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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...

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    $\begingroup$ 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 $P(A \cup B) \le P(A \cup B \cup C)$. $\endgroup$ – Greg Martin Jun 8 at 22:05
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    $\begingroup$ 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. $\endgroup$ – lasik43 Jun 8 at 22:15
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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 $p \le \frac12$.

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