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Let $A$ and $B$ be events associated with a probability space $(\Omega, \mathbb{A}, \mathbb{P})$. Prove that (a) if $\mathbb{P}(A/B) \geq \mathbb{P}(A)$, then $\mathbb{P}(B/A) \geq \mathbb{P}(B)$; (b) if $\mathbb{P}(A) = \mathbb{P}(B) = 3/4$, then $\mathbb{P}(A/B) \geq 2/3$.

I've been trying to sort this out for a while. But I can only get proof if I assume the odds are well defined. I do not think that's enough. Anyone have any suggestions?

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closed as off-topic by Xi'an, mdewey, kjetil b halvorsen, Peter Flom Jan 13 at 11:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This sounds like a homework of sorts, so please add the self-study tag and show more about your attempts and the reason why the basic definition of $\mathbb{P}(A|B)$ does not help. Have you considered using a Venn diagram? $\endgroup$ – Xi'an Jan 12 at 15:12
  • $\begingroup$ $P(A|B) \ge P(A)$ implies $P(A\cap B)/P(B) \ge P(A)$ implies $P(A\cap B) \ge P(A)(B)$ implies ... . $\endgroup$ – BruceET Jan 12 at 20:01
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Hint (a): Find the relation between $$\frac{\mathbb{P}(A|B)}{P(A)}\qquad\text{and}\qquad\frac{\mathbb{P}B|A)}{P(B)}$$

Hint (b): By drawing a Venn diagram, find the constraints on $\mathbb{P}(A\cap B^c)$ and $\mathbb{P}(B\cap A^c)$ resulting from $\mathbb{P}(A)=\mathbb{P}(B)=3/4$ and deduce a constraint on $\mathbb{P}(A\cap B)$

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