If $A, B$ are independent and $A, C$ are independent, how is pairwise/multual independence conncted to the dependence between $A$ and $B\cap C$? Toss 3 coins. A={first=second}, B={second=third}, C={third=one}
Then,
$$P(AB)=P(ABC)=\frac{1}{4}=P(A)P(B)$$
$$P(AC)=P(ABC)=\frac{1}{4}=P(A)P(C)$$
But,
$$\frac{1}{4} = P(ABC) \neq P(A)P(BC) \\\text{($A, B$ independent and $A, C$ independent $\not\Rightarrow A$ and $B\cap C$ independent)}\tag1$$
$$\frac{1}{4} = P(ABC) \neq P(A)P(B)P(C) \\\text{  (pairwise independence} \not\Rightarrow \text{mutal indepence}) \tag2$$
I wonder how (or if it is possible) to explain/relate $(1)$ and $(2).$
 A: Well, if $A$, $B$, and $C$ were mutually independent, then $A$ would also be independent from $B \cap C$. In general, if a bunch of events $A_1, A_2, \dots, A_n$ are mutually independent, and you use disjoint subsets of them to come up with complicated expressions (like $A_1 \cap A_3 \cup (A_7 - A_{12})$ and $(A_4 \cup A_6) - (A_2 \cup A_8)$) then those expression will be independent events.
So for $A$ and $B \cap C$ not to be independent, we want $A,B,C$ not to be mutually independent.
On the other hand, it's possible for $(A,B)$, $(B,C)$, and $(A,B\cap C)$ to be independent pairs, and still not have mutual independence.
Consider the experiment where we roll a fair die with the $6$ sides $\{\text a, \text b, \text c, \text{ab}, \text{bc}, \text{abc}\}$. Let $A$ be the event that the side rolled has the letter $\text a$ on it; define $B$ and $C$, similarly. Then:

*

*$\Pr[A] = \frac12$, $\Pr[B] = \frac23$, and $\Pr[A \cap B] = \frac13$, so $A$ and $B$ are independent.

*$\Pr[B] = \frac23$, $\Pr[C] = \frac12$, and $\Pr[B \cap C] = \frac13$, so $B$ and $C$ are independent.

*$\Pr[A] = \frac12$, $\Pr[B\cap C] = \frac13$, and $\Pr[A \cap B\cap C] = \frac16$, so $A$ and $B \cap C$ are independent.

*But $\Pr[\overline{A} \cap \overline{B} \cap \overline{C}] = 0$, which is the easiest way to see that $A,B,C$ are not mutually independent, because $\Pr[\overline A] \cdot \Pr[\overline B] \cdot \Pr[\overline C] = \frac12 \cdot \frac13 \cdot \frac12 \ne 0$.

(For an equivalent example, roll an ordinary fair die with sides $1$ through $6$. Let $A$ be the event "the value rolled is even". Let $B$ be the event "the value rolled is not a multiple of $3$". Let $C$ be the event "the value rolled is less than $4$".)
In summary: when $(A,B)$ and $(B,C)$ are independent pairs, mutual independence implies the independence of $A$ and $B \cap C$, but the independence of $A$ and $B \cap C$ does not imply mutual independence.
A: They are related in that if $A,B,C$ are mutually independent then $A$ and $B\cap C$ are independent:
$$P(A,B,C)=P(A)P(B,C).$$
And you just  found an example of pairwise independent events that violate an implication of mutual independence, i.e. $P(A,B,C)=P(A)P(B,C),$ and thus are not mutually independent.
