Unable to understand option 2 of this question Let $A$ and $B$ be events satisfying $P(A∩B)$ = $P(A)$ + $P(B)$.
Which of the following statements is/are true?

*

*If $A$≠ $B$, then $A$ and $B$ are mutually exclusive.


*If $A$ and $B$ are independent, then $P(A) = 0 $ or $P(B) = 0$.


*If $P(A) > P(B) > 0$, then $A$ and $B$ are not independent.


*If $A$ = $S$, the sample space, then $P(B) = 0$.

By process of elimination:
Option $1$ is correct since if $A$ and $B$ are mutually exclusive, $P(A ∩ B) = 0$.
Option $3$ is correct as if $P(A)>0$ and $P(B)>0$, $P(A)*P(B)>0$ meaning $P(A ∩ B) >0$. This is a contradiction thus not independent.
Option $4$ is correct as $P(B) > 0$ the initial statement given would be false.
Option $2$ is the one I'm having difficulty understanding, is it vacuously true(is logic language even used here or is it a normal if-then statement?) since it's established that $A$ and $B$ are exclusive?
 A: Answer for (2): The given equation gives $(1-P(A)) (1-P(B))=1$. If either $P(A) \neq 0$ or $P(B) \neq 0$ then the left is a product of two numbers one of which is less than $1$ and the other $\leq 1$ leading to a contradiction. Hence we must have $P(A)=P(B)=0$.
A: This seems more like an exercise to check the knowledge of definitions. I hope I am using the same definitions as you.
Option 1
Firstly, I believe that your argument for option 1 is not valid. Events are mutually exclusive if the intersection of both of them (as a subset of your probability space)  is empty. This indeed implies $P(A\cap B)=0$. But you need for your argument the opposite implication.
Now if you work over countable probability space, what I wrote in the previous paragraph is just some unnecessary technicality. But if you are in uncountable probability space things get more funnier.  Because intersection can be non-empty but still of measure zero.
Option 2 This option is indeed correct. But I would not say it is vacuously correct. Though if you have a mutual exclusiveness it is indeed easy, because you have $0=P(A\cap B)=P(A)+P(B)$.
Without the assumption of mutual exclusiveness, you go by $P(A\cap B)=P(A)P(B)$ by assumption of independence. Hence you have $ P(A)P(B)=P(A)+P(B)$.
Then you have two options. Firstly $P(A)=P(A)P(B)=P(A)+P(B)$ this directly implies $P(B)=0$ and also $P(A)=0$.  Or $P(A)<P(A)P(B)=P(A)+P(B)$ but this is contradiction.
