I've been working on a homework problem that I can't seem to be able to solve.
The question states:
Suppose 25 people attended a conference that contains 3 sessions. 15 people attended session #1; 18 people attended session #2; 12 people attended session #3. At least how many people attended all 3?
The inclusion-exclusion principle states
|A ∪ B| = |A| + |B| – |A ∩ B|
However, when I'm applying this problem to 3 sets, I obtained a negative number.
Let P(A ∪ B) ≤ 100%, P(A) = 60%, P(B) = 72%, P(C) = 0.48%.
P(A ∩ B) is at least 32%, P(A ∩ C) is at least 20%, P(B ∩ C) is at least 8%
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
After computing the result, I obtain 100% ≤ 120% + P(A ∩ B ∩ C). What am I doing wrong in applying this principle to 3 sets to find the intersection? The above percentages are unnecessary.
I computed the result: |A∪B∪C| ≤ 5 where |A∩B| ≥ 8, |A∩C| ≥ 2, |B∩C| ≥ 5, however, I don't quite understand the negative part where 25 ≥ |(A∩B)∪C| = |A∩B| + |C| - |A∩B∩C| (shorter way to find the minimum). The result just ends up as |A∩B∩C| + 25 ≥ 20 which would result in a negative unless I can somehow divide both sides by -1 which would change the equality?
Why do you set $P(A \cup B) = 100$%?
All you know is $P(A \cup B \color{red}{\cup C}) = 100$%, $P(A)=60$%, $P(B)=72$%, and $P(C) = 48$%
You will not be able to compute any of $P(A \cap B)$, $P(A \cup B)$, $P(A \cap C)$, $P(A \cup C)$, $P(B \cap C)$, and $P(B \cup C)$.
And in fact you also cannot know $P(A \cap B \cap C)$.
Nevertheless, how can you minimize $P(A \cap B \cap C)$? That is the question. Use your inclusion-exclusion formula to figure that out.
And by the way, there is no need to convert to probabilities/percentages here ... you can just work with the raw numbers if you use the formula:
$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C|- |B \cap C| + |A \cap B \cap C|$$