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 $$\begin{align*}P((A\cap B^c) \cup(A^c \cap B))&=P(A)-P(A\cap B)+P(B)-P(A\cap B)=\\&=P(A)+P(B)-2P(A\cap B)\end{align*}$$
A: Notice that: $$A\cup B= (A \cap B^C)\cup(A^C \cap B)\cup(A\cap B)$$ That is, $A \cup B$ is the disjoint union of three sets.
Now, remember that disjoint unions are nice, because we can distribute over them. (i.e. $A \cap B=\emptyset \implies P(A\cup B)=P(A)+P(B)$)
Combining that idea with the first line we get:
$$P(A\cup B)= P[(A \cap B^C)\cup(A^C \cap B)]+P(A\cap B)$$
Noting that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$, we can now rearrange the terms to get the desired result.
A: By the sum rule of probability, we have
$$P[(A\cap B^C)\cup (A^C\cap B)]=P(A\cap B^C)+P(A^C\cap B).$$  Then by noticing that $P(A\cap B^C)=P(A)-P(A\cap B)$ and $P(A^C\cap B)=P(B)-P(A\cap B)$ we arrive at:
$$=P(A)-P(A\cap B)+P(B)-P(A\cap B)=P(A)+P(B)-2P(A\cap B)$$ and we are done.
A: Ρ[(Α∩Β′)∪(Α′∩Β)]
= Ρ(Α∩Β′) + Ρ(Α′∩Β) − Ρ(Α∩Β′∩Α′∩Β) (Law of GRA)
= Ρ(Α∩Β′) + Ρ(Α′∩Β) − Ρ(Α∩Α′∩Β′∩Β) (Commutative)
= Ρ(Α∩Β′) + Ρ(Α′∩Β) − Ρ() (Complementary Law)
= Ρ(Α) − Ρ(Α∩Β) + Ρ(Β) − Ρ(Α∩Β) (Since, Ρ(Α) = Ρ(Α∩Β′) + Ρ(Α∩Β) and Ρ(Β) =  Ρ(Α′∩Β) + Ρ(Α∩Β))
= Ρ(Α) + Ρ(Β) − 2 Ρ(Α∩Β)
