# Intuition of mutually independent events

Three events $A, B$ and $C$ are said to be mutually independent, if $$\text{Pairwise independence}\implies\begin{cases} P(A\cap B)=P(A).P(B)\\ P(A\cap C)=P(A).P(C)\\ P(B\cap C)=P(B).P(C)\end{cases}\\ \text{ and }P(A\cap B\cap C)=P(A).P(B).P(C)$$ or $$\text{Pairwise independence}\implies\begin{cases} P(A|B)=P(A|C)=P(A)\\ P(B|A)=P(B|C)=P(B)\\ P(C|A)=P(C|B)=P(C)\end{cases}\\ \text{ and } P(A|B\cap C)=P(A)\\ P(B|A\cap C)=P(B)\\ P(C|A\cap B)=P(C)$$

I think I understand the meaning of pairwise independence, as each even does not affect the other event. But, why do we need the additional condition other than that of pairwise independence for mutual independence ?

How can events being independent of each other and being not independent with the intersection of other two events ?

Or why do we even need the condition that the events to be independent to the intersection of the other two events for being mutually independent ?

Note: I am not looking for any algebraic solution, just trying to make sense of my doubt

Suppose we flip two (distinguishable) coins. Let $A$ be the event that the first coin shows heads, $B$ the event that the second coin shows heads, and $C$ the event that both coins show the same. These events are pairwise independent but not mutually independent.
• Or if A is first coin shows heads. B is second coin shows tails. C is both coins the same. In the case $P(A \cap B \cap C)$ is zero rather than be greater than the product of the individual probabilities. – Simon Terrington Aug 21 '18 at 21:08
• @SimonTerrington Yup, both ways work: either $C$ (as I've defined it) or its complement is independent of each of $A$ and $B$ but not independent of $A \cap B$. – BallBoy Aug 21 '18 at 21:10