= Pairwise-independent three-event avoidance
{title2=$\mathbb P(\bigcap_{i=1}^3A_i^c)\geq(1-a)(1-2a)$}
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 $x\leq a^2$ gives the displayed lower bound. It is strictly positive for $0<a<1/2$. For $1/2\leq a<1$, 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 $a^2$.
Back to article page
