Suppose $\textbf{P}(A\cap B) = 0$. Prove $\textbf{P}(A) \leq \textbf{P}(B^{c})$ using the axioms of probability.
I know the axioms are:
- $\textbf{P}(A)\geq 0$.
- $\textbf{P}(\Omega) = 1$.
- If A and B are disjoint, $P(A\cup B) = P(A) + P(B)$.
I've tried messing around with each of these axioms to end up with the proof statement, but haven't been able to get to it. Does anybody know how to prove this using the axioms?
Thanks
$\endgroup$ 22 Answers
$\begingroup$$$P(A)=P(A\cap B) + P(A\cap B^c)= P(A\cap B^c)\leq P(B^c)$$The first equality uses $A=(A\cap B)\cup (A\cap B^c)$, and Axiom 3. (union of disjoints sets).
The last inequality follows from the more general $X\subset Y \implies P(X)\leq P(Y)$, which is a consequence of $Y=X\cup(Y\setminus X)$ and Axiom 3.
$\endgroup$ 1 $\begingroup$Note that $$P(B^\complement)-P(A)=1-P(B)-P(A)=1-P(A\cup B)\ge0,$$where the second $=$ uses $P(A\cap B)=0$.
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