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:
 * $$\forall \omega \in \Omega: \Pr \left({\omega}\right) = \frac 1 4$$

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 $$\Pr \left({A}\right) = \Pr \left({B}\right) = \Pr \left({C}\right) = \frac 1 2$$.

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

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 $$\Pr \left({A \cap B \cap C}\right) = \Pr \left({\left\{{1}\right\}}\right) = \frac 1 4$$.

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

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