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?
 A: $P(ABC)=P(A)P(B)P(C)$ 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.
A: 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 $$P(ABC) \neq P(A)P(B)P(C).$$
A: Here is a quick offhand example. Toss a fair coin once. Let $A$ be the event "head," $B$ the event "tail" and $C$ the event "head and tail." (Of course $C$ has probability $0$.)  That counterexample is trivial and boring, so we produce a non-trivial one.
Let $A$ and $B$ be events such that $P(A)=P(B)=1/2$ and $P(A\cap B)=1/5$. It is clear that $A$ and $B$ are not independent and that $P(A\cup B)=4/5$. Let $C=A\cup B$. Then 
$$P(A\cap B\cap C)=P(A\cap B)=\frac{1}{5}\qquad\text{and}\qquad P(A)P(B)P(C)=\frac{1}{5}.$$
A: Your second equality seems fishy. What if $P(A_3)=0$ and $A_1=A_2$, with $P(A_1)=P(A_2)=P(A_1\cap A_2)=\frac{1}{2}$. This certainly is an example of why you need the full definition.
