Questions on symmetric difference of events

Question

From a comment on my math overflow question:

No, $P(A\bigtriangleup B)=0$ means $A$ and $B$ are essentially the same except in situations that almost surely do not happen. $P(A)=P(B)$ says much less.

I lost my notes, but I think I was able to show symm difference has prob 0 if $A=B$ or if $P(A) = P(B) = 0$ or $1$. I recall I wasn't able to show it if $P(A) = P(B)$. So apparently

$$ P(A)=P(B) \to P(A\bigtriangleup B)=0 \tag{*}$$

is false.

1. It's possible A and B are disjoint. That's a counterexample to $(*)$?

2. What if $1 > P(A) = P(B) = p > 0$ but $P(A \cap B) = p$? That proves $(*)$?

3. What if merely that A and B are not disjoint?

Answer 1

For (1), take $A,B$ disjoints with $A\cup B=\Omega$. Then $P(A)=P(B)=1/2$ and $P(A\Delta B)=P(\Omega)=1$.

For (2), $P(A\Delta B)=P(A\cup B)-P(A\cap B)=P(A)+P(B)-P(A\cap B)-P(A\cap B)=2p-2p=0$ if $P(A)=P(B)=P(A\cap B)$ (please note that we don't use $0<p<1$).

For (3), we only have $P(A\Delta B)=P(A)+P(B)-2P(A\cap B)$. Thus, we need $\frac{P(A)+P(B)}{2}=P(A\cap B)$