A homework problem involving counting using a special case of inclusion/exclusion

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?

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Could you please use an even less useful title for your question? – Tobias Kienzler Oct 29 '10 at 13:52
I have the opposite reaction -- :-) -- to that of Tobias Kienzler: please choose a better title for your question. This one could apply to any question, so is absolutely useless. – Pete L. Clark Nov 28 '10 at 6:56

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%.

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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 .

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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?

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