TITLE: Probability of the union of $3$ events?
QUESTION [14 upvotes]: I need some clarification for why the probability of the union of three events is equal to the right side in the following:
$$P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(E\cap G)-P(F\cap G)+P(E\cap F\cap G)$$
What I don't understand is, why is the last term(intersection of all) added back just once, when it was subtracted three times as it appears from a Venn Diagram?
Here on page 3, this is explained but not in enough details that I can understand it:
http://www.math.dartmouth.edu/archive/m19w03/public_html/Section6-2.pdf

REPLY [6 votes]: An interesting way to find the answer is by looking at the following Venn diagram:

The numbers in red indicate how many times a certain section has been included/added by using $P(A) + P(B) + P(C)$. The sections which feature more than once must be subtracted from the former to avoid double-counting and obtain the final answer.

The blank sections are added once. $\implies$ No subtraction required.

The dashed (||) sections represent $P(A \cap B \cap C^c)$, $P(A \cap B^c \cap C)$, and $P(A^c \cap B \cap C)$. Each term is included twice, $\implies$ hence each must be subtracted once.

The dotted (::) section represents $P(A \cap B \cap C)$. Appears thrice $\implies$ must subtracted twice.


This yields:
$$P(A \cup B \cup C ) = P(A) + P(B) + P(C) - P(A \cap B \cap C^c) - P(A \cap B^c \cap C) - P(A^c \cap B \cap C) - 2P(A \cap B \cap C).$$
Observing from the figure that:
$$P(A \cap B \cap C^c) = P(A \cap B) - P(A \cap B \cap C),$$
$$P(A \cap B^c \cap C) = P(A \cap C) - P(A \cap B \cap C),$$
$$P(A^c \cap B \cap C) = P(B \cap C) -P(A \cap B \cap C),$$
Replacing these values in the above results in the final formula, given by:
$$P(A \cup B \cup C ) = P(A) + P(B) + P(C) - P(A \cap B) -  P(A \cap C) - P(B \cap C) + P(A \cap B \cap C).$$