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Events turn Boolean operations into operations on sets. For a die with sample space $U=\{1,2,3,4,5,6\}$, let $A=\{2,4,6\}$ be the event “even outcome” and $B=\{4,5,6\}$ the event “outcome at least four”. At outcome $u$, use the indicator functions $P=1_A(u)$ and $Q=1_B(u)$. ThenThus logical conjunction becomes set intersection, logical disjunction becomes set union, exclusive or becomes symmetric difference, and logical negation becomes the complement of a set within $U$. Outcome four satisfies both events, so inclusive logical disjunction includes it and exclusive or excludes it. If the die is fair, these four events have probabilities $1/3$, $2/3$, $1/3$, and $1/2$, respectively. No assumption of independent events is needed for the set identities; indeed $A$ and $B$ here are not independent, since $1/3\ne (1/2)(1/2)$.

The Borel-Cantelli lemmas turn a series of event probabilities into a statement about an entire infinite sequence of observations. A summable sequence of probabilities forces only finitely many occurrences almost surely, regardless of independence. A divergent series forces infinitely many occurrences when the events are mutually independent events. The precise event being measured is the limsup of events.

Imagine alarms with probabilities $p_n=1/(n+1{)}^2$ or $p_n=1/(n+1)$, starting at $n=1$. Both individual probabilities tend to zero. Nevertheless, the first alarm sequence eventually stops almost surely, whereas the second keeps returning almost surely if its events are mutually independent events. Reusing one uniform random variable for every alarm instead produces the shared-uniform counterexample to the second Borel-Cantelli lemma. Thus the individual probabilities alone do not determine long-run recurrence.

If the events $A_n$ are mutually independent events and ${\sum }_{n=1}^{\infty }\mathbb{P}(A_n)=\infty$, then $\mathbb{P}({\mathrm{limsup}}_n{A}_n)=1$. For an infinite family, mutual independence means every finite subfamily satisfies the factorization defining independent events.^[ref]

Proof outline: for $M\ge m$, independent events and $1-x\le e^{-x}$ giveContinuity of probability makes each infinite tail union have probability one; their countable intersection has probability one as well.

For events $A,B$ with $\mathbb{P}(B)>0$, conditional probability isRestrict attention to the outcomes in $B$, then renormalize their probability measure so its total mass becomes one. The resulting rule $A\mapsto \mathbb{P}(A\mid B)$ is itself a probability measure. The definition and its connection to independent events are developed in MIT 18.05, Reading 3: Conditional Probability, Independence and Bayes' Theorem.

For the two-dice model, let $A$ mean the sum is at least nine and $B$ mean the first result is six. Of the six outcomes in $B$, four belong to $A$: the second die can be 3, 4, 5, or 6. Thus $\mathbb{P}(A\mid B)=2/3$. Reversing the information gives a different answer: four of the ten outcomes in $A$ have first result six, so $\mathbb{P}(B\mid A)=2/5$. Both ratios use the same overlap; their denominators describe different information.

The definition requires positive conditioning mass. For a continuously distributed random variable $X$, it cannot directly define $\mathbb{P}(A\mid X=x)$ when $\mathbb{P}(X=x)=0$. A regular conditional distribution or conditional density supplies the appropriate extension, with the usual almost everywhere qualification. Nor does conditioning by itself justify causal inference: changing which outcomes are considered is different from physically changing how the experiment is generated.

Two events are disjoint when they cannot both occur. For a finite family of pairwise disjoint events, the probability of their union is the sum of their probabilities, so that sum is at most one. This condition differs from the condition defining independent events.

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, $A=\{1,\dots ,8\}$ and $B=\{5,\dots ,12\}$ 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.

For a finite family of events, the probability of their union is the alternating sum of their intersection probabilities, with the same pattern as the inclusion-exclusion principle. Expand the indicator function of the union and apply linearity of expectation; this does not require independent events.

Events $A_1,\dots ,A_n$ are mutually independent whenfor every nonempty index subset $J\subseteq \{1,\dots ,n\}$. For two events and $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning whether $B$ occurred does not change the chance of $A$.

For independent uniform dice, the events that the first result is an even integer and that the second result is an even integer each have probability $1/2$; their intersection has probability $9/36=1/4$. In contrast, "first result is an even integer" and "first result is an odd integer" are disjoint events of positive probability, so they cannot be independent events. Learning one occurred completely rules out the other.

Pairwise independence is weaker than mutual independence. Take two independent fair bits $X,Y$. The three events $X=0$, $Y=0$, and $X=Y$ each have probability $1/2$ and every pair has intersection probability $1/4$. Their triple intersection is the single outcome $(0,0)$, of probability $1/4$ rather than $1/8$. Checking pairs alone misses the joint constraint.

A permissive gate allows conduction when it is open. Suppose an ion channel conducts only when all $q$ gates are open, where $q$ is a positive integer. If the gate-open events are mutually independent events and each has the same open probability $p$ at the time considered, thenThis follows from the multiplication rule for independent events. Identical transition rates alone do not guarantee independence: the initial gate states must also be independently prepared, and the model must not couple their transitions. Gates sharing a fluctuating membrane potential can be correlated; the simple product requires an appropriate independence assumption, for example independent gates at an externally fixed voltage.

At $p=0.5$, channels requiring $q=1,2,4$ permissive gates have open probabilities $0.5$, $0.25$ and $0.0625$. For four independent gates there are $2^4=16$ possible open/closed configurations, equally probable at $p=0.5$; only the all-open configuration conducts. Thus $0.5$ is the gate-open probability, while $0.0625$ is the whole-channel open probability. Neither is a fractional state of one observed channel. Among $160$ independently prepared identical four-gate channels, the expected value of the open count is $160/16=10$, although an actual count fluctuates.

For unequal independent gate-open probabilities $p_j$, use $P_{\mathrm{open}}={\prod }_j{p}_j$ rather than $p^q$. This is the distinction behind the Hodgkin-Huxley model's $m^{3}h$: $m$ and $h$ are different activation and inactivation-availability variables, rather than one common $p$. Its $n^4$ uses four identical activation factors. Independence and the all-open requirement are model assumptions; correlated gates or alternative conducting configurations require a different calculation. The simple product is useful because a modest gate-open probability can produce sharply suppressed whole-channel availability without any additional gate interaction.

On $\{0,1{\}}^{\mathbb{N}}$, the event of infinitely many heads isIt says that after every index there is another head. For each $n$, the same intersection can start at $m=n$, so $A\in \sigma (X_n,X_{n+1},\dots )$; therefore $A$ is a tail event. This verifies measurability directly using countable unions and intersections.

Assume now that the coordinates are fair independent Bernoulli trials. For fixed $m$, let $B_{m,r}$ be the event of at least one head among $X_m,\dots ,X_{m+r-1}$. Its complement requires $r$ tails, soThe last equality follows from continuity of probability. Now $A={\bigcap }_{m\ge 1}{B}_m$, a countable intersection of almost sure events, and hence $\mathbb{P}(A)=1$. This determines which of the two possibilities in Kolmogorov's zero–one law occurs. Equivalently, the Borel-Cantelli second lemma applies to the independent events $\{X_j=1\}$.

Nevertheless, no finite prefix certifies membership in $A$: every finite record can be extended by all tails or by alternating heads and tails. Under the fair independent model, conditioning on any exact finite prefix leaves $\mathbb{P}(A)=1$, since $A$ is independent of the initial sigma-algebra. The excluded all-tails continuation has probability zero, but remains a possible member of the sample space. Logical determination by a finite record and a conclusion holding almost surely are different statements.

Let $(B_i)$ be a countable measurable partition of the sample space, with positive probabilities. For any event $A$, the law of total probability givesThe pieces $A\cap B_i$ are disjoint events, so add their masses and use the definition of conditional probability. Zero-mass partition pieces can be omitted without changing the answer.

For the coin selection in probability, let $B$ mean the $3/4$-head coin was selected and let $H$ mean the first toss was heads. Then $\mathbb{P}(H)=\mathbb{P}(H\mid B)\mathbb{P}(B)+\mathbb{P}(H\mid B^c)\mathbb{P}(B^c)=11/20$. Each conditional answer describes one possible hidden situation; its weight describes how often that situation occurs under the model. This decomposition is useful whenever a simple conditional model sits behind a more complicated marginal probability distribution.

The law of total probability combines local conditional probabilities into one overall probability. The measurable partition must cover the sample space with pairwise disjoint events: every elementary outcome belongs to exactly one part. Its probabilities are the weights of a weighted mean, and they sum to one.^[ref]

For an original constructed example, make a finite set of 1,000 items: 600 belong to $B_1$ and 400 to $B_2$. Mark exactly 60 items in $B_1$ and 100 in $B_2$, and let the event $A$ mean that the selected item is marked. Select an item using the uniform distribution on a finite set. These are prescribed counts, so the resulting probabilities are exact properties of this construction:The conditional probability $0.1$ uses only the 600 items in $B_1$ as its denominator. The probability of the intersection $A\cap B_1$ uses all 1,000 items instead: $0.6\times 0.1=0.06$. Similarly, $\mathbb{P}(A\cap B_2)=0.4\times 0.25=0.10$. Adding the two disjoint events gives

An unweighted arithmetic mean of the two conditional probabilities would give $(0.1+0.25)/2=0.175$. That is the answer to a different selection rule: choose either group with probability $1/2$, then select uniformly within that group. Here the first group contains more items, so selecting uniformly among all items gives it weight $0.6$. In fact, if its weight were $w$, the overall probability would be $0.1w+0.25(1-w)=0.25-0.15w$. Moving the weight from $1/2$ to $0.6$ lowers the result by $0.015$, explaining exactly why $0.175$ is too large for the original model.

The same construction is a mixture distribution: first select a group with its mixture weight, then select an item within that group. Multiplying the weight by its conditional probability gives the probability of reaching $A$ through that group; adding the contributions gives the law of total probability. For example, an item in $B_1$ has selection probability $0.6/600=1/1000$, and an item in $B_2$ has selection probability $0.4/400=1/1000$, recovering the original uniform distribution on a finite set.

No assumption of independent events is needed. In this construction $\mathbb{P}(A\mid B_1)=0.1$ differs from $\mathbb{P}(A)=0.16$, so $A$ and $B_1$ are not independent events. Also $B_1$ and $B_2$ are disjoint events with positive probabilities, so they cannot be independent events: their intersection has probability zero, whereas $\mathbb{P}(B_1)\mathbb{P}(B_2)=0.24$. The measurable partition supplies exclusive alternatives, not an independence assumption.

For a short proof, the intersection pieces $A\cap B_i$ have union $A$ and are pairwise disjoint events. Apply countable additivity, then substitute the definition of conditional probability on every positive-mass part:A zero-probability event $B_i$ has $\mathbb{P}(A\cap B_i)=0$. Omit it: writing $\mathbb{P}(A\mid B_i)$ would divide by zero. Countability ensures that the union of the omitted zero-probability events also has probability zero. Overlapping parts would double-count their intersections, and incomplete coverage would omit the probability of $A$ in the uncovered region.^[ref]

Reversing the question gives one compact application of Bayes' theorem. Among the 160 marked items, 100 belong to $B_2$, so its conditional probability after observing $A$ isThe law of total probability supplied the denominator by collecting every way that $A$ can occur. Its usefulness lies in choosing a measurable partition where the conditional probabilities are simple, then combining them without losing the relative sizes of the alternatives.

For events $A_n$ in a probability space, the infinitely-often event isIts complement of an event is ${\bigcup }_{m\ge 1}{\bigcap }_{n\ge m}{A}_n^c$: after some outcome-dependent index, no more events occur.^[ref]

Deleting finitely many events leaves the limsup of events unchanged. It is therefore a tail event of the sequence of indicator random variables $1_{A_n}$, whether or not those random variables are independent. For independent events, the Kolmogorov zero-one law restricts its probability to zero or one; the Borel-Cantelli lemmas determine which alternative holds from the series of probabilities. Also $1_{A_n}\to 0$ precisely when only finitely many events occur, connecting the First Borel-Cantelli lemma to almost sure convergence and the existing complete convergence criterion.

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.

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 $a^2$.

Normalize the total channel density to one. For independent permissive gates in an ion channel, the probability that the $M$ gate is open is $m$, and the probability that the $H$ gate is open is $h$. The joint probability is the product of the appropriate open or closed probabilities, givingIn particularThe original PDF really prints $j$, rather than $1-j$, on the final factor. That expression fails normalization of a probability distribution and does not give the closed-$H$ probability at $j=0$; the displayed correction is necessary. If $S_{ij}$ are literal densities with conserved total density $N$, multiply each expression by $N$.

The product additionally assumes initially independent events and independent gate transitions. The transition scheme alone permits initially correlated gates and does not force this factorization at every time. The all-closed initial preparation used below satisfies the required product assumption.

Two independent events satisfy $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$. If $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning that $B$ occurred does not change the probability of $A$. The product definition also covers zero-probability events without dividing by zero.

The 36 ordered outcomes are equiprobable. For $2\le s\le 12$, let $c_s=6-|s-7|$ count outcomes with sum $s$. Then $\mathbb{P}(A_s)=c_s/36$ and $\mathbb{P}(B_i)=1/6$ for $1\le i\le 6$. Their joint probability is $1/36$ if $1\le s-i\le 6$, and zero otherwise.

If the joint probability is zero, independence is impossible in these ranges because both marginal probabilities are positive. If it is $1/36$, the independent events condition requires $1/36=c_s/216$, or $c_s=6$. This happens only for $s=7$, and then all six values of $i$ allow the partner $7-i$. Therefore within the possible sums and facesIf arbitrary integer labels outside these ranges are included, an impossible sum or face defines an empty event, which is trivially independent of every event.

Events $(A_n)$ are independent events if for every finite list of distinct indices $i_1,\dots ,i_k$, $\mathbb{P}({\bigcap }_j{A}_{i_j})={\prod }_j\mathbb{P}(A_{i_j})$. Random variables are independent random variables if for every such list and Borel sets $B_j\subseteq \mathbb{R}$, $\mathbb{P}(X_{i_j}\in B_j\text{ for all }j)={\prod }_j\mathbb{P}(X_{i_j}\in B_j)$. This is equivalent to independence of their generated sigma-algebras; pairwise independence alone is insufficient.

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 $0<a<1/3$.

Independence or another ergodicity hypothesis is necessary. Read literally for an arbitrary translation-invariant site percolation measure, the assertion is false: the mixture assigning probability $1/2$ to the all-open configuration and $1/2$ to the all-closed configuration is translation-invariant, but the invariant event of an infinite percolation cluster has probability $1/2$. We prove the zero–one law for the invariant independent law used in part (ii); the proof also works for the stated finite-range independent alternative.

Let $\Gamma$ be the group of translations, or the specified group of graph automorphisms, and let $A$ be invariant under it, up to null sets. An event determined by the states in a finite vertex set is a cylinder set. For every $\epsilon >0$, approximation in the product sigma-algebra provides a cylinder set $B$, supported on a finite set $F$, with $\mathbb{P}(A△B)<\epsilon$. Choose $g\in \Gamma$ so that $F$ and $gF$ are disjoint. Then $B$ and $gB$ are independent events, have equal probabilities, and $\mathbb{P}(A△gB)<\epsilon$. HenceandLetting $\epsilon \downarrow 0$ proves the zero–one law $\boxed{\mathbb{P}(A)\in \{0,1\}}$. This is translation ergodicity of Bernoulli percolation.

For a quasi-transitive graph, the required displacement exists: put $F$ in a ball about a vertex $v$; its orbit is infinite, so an automorphism can move $v$, and therefore that whole ball, beyond any prescribed finite neighborhood. A $k$-independent site percolation law means that the sigma-algebras generated by vertex sets at graph distance greater than $k$ are independent. Move $F$ far enough that $\mathrm{dist}(F,gF)>k$ and the same proof applies. Thus finite-range independence does give the zero–one law, although it will not give uniqueness.