Set Theory
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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. 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(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
Number of people in exactly one set = P(A) + P(B) + P(C) – 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)
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
To add to TwinSplitter's forumla list: No of people in atleast 1 set =
P(A) + P(B) + P(C) - 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 __________________________________________________________
SET THEORY
1. If there are three sets A, B, and C, then (1+2+3+X+Y+Z+O) P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC) 2. Number of people in exactly one set = ( 1+2+3) P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC) 3. Number of people in exactly two of the sets = ( X+Y+Z) P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC) 4. Number of people in exactly three of the sets = O P(AnBnC) 5. Number of people in two or more sets = ( X+Y+Z+O) P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC) 6. Number of people only in A = 1 P(A) – P (AnB)- P(AnC) + P ( AnBnC) P(A) = 1+X+Y+O P( AnB) = X+O
Number of people in exactly one set = P(A) + P(B) + P(C) – 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)
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
To add to TwinSplitter's forumla list: No of people in atleast 1 set =
P(A) + P(B) + P(C) - 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 __________________________________________________________
SET THEORY
1. If there are three sets A, B, and C, then (1+2+3+X+Y+Z+O) P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC) 2. Number of people in exactly one set = ( 1+2+3) P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC) 3. Number of people in exactly two of the sets = ( X+Y+Z) P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC) 4. Number of people in exactly three of the sets = O P(AnBnC) 5. Number of people in two or more sets = ( X+Y+Z+O) P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC) 6. Number of people only in A = 1 P(A) – P (AnB)- P(AnC) + P ( AnBnC) P(A) = 1+X+Y+O P( AnB) = X+O
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