Originally Posted by
Laurent I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that $\displaystyle P(A\cap B)=P(A)P(B)$ (same with A,C and B,C), and $\displaystyle P(A\cap B\cap C)=P(A)P(B)P(C)$.
Then the property is proved as follows (for instance):
$\displaystyle P(A\cap(B\cup C))=P((A\cap B\cap C)\cup(A\cap B^c\cap C)\cup (A\cap B\cap C^c))$, and the three (main) events on the right-hand side are disjoints, hence
$\displaystyle P(A\cap(B\cup C))=P(A\cap B\cap C)+P(A\cap B^c\cap C)+P(A\cap B\cap C^c)$,
and you can compute the latter probabilities using $\displaystyle A\cap B^c\cap C=(A\cap C)\setminus (A\cap B\cap C)$ (and similarly for the last term) and the assumptions (recalled above).
You should get $\displaystyle \cdots=P(A)P(B\cup C)$, or in fact $\displaystyle P(A)(P(B)+P(C)-P(B)P(C))$, which is the same...