Pairwise Independence does not imply Independence

Theorem
Just because all the events in a family of events in a probability space are pairwise independent, it does not mean that the family is independent.

Proof
Consider throwing a fair four-sided die.

This gives us an event space $\Omega = \left\{{1, 2, 3, 4}\right\}$, with each $\omega \in \Omega$ equally likely to occur:
 * $\forall \omega \in \Omega: \Pr \left({\omega}\right) = \dfrac 1 4$

Consider the set of events $\mathcal S = \left\{{A, B, C}\right\}$ where $A = \left\{{1, 2}\right\}, B = \left\{{1, 3}\right\}, C = \left\{{1, 4}\right\}$.

We have that $\Pr \left({A}\right) = \Pr \left({B}\right) = \Pr \left({C}\right) = \dfrac 1 2$.

We have that $\Pr \left({A \cap B}\right) = \Pr \left({A \cap C}\right) = \Pr \left({B \cap C}\right) = \Pr \left({\left\{{1}\right\}}\right) = \dfrac 1 4$.

Thus:
 * $\Pr \left({A}\right) \Pr \left({B}\right) = \Pr \left({A \cap B}\right)$
 * $\Pr \left({A}\right) \Pr \left({C}\right) = \Pr \left({A \cap C}\right)$
 * $\Pr \left({B}\right) \Pr \left({C}\right) = \Pr \left({B \cap C}\right)$

Thus the events $A, B, C$ are pairwise independent.

Now, consider $\Pr \left({A \cap B \cap C}\right) = \Pr \left({\left\{{1}\right\}}\right) = \dfrac 1 4$.

But $\Pr \left({A}\right) \Pr \left({B}\right) \Pr \left({C}\right) = \dfrac 1 8 \ne \Pr \left({A \cap B \cap C}\right)$.

So, although $\mathcal S$ is pairwise independent, it is not independent.