# Property of pairwise independent but not independent events

Let $$(\Omega, \mathcal A, P)$$ be a probability space and $$A,B,C \in \mathcal A$$ with the properties
a) $$A,B,C$$ are pairwise independent,
b) $$A\cap B \cap C = \emptyset$$,
c) $$P(A) = P(B) = P(C) =: p$$.
Then $$p \leq \frac{1}{2}$$.

I tried to use the standard trick and expand with the inclusion-exclusion principle: $$0 = P(A\cap B \cap C) = P(A \cup B \cup C) + P(A\cap C) + P(A\cap B) + P(B\cap C) - P(A) - P(B) - P(C) = P(A\cup B \cup C) + 3p^2 - 3p$$ and hence $$P(A\cup B \cup C) = 3p - 3p^2$$. Here I got stuck; using $$0 \leq P(A \cup B \cup C) \leq 1$$ didn't get me anywhere. I am kind of skeptical of the statement but also could not find a counterexample. Any help appreciated...

• Hint: you know more than $P(A \cup B \cup C) \le 1$: you also know that $P(A) \le P(A \cup B \cup C)$ and (more conveniently) that $P(A \cup B) \le P(A \cup B \cup C)$. – Greg Martin Jun 8 at 22:05
• Oh of course, and then $P(A\cup B) = P(A) + P(B) - P(A\cap B)$... stupid of me not to see it. Thanks! You can post an answer if you want. – lasik43 Jun 8 at 22:15

## 1 Answer

We also know that $$P(A\cup B) \le P(A\cup B\cup C)$$, which leads to the inequality $$2p-p^2 \le 3p-3p^2$$; this is equivalent to $$2p^2 \le p$$, which implies that $$p \le \frac12$$.