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Pairwise independent events describe a subtle kind of uncertainty: learning that one event occurred leaves each other event's probability unchanged, yet learning that two occurred can reveal a third completely. Checking every pair therefore answers a weaker question than checking the whole collection. The parity construction of pairwise independent bits makes this distinction visible with just four equally likely outcomes.

A family of events $A_i$ is pairwise independent whenIf $\mathbb{P}(A_j)>0$, this is equivalent to the conditional probability identity $\mathbb{P}(A_i\mid A_j)=\mathbb{P}(A_i)$. The product definition also handles events of probability zero. Independent events, meaning mutually independent events, require the same product rule for every finite subfamily, including triples. For three events, the three pair equations and the triple equation are four separate requirements. ^[ref].

For a worked example, take independent fair bits $U,V$: the sample space is $\{00,01,10,11\}$ with probability $1/4$ per outcome. Set $X=U$, $Y=V$, and $Z=U\oplus V$, using exclusive or, which is one precisely when its inputs differ. The events $A=\{X=1\}$, $B=\{Y=1\}$, and $C=\{Z=1\}$ each contain two outcomes. Their pair intersections areThus each event has probability $1/2$ and each pair intersection has probability $1/4=(1/2{)}^2$: these are pairwise independent events. But $X=Y=1$ forces $Z=0$, soThe three are not independent events. This is the complementary-third-event version of the two-toss example in Tsitsiklis and Jaillet, MIT Introduction to Probability, Lecture 3, slides 7–8.

Another useful diagnostic preserves a special common-intersection case. Suppose three events have common probability $s$ and every pair and triple intersect in the same event, of probability $q$. Pairwise independence requires $q=s^2$, while mutual independence additionally requires $q=s^3$. Both hold only for $s=0$ or $s=1$. For a concrete nontrivial instance, assign probability $1/4$ to each of four outcomes, let all three events contain the first outcome, and give each event one different additional outcome. Then $s=1/2$ and $q=1/4$: the pair equations hold, while the triple equation fails. Unlike the exclusive or example above, this triple intersection is nonempty.

Pairwise independent events satisfy $\mathbb{P}(A_i\cap A_j)=\mathbb{P}(A_i)\mathbb{P}(A_j)$ for every distinct pair. Independent events satisfyfor every nonempty subset $I$ of the indices. For three events this includes the triple intersection, not just the three pairwise intersections.

Write $A_i=B_i\cup C$. Disjointness gives $\mathbb{P}(A_i)=p+q$, while every intersection of two or three distinct $A_i$ is exactly $C$. Thus pairwise independence is equivalent to $q=(p+q{)}^2$. Since $p+q\ge 0$,For $0\le q\le 1/16$, this is an admissible nonnegative $p$: putting $u=\sqrt{q}\in [0,1/4]$ gives $3p+q=3u-2u^2\le 5/8<1$. The four disjoint events can therefore have the stated probabilities, with the remaining probability assigned to their complement. This also covers $q=0$.

Full independence would additionally require $q=(p+q{)}^3$. Together with $q=(p+q{)}^2$, this forces $p+q$ to be zero or one. The first case gives $p=q=0$; the second gives $q=1,p=0$. Neither allows positive $p,q$. There are no such mutually independent events with $p>0$ and $q>0$.