$A_1,...,A_n$ are independent $\iff \forall 1 \leq k\leq n. \mathbb{P}\left(\bigcap_{i=1}^{k}A_{i}\right)=\prod_{i=1}^{k}\mathbb{P}\left(A_{i}\right)$ Prove/Disprove: $ A_1,...,A_n $ are independent events If and only If for all $ 1 \leq k \leq n $ it occurs that   $\mathbb{P}\left(\bigcap_{i=1}^{k} A_{i}\right)=\prod_{i=1}^{k} \mathbb{P}\left(A_{i}\right)$
Attempt: I think the above statement is false. the "$ \rightarrow $" implication is correct since the result will follow by definition, but the reverse is not necessarily true but I can't seem to make up a counter example. I tried making a dice counterexample but It think that was bad idea since most dice problems have independency rooted in them. So I tried the following example based on other examples I've encountered:
Let there be a jar with $15$ balls, $5$ black,$5$ white,$5$ red. Let there be persons $a,b,c$.
$c$ takes out a ball uniformly ( we don't know what ball he takes out ) and throws it into the garbage, then, $ a $ takes out a black ball and returns it and  $b $ takes out a black ball and returns it.
Denote $ A$ as the event person $a $ pulled out a black ball,$ B$ as the event person $b $ pulled out a black ball,$ C$ as the event person $c $ pulled out a black ball.
Note that,
$ P(A)=P(B) = 1/3 $
$ P(A) = P(A|C)P(C) + P(A|C^C)P(C^C) = (4/14)\cdot (1/3) + (5/14)\cdot (2/3) = 1/3 $
$ P(A\cap B \cap C) = P(B\cap C)P(A|B \cap C) $ ... ( I don't know If it is worth continuing, I don't see if my example helps me in the refutation)
My attempt obviously will not work, I need a different example that will work... but I can't think up of anything, can you please help? I'm stuck on this problem for a long time.

The statement is true for $n=2$ but not for $n\geqslant 3$ : take $A_1=\emptyset$, and $A_2=A_3=\dots=A_n=A$, a set of probability belonging to $(0,1)$.
