MHB Understanding the Inclusion-Exclusion Principle for Multiple Events

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The inclusion-exclusion principle calculates the probability of the union of multiple events by accounting for overlaps. For four events, the formula is P[A ∪ B ∪ C ∪ D] = P[A] + P[B] + P[C] + P[D] - P[A ∩ B] - P[A ∩ C] - P[A ∩ D] - P[B ∩ C] - P[B ∩ D] - P[C ∩ D] + P[A ∩ B ∩ C] + P[A ∩ B ∩ D] + P[A ∩ C ∩ D] + P[B ∩ C ∩ D] - P[A ∩ B ∩ C ∩ D]. In general, the number of terms in the probability of the union of n events is given by 2^n - 1, which includes all possible intersections. The discussion highlights the complexity of calculating probabilities as the number of events increases. Understanding this principle is crucial for accurate probability assessments in various applications.
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The inclusion-exclusion principle shows that for two events $P[A \cup B]=P[A]+P-P[A \cap B]$ and for three events $P[A \cup B \cup C] = P[A] + P + P[C] - P[A \cap B] - P[A \cap C] - P[B \cap C] + P[A \cap B \cap C]$.

Using these as a guideline what is $P[A \cup B \cup C \cup D]$? In general, how many terms will be in $$P\left[ \bigcup_{i=1}^{n} A_i \right]$$?
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Congratulations to the following members for their correct solutions:

1) Ackbach

Solution (from Ackbach):
The guideline yields the pattern of having to subtract, then add, then subtract, so as not to count things more than once. Based on that, we have that
\begin{align*}
P(A \cup B \cup C \cup D)&=P(A)+P(B)+P(C)+P(D) \\
& \quad -P(A \cap B)-P(A \cap C)-P(A \cap D)-P(B \cap C)-P(B \cap D)-P(C \cap D) \\
& \quad +P(A \cap B \cap C)+P(A \cap B \cap D)+P(A \cap C \cap D)+P(B \cap C \cap D) \\
& \quad -P(A \cap B \cap C \cap D).
\end{align*}
The pattern here for the number of terms is that
$$P \left[ \bigcup_{j=1}^{n}A_{j} \right]$$
has $2^{n}-1$ terms. Technically, it has $2^{n}$ terms, one for each possible subset; but $P[ \varnothing]=0$, so it'll never contribute to the sum.
Alternate solution:
We can see that $P[A \cup B \cup C \cup D]$ has 4 terms, then 3, then 3 then 1 if you partition the terms into the amount of intersections. Written another way, it has $$\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}$$ terms. If we generalize this we can see that the number of terms in $$P \left[ \bigcup_{i=1}^{n}A_{i} \right]$$ is $$\sum_{i=1}^{n}\binom{n}{i}$$.
 
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