= Equality in Boole's inequality
{title2=$\mathbb P(A_i\cap A_j)=0\quad(i\ne j)$}
For finitely many <events>, or a <countable> family with finite sum of individual <probabilities>, equality in the uncapped <union bound> holds exactly when every pairwise <intersection> has <probability> zero. Equivalently, at most one <event> occurs <almost surely>. Pairwise <disjoint events> give equality, but literal disjointness is stronger than necessary: shared outcomes of <probability> zero do not change either side. For the capped bound, if the sum exceeds one, equality instead means that the <union> is an <almost sure event>; overlap can then have positive <probability>.
The cap and the disjoint case are both sharp. If the individual <probabilities> sum to at most one, put the <events> on separate pieces of a <sample space> to attain their sum. The twelve-outcome disjoint example does this with two <events> of <probability> $1/3$. If the sum exceeds one, a covering <union> can attain the cap: on the same twelve-outcome <sample space>, $A=\{1,\ldots,8\}$ and $B=\{5,\ldots,12\}$ have sum $4/3$ and exact <union> <probability> one. These are sharpness examples, not an assertion that every overlapping family attains the bound. Positive-<probability> <disjoint events> are also not <independent events>, because their joint <probability> is zero while the product of their individual <probabilities> is positive.
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