Probability complement question How is it that $$P(A \cap B^c) + P(A^c \cap B) \le 1?$$ I know that this statement is true and have been trying to prove it with Venn diagrams, but I am stuck. 
I am able to show it is true for A in B and when B is in A, but what about when A and B and independent, disjoint events?
Any ideas?
 A: Simply because $P(A)+P(A^c) = 1$. You know that $A\cap B^c\subseteq A$ and $A^c\cap B\subseteq A^c$ so
$$
P(A\cap B^c)+P(A^c\cap B)\leq P(A)+P(A^c) = 1.
$$
Even 'stronger result' is impled: $P(A\cap C)+P(A^c\cap D)\leq 1$.
A: Can you show that for two disjoint events $C,D$, we have $P(C) + P(D) \le 1$?
Then just note that $A \cap B^c$ and $A^c \cap B$ are disjoint, which you should be able to see from your Venn diagram.  Indeed, one is contained in $A$ and the other in $A^c$.
A: Let $\Omega$ be the whole space.  Then
$$
\Omega = (A\cap B^c) \cup (B \cap A^c) \cup (A^c \cap B^c) \cup (A\cup B)^c
$$
and these four sets are disjoint, or, if you prefer that language, mutually exclusive.  Only two of them are included among those whose probabilities you added. 
A: Hint 1: For every sets $C$ and $D$, if $C\cap D=\varnothing$, then $\mathrm P(C)+\mathrm P(D)\leqslant1$. 
Hint 2: For every sets $A$ and $B$, $(A\cap\overline{B})\cap(B\cap\overline{A})=\varnothing$.
Hoping you can prove Hints 1 and 2 and proceed from there.
