Probability of P(A|B) given P(A), P(B)

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Given $P(A)=0.9$ and $P(B)=0.8$.

How to prove that $P(A|B)\ge 0.875$ ?

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2 Answers

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$$P(A|B)=\frac{P(A\cap B)}{ P(B)}$$ $$=\frac{P(A)+P(B)-P(A\cup B)}{ P(B)}$$ $$=\frac{0.9+0.8-P(A\cup B)}{ 0.8}$$ $$\ge \frac{0.9+0.8-1}{ 0.8}$$ $$\ge 0.875$$

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Because $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$, and $\Pr(A\cup B)\le 1$, we have $\Pr(A\cap B)\ge 0.7$.

Now use the fact that $\Pr(A\cap B)=\Pr(A\mid B)\Pr(B)$. We get $\Pr(A\mid B)\ge \frac{0.7}{0.8}$.

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