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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.

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    $\begingroup$ Have you tried drawing a picture? I'd recommend that. $\endgroup$
    – Tyler
    Mar 3, 2014 at 19:31
  • $\begingroup$ Hint: $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ $\endgroup$ Mar 3, 2014 at 19:35
  • $\begingroup$ I already know that but I'm not exactly sure how to go about converting $ A \cap B^c $ into $ A \cup B $ $\endgroup$
    – petrov
    Mar 3, 2014 at 19:38
  • $\begingroup$ Okay: try to write $A \cup B$ as a union of three disjoint sets and see what you get. $\endgroup$ Mar 3, 2014 at 19:39
  • $\begingroup$ I got: $ P(A \cap B^C) + P(A^c \cap B) - P[(A \cap B^c) \cap (A^c \cap B)]$ $\endgroup$
    – petrov
    Mar 3, 2014 at 19:41

4 Answers 4

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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 $$\begin{align*}P((A\cap B^c) \cup(A^c \cap B))&=P(A)-P(A\cap B)+P(B)-P(A\cap B)=\\&=P(A)+P(B)-2P(A\cap B)\end{align*}$$

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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.

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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.

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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(𝜙) (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)

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