Q:

Probability Proof $ P[(A \cap B^c) \cup (A^c \cap B)] = P(A) + P(B) - 2P( A \cap B) $

How would I go about proving this statement:
$ P[(A \cap B^c) \cup (A^c \cap B)] = P(A) + P(B) - 2P( A \cap B) $
Describe in English the event where the probability is computed by the expression on the LHS of the equation.

A:

The event on the LHS in english can be described as "The event that either A but not B or B but not A occurs". 
This event is the union of two disjoint sets $A \cap B^c$ and $A^c \cap B$. They are disjoint because $$(A\cap B^c) \cap(A^c \cap B)=(A\cap A^c)\cap (B\cap B^c)=\emptyset\cap\emptyset=\emptyset$$The probability of the union of two disjoint sets can be written as the sum of their probabilities, that is, eq$(1)$: $$P[(A \cap B^c) \cup (A^c \cap B)] = P(A\cap B^c) + P(A^c\cap B)$$ Now, since $$P(A)=P(A\cap B^c)+P(A\cap B)$$ we have that, eq$(2a)$: $$P(A\cap B^c)=P(A)-P(A\cap B)$$ Similarly, since $$P(B)=P(A^c\cap B)+P(A\cap B)$$ we have that, eq$(2b)$: $$P(A^c\cap B)=P(B)-P(A\cap B)$$ Combining equations $(2a)$ and $(2b)$ and substituting into equation $(1)$ we get the required result, that is $$\begin{align*}P((A\cap B^c) \cup(A^c \cap B))&=P(A)-P(A\cap B)+P(B)-P(A\cap B)=\\&=P(A)+P(B)-2P(A\cap B)\end{align*}$$

