assume that
and
the question is, does
????????
I don't think so.
Try tossing a fair die.
Let
and
I get 1/6 =1/6 for the first two
but then I get 1/6 and 1/4 on the last one.
I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that (same with A,C and B,C), and .
Then the property is proved as follows (for instance):
, and the three (main) events on the right-hand side are disjoints, hence
,
and you can compute the latter probabilities using (and similarly for the last term) and the assumptions (recalled above).
You should get , or in fact , which is the same...