Find $P(A \cup B \cup C)$ given they are pairwise independent but not mutually independent Find an upper bound for $P(A \cup B \cup C)$ given they are pairwise independent but not mutually independent. A, B and C are fixed and I know to use the inclusion-exclusion rule to get their union, but I don't know how to get the intersection since they are pairwise independent but not mutually independent
 A: The question is not 100% clear, but I will try to answer it assuming that the probabilities of $A$, $B$ and $C$ are fixed, and we are trying to see how high $P(A\cup B\cup C)$ can go.
Let's suppose that $a=P(A)\ge b=P(B)\ge c=P(C)$. From inclusion-exclusion principle:
$$\begin{array}{rcl}P(A\cup B\cup C)&=&P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)\\&\le&P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(B\cap C)\\&=&a+b+c-ab-ac\end{array}$$
On the other hand, this limit can be achieved: see the following Venn diagram:

(note that $a-ab-ac+bc\ge a^2-ab-ac+bc=(a-b)(a-c)\ge 0$).
A: I originally posted this as an edit but was asked to repost as an answer, so here we go!
The bound derived by @ Stinking Bishop is indeed tight, provided the condition $b+c \leq 1$ is satisfied. To see this, note that $a+b+c-ab-ac=1-(1-a)(1-b-c)$, which is a valid probability bound  ($\leq 1$) if and only if $b+c \leq 1$ or the smallest two probabilities add up to at most one (assuming $a<1$ otherwise the union probability is trivially one).
So it would be more precise to say that the tight bound on the union of three pairwise independent (but not mutually independent) Bernoulli events, given their fixed marginal probabilities is $$\min\left(a+b+c-a(b+c),1\right), \quad \mbox{where}\quad 1 > a \geq b \geq c \geq 0$$
Interestingly, this result can be generalized beyond three events to an arbitrary number of events. The wikipedia page on pairwise independence shows that the tight bound on the union of $n$ pairwise independent Bernoulli events $A_1,\ldots, A_n$ with given probabilities $p_1,p_2,\ldots,p_n$ (assumed to be arranged in order of increasing probabilities) is given as:
$$\max P({\cup}_i A_{i})  =  \min\left(\sum_{i=1}^n p_{i}-p_{n}\left(\sum_{i=1}^{n-1} p_{i}\right),1\right)$$
which is at most one when the sum of the smallest $n-1$ probabilities is at most one.
The proof is non-trivial and can be found here.
