, 2005 [quote name=If there are three sets A, B, and C, then P(AnBnC) = P(A) + P(B) + P© – P(AnB) – P(AnC) – P(BnC) + P(AnBnC) =/QUOTE] shouldnt this be a UNION of sets???:hmm: Quote Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 30, 2005 Share Posted April 30, 2005 sorry about that ish, I just fixed it:) Quote Link to comment Share on other sites More sharing options...
ish Posted April 30, 2005 Share Posted April 30, 2005 these formulas are really good.Thanks for the trouble. :) Quote Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 30, 2005 Share Posted April 30, 2005 Let X be P(AnB) + P(AnC) + P(BnC). Using the formulas above, we have 85 = 50+30+20-X+5 X = 20 Number of people having two or more = X - 2(AnBnC) = 20 - 10 = 10 Answer is 10 Quote Link to comment Share on other sites More sharing options...
hem Posted May 28, 2005 Share Posted May 28, 2005 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! Quote Link to comment Share on other sites More sharing options...
ramanathansankar Posted November 23, 2005 Share Posted November 23, 2005 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. Quote Link to comment Share on other sites More sharing options...
bragss2 Posted November 24, 2005 Share Posted November 24, 2005 This discussion regarding formula should rest now Pls see the attachment[ATTACH]1512[/ATTACH] Quote Link to comment Share on other sites More sharing options...
g.bianchi Posted January 14, 2008 Share Posted January 14, 2008 hi! I'm new and I registered to get this files, but it seems I can't :( I'm studying GMAT. could you help me? Quote Link to comment Share on other sites More sharing options...
krsna83_pp Posted March 19, 2008 Share Posted March 19, 2008 thanks for the attachment.Really conjoins all the pieces Quote Link to comment Share on other sites More sharing options...
AliceWonderful Posted July 1, 2008 Share Posted July 1, 2008 Can someone e-mail me that attachment? I can't download it even though I've been posting for a while now. I'm not sure why. My e-mail is adelynortlieb@gmail.com . I would like the set theory doc. It would only take you a second, but it would save me a TON of trouble. Thanks! Quote Link to comment Share on other sites More sharing options...
ranjeet_1975 Posted July 1, 2008 Share Posted July 1, 2008 Easy with Venn Diagram (50-5-x-z)+(30-5-x-y)+(20-5-z-y)+5+x+y+z = 85 x+y+z = 5 Ans should be 10 Quote Link to comment Share on other sites More sharing options...
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