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?

When a tetrahedral die is rolled, the outcome is the face on the bottom
of the die when it comes to rest.  Suppose the four faces are
marked $2,3,5,30$, and these numerical outcomes occur with 
probabilities $\frac{11}{24}, \frac{7}{24}, \frac{5}{24}$ and $\frac{1}{24}$
respectively.
Let $A$, $B$, and $C$ denote the events that the outcome is a 
multiple of $2$, $3$, and $5$ respectively.  Then,
$$\begin{align*}
P(A) &= P\{2,30\} = \frac{1}{2}\\
P(B) &= P\{3,30\} = \frac{1}{3}\\
P(C) &= P\{5,30\} = \frac{1}{4}\\
P(ABC) &= P\{30\} = \frac{1}{24} = P(A)P(B)P(C)
\end{align*}$$
but $AB = AC = BC = ABC$ and thus 
$$P(AB) \neq P(A)P(B), \quad P(AC) \neq P(A)P(C), \quad P(BC) \neq P(B)P(C)$$

On the other hand, if it is a fair die, then $P(A)=P(B)=P(C) = \frac{1}{2}$
and since $P(ABC) = \frac{1}{4}$, we have that
$$P(AB) = P(A)P(B), \quad P(AC) = P(A)P(C), \quad P(BC) = P(B)P(C)$$
but $$P(ABC) \neq P(A)P(B)P(C).$$
