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With a cyclically oriented independence digraph of events, each event is independent of one of the other two. Independence is symmetric, so all three pairs are independent events. This does not imply mutual independence. Put $x=\mathbb{P}(A_1\cap A_2\cap A_3)$. The inclusion-exclusion principle givesbecause $x\le \mathbb{P}(A_1\cap A_2)=a^2$. Thus the oriented case guarantees positive avoidance exactly when $0<a<1/2$.

For sharpness, for any $1/2\le a<1$ prescribe the law of the three event indicators by assigning mass $(1-a{)}^2$ to each of the three singleton success patterns, $(1-a)(2a-1)$ to each doubleton, $1-3a+3a^2$ to the tripleton, and zero to the zero-success pattern. All masses are nonnegative and their sum is one. Each marginal isand each pair intersection has mass $(1-a)(2a-1)+(1-3a+3a^2)=a^2$. These events satisfy the oriented independence conditions and have zero avoidance probability.

The unoriented triangle imposes no independence conditions at all: every vertex is adjacent to every other. The union bound gives positive avoidance if $3a<1$. To see that this is sharp for every $a\ge 1/3$, use uniform measure on the circle of circumference one and take three arcs of length $a$ starting at $0,1/3,2/3$. They cover the circle. Hence the unoriented case guarantees positive avoidance exactly when $0<a<1/3$.