# Conditional Probability: $\mathbb{P}(B/A) \geq \mathbb{P}(B)$ [closed]

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?

## 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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• 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? – Xi'an Jan 12 at 15:12
• $P(A|B) \ge P(A)$ implies $P(A\cap B)/P(B) \ge P(A)$ implies $P(A\cap B) \ge P(A)(B)$ implies ... . – BruceET Jan 12 at 20:01

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 $$\mathbb{P}(A\cap B)$$