Event independence I have the following in my lecture notes:
$$\Omega = \left\{ E_r, E_g, E_b, E_m \right\}$$
The above is for the colour of the faces of a tetrahedric die: red, green, blue, mixed (red and green and blue). Define $$R = \left\{ E_r,E_m\right\},\,\,\, G = \left\{ E_g, E_m \right\}, \,\,\, B = \left\{ E_b, E_m \right\}$$
It says that the above are pairwise independent but not collectively. I am probably wrong in my understanding here. Pairwise independent I thought is that the elementary events within these composite events are independent, i.e. if you take $R$ then $E_r$ and $E_m$ are independent. But apparently this understanding is flawed, because my lecturer follows to say that $$P(RG) = P(R)\cdot P(G)$$ which implies that $R$ and $G$ are independent. 
But I can't see how this is the case, as $R$ and $G$ have an elementary event in common, namely $E_m$. What am I missing here?
 A: $P(R\cap G)=P(\{E_r,E_m\}\cap\{E_g,E_m\})=P(\{E_m\})=\frac14$
$P(R)\times P(G)=\frac24\times\frac24=\frac14$
The outcomes are equal so events $R$ and $G$ are independent.
Likewise it can be shown that $R$ and $B$ are independent and that $G$ and $B$ are independent. That means: $R,G,B$ are pairwise independent.

$P(R\cap G\cap B)=P(\{E_m\})=\frac14$
$P(R)P(G)P(B)=\frac24\frac24\frac24\neq\frac14$
This shows that $R,G,B$ are not dependent.

Events are subsets of $\Omega$. Observe that e.g $E_r$ is not a subset but an element of $\Omega$.
A: The events $R,G,B$ are pairwise independent iff any two of them (or any pair of them) are (is) independent. Mathematically $$P(RG)=P(R)P(G),\, P(RB)=P(R)P(B), \,P(GB)=P(G)P(B)$$ The events $R,G,B$ are independent or mutually independent (or collectively as you say) if any two of them and if any three of them are independent when taken together.
The following holds 

mutually independent $\implies \\ \not \Longleftarrow$ pairwise independent

Now, independence is very different than disjoint-ness. As your example shows two events can be independent even if they are not disjoint! Independence does not imply disjoint and disjoint does not imply independent.


*

*Two events $R,G$ are independent iff: $P(RG)=P(R)P(G)$. Independence means that "knowing $R$ occurs, you do not know more or less about $G$ occured or not. It is equally probable that $G$ occurs and you can discard the information that $R$ occured."  (very informaly said).

*Two events $R,G$ are disjoint iff: $P(RG)=0$. Disjoint-ness means that $R$ and $G$ cannot occur together. So, actually if two events (that occur with positive probability) are disjoint they cannot be independent(!) because knowing that the one occured implies directly that the other did not occur.


It is an (easy) exercise to show, that

If $R,G$ are disjoint, then they are independent iff $P(R)=0$ or $P(G)=0$.

