# 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?

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I adjusted your notation. Please double-check that it matches what you wanted to say. – Nate Eldredge Oct 14 '11 at 17:24

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: $P(A\cap C)+P(A^c\cap D)\leq 1$.

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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 $A^c$.

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This is probably the best answer so far. The only way it can happen that $P(C)+P(D) > 1$ is if $C$ and $D$ have quite a large intersection. If they have no intersection at all, you're done. – Michael Hardy Oct 14 '11 at 19:09
....actually I should say if at least one of $C,D$ is quite large and they intersect. – Michael Hardy Oct 16 '11 at 16:44

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.

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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.

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