TITLE: Probability proof that $P(A|B)\geq \frac{2}{3}$
QUESTION [1 upvotes]: Prove that if $P(A)=P(B)=\frac{3}{4}$ then $P(A|B)\geq \frac{2}{3}$.

I know that $P(A|B)=\frac{P(A\cap B)}{P(B)}$, but I can't get problem this any further. Where should I start?

REPLY [6 votes]: Remeber that $P(...)\leq 1$, so: $$\frac{P(A\cap B)}{P(B)} ={P(A)+P(B)-P(A\cup B)\over P(B)}\geq {2\cdot {3\over 4}-1\over {3\over 4}}={2\over 3}$$
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