Probability - Show inequality $P(A \cap B) \ge P(A) + P(B)-1$

Question

My task is to show that: $$P(A \cap B) \ge P(A) + P(B)-1$$ The rule inclusion-exclusion is: $$P(A\cup B) = P(A) + P(B) - P(A \cap B)$$ If I rewrite this rule, I get $$P(A\cap B) = -P(A \cup B) + P(A) + P(B)$$ Am I on the right track, and if so, how do I proceed?

Answer 1

Yes you are on the right track. Now, you should that any probability is $\le 1$. So $P(A\cup B)\le 1\implies -P(A\cup B)\ge -1$. Hence your answer.

Answer 2

Yes, you are completely on the right track. Now, if you get easily confused by the direction of inequalities you could also write $$ P(A\cup B) = P(A) + P(B) - P(A \cap B) $$ since in $(\Omega,P,\mathcal{F})$ we have $$ \forall A\in\mathcal{F}: 1\geq P(A) $$ as $$ 1\geq P(A\cup B) = P(A) + P(B) - P(A \cap B) $$ so $$ 1\geq P(A) + P(B) - P(A \cap B)\Leftrightarrow P(A \cap B)\geq P(A) + P(B)-1 $$