Prove that if $P(A) = P(B) = \frac23$ , then $P(A|B) ≥ \frac12$ Prove that if $P(A) = P(B) = \dfrac23$ , then $P(A|B) ≥ \dfrac12$.

Well I thought that because $P(A) + P(B)&gt; 0$, then they are independent. So I used $P(A|B)= P(A)$ which I can use due to independence. However I have doubts because to be independent they don't necessarily have to be bigger than $0$ but I see no other reason or way to prove that that given probability is bigger than $1/2$.
Hint: Here are the relations which you should use


$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$P(A\cap B)=P(A|B)\cdot P(B)$
$P(A\cup B)\leq 1$


All it is left is to put them together.
We have P(A) = P(B) = 2/3. 

Take P(A|B) and P(A|not B).

We know that P(A) = P(B) * P (A|B) + P(not B) * P(A|not B). 

2/3 = 2/3 * P(A|B) + 1/3 * P(A|not B)

We have 0 ≤ P (A|not B) ≤ 1. Therefore 

2/3 * P(A|B) + 1/3 * 0 ≤ 2/3 ≤ 2/3 * P(A|B) + 1/3 * 1

P(A|B) ≤ 1 ≤ P(A|B) + 0.5

0.5 ≤ P(A|B) ≤ 1.

(Intuitively: In 2/3rds of all cases we have B, in 1/3rd we have not B. If we don't have A in at least half the cases where B, the numbers of cases with A don't add up to 2/3rds, even if we have A in all cases where not B. )