Question about the independence definition. Why does the independence definition requires that every subfamily of events $A_1,A_2,\ldots,A_n$ satisfies $P(A_{i1}\cap \cdots \cap A_{ik})=\prod_j P(A_{ij})$ where $i_1 &lt i_2 &lt \cdots &lt i_n$ and $j &lt n$.
My doubt arose from this: Suppose $A_1,A_2$ and $A_3$ such as $P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2)P(A_3)$.
Then
$$P(A_1\cap A_2)=P(A_1\cap A_2 \cap A_3) + P(A_1\cap A_2 \cap A_3^c)$$
$$=P(A_1)P(A_2)(P(A_3)+P(A_3^c))=P(A_1)P(A_2).$$
So it seems to me that if $P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2)P(A_3)$ then $P(A_i\cap A_j)=P(A_i)P(A_j)$, i.e., the biggest collection independence implies the smaller ones. Why am I wrong? The calculations seems right to me, maybe my conclusion from it are wrong?

$P(ABC)=P(A)P(B)P(C)$ does not imply that $P(ABC^C)=P(A)P(B)P(C^C)$, which it seems you're using. Consider, for instance, $C=\emptyset$.
However, see this question.


Another example:
Let $S=\{a,b,c,d,e,f\}$ with $P(a)=P(b)={1\over8}$, and $P(c)=P(d)=P(e)=P(f)={3\over16}$.
Let $A=\{a,d,e\}$, $B=\{a,c,e\}$, and $C=\{a,c,d\}$.
Then
$\ \ \ \ \ \  P(ABC)=P(\{a\})={1\over8}$
and
$\ \ \ \ \ \ P(A)P(B)P(C)= {1\over2}\cdot{1\over2}\cdot{1\over2}={1\over8}$.
But
$\ \ \ \ \ \ P(ABC^C)=P(\{e\})= {3\over16}$
while
$\ \ \ \ \ \ P(A)P(B)P(C^C) = {1\over2}\cdot{1\over2}\cdot{1\over2}={1\over8}$.
In fact no two of the events $A$, $B$, and $C$ are independent.
