General Addition Rule for Probability extended to 4 events? I just started statistics and need to use the general addition rule. I know what it looks like for $3$ events:
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) - (2 * P(A \cap B \cap C)) + P(A'\cup B' \cup C').$$
But I'm confused to exactly how it extends to 4 events?
Thanks.
 A: There's this thing called the inclusion-exclusion identity, for any number $n$, this is the formula. 

for a proof of how this is works, you can check out this link at wikipedia 
A: The general formula is
$P\left( \bigcup\limits_{i=1}^{n} A_i\right) = \sum\limits_{i=1}^{n} P(A_i) - \sum\limits_{i, j : i < j} P(A_i \cap A_j) + \sum\limits_{i, j, k : i < j < k} P(A_i \cap A_j \cap A_k) - ... + (-1)^{n-1} P\left(\bigcap\limits_{i=1}^{n} A_i\right)$
When $n = 4$, we get
$P\left( \bigcup\limits_{i=1}^{4} A_i\right) = \sum\limits_{i=1}^{4} P(A_i) - \sum\limits_{i, j : i < j} P(A_i \cap A_j) + \sum\limits_{i, j, k : i < j < k} P(A_i \cap A_j \cap A_k) -  P\left(\bigcap\limits_{i=1}^{4} A_i\right)$
If we further assume the events $\{A_i\}$ to be exchangeable, we get
$P\left( \bigcup\limits_{i=1}^{n} A_i\right) = n \times P(A_1) - {n \choose 2} \times P\left(\bigcap\limits_{i=1}^{2} A_i\right) +  {n \choose 3} \times P\left(\bigcap\limits_{i=1}^{3} A_i\right) - ... + (-1)^{n-1} P\left(\bigcap\limits_{i=1}^{n} A_i\right)$
For $n = 4$ with exchangeable events, this gives
$P\left( \bigcup\limits_{i=1}^{4} A_i\right) = 4 \times P(A_1) - 6 \times P\left(\bigcap\limits_{i=1}^{2} A_i\right) + 3 \times P\left(\bigcap\limits_{i=1}^{3} A_i\right) - P\left(\bigcap\limits_{i=1}^{4} A_i\right)$
