Does Conditional Probability Increase with Dependence?

[mathmari]

Hey!

Suppose that for the events $A,B$ of an experiment it holds that $P(A|B)>P(A)$ ( $P(A)=P(B)=\frac{2}{3}$ ) then what of the following holds?

1. $P(B|A)>P(B)$ ( $P(A|B)\geq \frac{1}{2}$ )
2. $P(B|A)>P(B)$ ( $P(A|B)\leq \frac{1}{3}$ )
3. $P(B|A)<P(B)$ ( $P(A|B)\geq \frac{1}{3}$ )
4. $P(B|A)<P(B)$ ( $P(A|B)\geq \frac{1}{6}$ )
   

I have done the following:

$P(A|B)=\frac{P(BA)}{P(B)}$ and since $P(A|B)>P(A) \Rightarrow \frac{P(BA)}{P(B)}>P(A) \Rightarrow P(BA)>P(B)P(A)$.

Then $P(B|A)=\frac{P(AB)}{P(A)}>\frac{P(B)P(A)}{P(A)}=P(B)$, right? (Wondering)

When $P(A)=P(B)=\frac{2}{3}$, then $P(A|B)>P(A)\Rightarrow P(A|B)>\frac{2}{3}$.

How can we continue?

 

[I like Serena]

Hey mathmari! (Smile)

Don't we already have enough information to figure out which answers are true or false? (Wondering)

 

[mathmari]

Hey mathmari! (Smile)

Don't we already have enough information to figure out which answers are true or false? (Wondering)

(Thinking)

The first one is correct, or not? (Wondering)

 

[I like Serena]

(Thinking)

The first one is correct, or not? (Wondering)

Hmm... let's see...

We found that $P(A|B) > \frac 23$.
Does that imply that $P(A|B) \ge \frac 12$ or not? (Wondering)

 

[Siron]

In addition to I Like Serena's post, making use of Bayes rule:
$$P(B \ | \ A) = \frac{P(A \ | \ B) P(B)}{P(A)} = P(A \ | \ B) > P(A) = P(B).$$
Hence, first answer is indeed correct.