In a survey of political preference, 78% of those asked were in favor of at-least one of the proposals: I, II and III. 50% of those asked favored proposal I, 30% favored proposal II, and 20% favored proposal III. If 5% favored all the three proposals, what % of those asked favored more than one of the three proposals?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
The simplest way is probably to draw Venn diagrams and stuff, but here's how you do it using the inclusion-exclusion principle.
For any three sets $A,B,C$, we have $$|A \cup B \cup C| = S_1 - S_2 + S_3,$$
where $S_1 = |A|+|B|+|C|$, $S_2 = |A \cap B| + |B \cap C| + |C \cap A|$, and $S_3 = |A \cap B \cap C|$.
Letting $A,B,C$ denote the obvious sets, you are given the values of $|A \cup B \cup C|$, $S_1$ and $S_3$ so you can solve for $S_2$.
Now you are asked to find $x = |(A \cap B) \cup (B \cap C) \cup (C \cap A)|$, which we can expand with the same formula (using the fact that $(A \cap B) \cap (B \cap C) = A \cap B \cap C$ etc.) as $$x = |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| + 3|A \cap B \cap C| = S_2 - 2S_3.$$
If you do this right, and assuming I've done it right, the answer should be 17%.
hmm, I got 17, not 27...
If we count |A|+|B|+|C|, then we overcount intersections of any 2 sets twice, and intersections of all three 3 times. Then, if we subtract $|A \cup B \cup C|$, we get the intersections of any 2 sets once and 3 sets 2 times. (most easily seen by drawing a Venn diagram). Subtracting the 3-set intersection once more gives 100-78-5=17 .
We have (using ab for those who favor I and II, n for those who favor none, etc) a+b+c+2ab+2ac+2bc+3abc=? a+b+c+2ab+2ab+2ac=? a+b+c+ab+ac+bc+abc=? a+b+c+ab+ac+bc=? and so?