Pairwise-independent three-event avoidance ($\mathbb{P}({\bigcap }_{i=1}^3{A}_i^c)\ge (1-a)(1-2a)$)

For three pairwise independent events of common probability $a$, write $x=\mathbb{P}(A_1\cap A_2\cap A_3)$. Inclusion-exclusion principle gives avoidance probability $1-3a+3a^2-x$, and $x\le a^2$ gives the displayed lower bound. It is strictly positive for $0<a<1/2$. For $1/2\le a<1$, a symmetric atomic law with no zero-success atom has mass $(1-a{)}^2$ on each singleton success pattern, $(1-a)(2a-1)$ on each doubleton, and $1-3a+3a^2$ on the tripleton. These masses sum to one, have marginal $a$, and have pair intersections $a^2$.