What can be deduced from $P(A or B) = P(A) + P(B)$? $P(A \cup B) = P(A) + P(B)$
(1) If $A$ and $B$ are independent, then $P(A) = 0$ or $P(B) = 0$.
$\quad$ True, $P(A \cap B)= 0 =P(A)\cdot P(B)$
(2) If $A$ = S(sample space), then $P(B) = 0$
$\quad$ True, since $P(A \cap B) = 0$
(3)  If $P(A) >P(B) > 0$, then $A$ and $B$ are not independent
$\quad$ True, if both are positive, then $P(A \cap B) > 0$, which contradicts $P(A \cap B) = 0$
(4) If $P(A) \neq P(B)$, then $A$ and $B$ are mutually exclusive
$\quad$ True, this is vacuously true since $P(A \cap B)= 0$, $A$ and $B$ are already mutually exclusive
Is my reasoning correct? I am unsure about (4), is there a counterexample that disproves (4)?
 A: Well, we know $\Pr(A\cap B)=0$, but that doesn't necessarily mean $A\cap B=\emptyset$.  Choose a real number $X$ uniformly at random from $[0,1]$. Let $A$ be the event $X\in[0,\frac23]$ and let $B$ be the event $X\in[\frac23,1]$.  $\Pr(A\cap B)=0$, but $A\cap B$ is the event that $X=\frac23$.
A: I think the third one is incorrect, if you choose $B=A^{c}$ then they are dependent, and you can build an example where both of probabilities is positive, but of course still $P(A\cap B)=0$
As for the fourth one, I think the counter example comes from not countable sample spaces, where you can consider:
A is distributed uniformly over [0,1] ($P(A)=1$)
and B is the rational numbers set ($\mathbb{Q}$) (*)
Then you get $$P(A\cup B)= P(A)+P(B) =1$$
and $1=P(A)\neq P(B)=0$
but of course $B\subseteq A$
(*) or any other countable set
A: In general,
$
P(A\cup B)=P(A)+P(B)-P(A\cap B).
$

*

*(1) is correct, as you observed that the assumption $P(A\cup B)=P(A)+P(B)$ is equivalent to $P(A\cap B)=0$, and independence says $P(A\cap B)=P(A)P(B)$.

*(2) is correct since $P(B)=P(A\cup B)=0$.

*(3) is correct since it is the contrapositive of (1).

*(4) No, $P(A\cap B)=0$ does not imply that $A\cap B=\emptyset$. A counterexample: $A=\{0\le X\le 0\}$ and $B=\{0\le X<\infty\}$ where $X$ is a random variable with normal distribution.  Then in this example $P(A)\ne P(B)$ and $P(A\cap B)=0$. But $A\cap B\ne\emptyset$.

A: We have P(A or B)=P(A)+P(B)-P(A and B)
So, P(A or B)=P(A)+P(B) implies P(A and B)=0
As already answered here, note that it does not mean (A and B) is an impossible event.
