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$P(A\mid B)\leq \frac{a+b-1}{b}$ where $P(A)=a,P(B)=b$. I found this example problem on notes I took in class but I forgot to copy down how the prof proved the proposition. Maybe I miscopied the proposition, since it seems false, as it implies $P(AB)\leq P(A)+P(B)-1$ which is clearly false, say, for two independent events with $P(A)=P(B)=0.1$. |
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The $\le$ should be replaced by $\ge$. Apart from that (!) the result is correct. If $\Pr(B)\ne 0$, we have $\Pr(A|B)=\dfrac{\Pr(A\cap B)}{\Pr(B)}$. Recall that $$\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B).$$ Since $\Pr(A\cup B)\le 1$, we have $\Pr(A)+\Pr(B)-\Pr(A\cap B)\le 1$ and therefore $\Pr(A\cap B)\ge \Pr(A)+\Pr(B)-1$. In the notation of your post, if $b\ne 0$, we have $$\Pr(A|B)\ge \frac{a+b-1}{b}.$$ |
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