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For disjoint pairs $A_i,B_i$ such that $A_i\cap B_j\ne \varnothing$ whenever $i\ne j$, the displayed sum is at most one. In a uniformly random permutation, the events that all of $A_i$ precedes all of $B_i$ are disjoint events, and their probabilities are the summands. Taking $B_i$ to be the complement of the members $A_i$ of an antichain recovers the LYM inequality.

Boole's inequality, also called the union bound, bounds the probability that at least one of several events occurs by adding their individual probabilities. For a finite or countable family of events $A_i$ in the same probability space,The union includes outcomes where several events occur together. There is no independence assumption: adding the individual probabilities counts such outcomes repeatedly, which can only make the sum larger. The cap at one uses the total mass of a probability measure. For infinitely many events, the sum is a nonnegative series, possibly infinite; an infinite sum gives only the bound one. This is countable subadditivity of a measure specialized to a probability measure. See Probability: Theory and Examples (fifth edition) by Rick Durrett (Section 1.1, Theorem 1.1.1(ii), printed p. 2).

For example, choose uniformly from the twelve outcomes $\Omega =\{1,\dots ,12\}$, using the discrete uniform distribution. Put $A=\{1,2,3,4\}$. If $B=\{5,6,7,8\}$, these are disjoint events and their union has probability $8/12=2/3$, exactly the union bound. If $B=\{3,4,5,6\}$, the union has only six outcomes, so its exact probability is $1/2$, although both individual probabilities are still $1/3$ and the bound is still $2/3$. If $B=A$, the exact probability falls to $1/3$ with the same bound. These examples isolate overlap in the union bound by keeping the individual probabilities fixed.

The union bound is useful when each event is easy to control but their joint behavior is difficult to determine. A short proof uses indicator functions: at every outcome,Taking expected values and using linearity of expectation proves the finite bound.^[ref] For a countable family, the finite unions increase to the full union; continuity of probability passes the finite inequalities to the limit. The closure of a sigma-algebra under countable unions ensures that this limiting union is an event. For the continuity of probability step, see Probability: Theory and Examples (fifth edition) by Rick Durrett (Section 1.1, Theorem 1.1.1(iii), printed p. 2).

Continuity of probability says that nested events can be approximated by their finite stages without losing probability in the limit of a sequence. In a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, let $A_n,B_n\in \mathcal{F}$. ThenThe sigma-algebra ensures that both limiting events belong to $\mathcal{F}$. These are continuity from below of a measure and continuity from above of a measure. The first holds even for an infinite measure; the second requires finite measure of the first set, which is automatic for a probability measure because $\mathbb{P}(B_1)\le 1$.^[ref]

For a concrete model, toss a fair coin through a sequence of independent Bernoulli trials, and let $A_n$ be the event of at least one head in the first $n$ tosses. Its complement $B_n=A_n^c$ is the event of $n$ initial tails. On the sample space of infinite head-or-tail sequences,Every finite waiting horizon has positive tail probability, yet the event of waiting forever has probability zero. That intersection is not empty: it contains the all-tails sequence. Thus a zero-probability event need not be impossible as an outcome, and eventual success almost surely does not supply a deterministic deadline. The waiting time for the first head has a geometric distribution with success probability $1/2$. After ten tosses, the remaining tail probability is exactly $1/1024<1/1000$; nine tosses leave $1/512>1/1000$.

A proof outline exposes what makes countable additivity stronger than finite additivity of a set function. For increasing events, put $D_1=A_1$ and $D_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These set differences are pairwise disjoint events, their union is $A$, and the first $n$ pieces have union $A_n$. ConsequentlyFor decreasing events, their complements increase to $B^c$; applying the increasing result and the complement rule of probability gives $\mathbb{P}(B_n)=1-\mathbb{P}(B_n^c)\to 1-\mathbb{P}(B^c)=\mathbb{P}(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the counting measure of each set $\{n,n+1,\dots \}$ of natural numbers is infinite, although their intersection is empty and has measure zero.

Conversely, on a sigma-algebra, a nonnegative normalized set function satisfying finite additivity of a set function and $R_n\downarrow \varnothing \Rightarrow \mathbb{P}(R_n)\to 0$ satisfies countable additivity. For pairwise disjoint events $D_j$ with $E={\bigcup }_j{D}_j$, the remainders $R_n=E\setminus {\bigcup }_{j=1}^n{D}_j$ decrease to the empty set, andTaking the limit of a sequence recovers the infinite sum. This equivalence explains why continuity of probability is part of the foundations rather than a separate assumption about smooth curves.

One useful application controls infinitely many possible errors. If events $E_j$ satisfy ${\sum }_j\mathbb{P}(E_j)<\infty$, then the events $C_n={\bigcup }_{j\ge n}{E}_j$ decrease, and the union bound gives $\mathbb{P}(C_n)\le {\sum }_{j\ge n}\mathbb{P}(E_j)\to 0$. Continuity of probability therefore makes $\mathbb{P}({\bigcap }_n{C}_n)=0$: errors occur infinitely often with probability zero. This is the Borel-Cantelli first lemma, and it requires no independence. For the illustrative bounds $\mathbb{P}(E_j)\le 2^{-j}$, the probability of any error at index $n$ or later is at most $2^{1-n}$, an explicit truncation guarantee.

The word continuity here concerns nested sets, not the absence of atoms in a probability distribution. For any real-valued random variable $X$, the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Hence its cumulative distribution function is right-continuous even when it has jumps, as for a geometric distribution. The nesting hypothesis matters: an arbitrary sequence of events need not admit either displayed interchange of limit and probability.

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.

A failure-probability budget assigns nonnegative numbers ${\delta }_i$ to a finite or countable family of failure events $F_i$ in a common probability space, with verified bounds $\mathbb{P}(F_i)\le {\delta }_i$. The union bound guarantees that the probability of any failure is at most $\min \{1,{\sum }_i{\delta }_i\}$. By De Morgan's laws and the complement rule of probability, the probability that every requirement succeeds obeysThus ${\sum }_i{\delta }_i\le \delta$ guarantees simultaneous success with probability at least $1-\delta$, even when all failure events depend on the same observations. The numbers ${\delta }_i$ are upper allowances; adding them is not a calculation of the exact failure probability.

For three requirements, give each a failure-probability budget ${\delta }_i=1/20$. The total budget is $3/20$, so simultaneous success has probability at least $17/20$. On a sample space of twenty equally likely outcomes, let $F_1=\{1\}$, $F_2=\{2\}$, and $F_3=\{3\}$. These disjoint events attain the budget and the success bound. If instead all three failure events equal $\{1\}$, their individual probabilities are unchanged, but exact failure has probability $1/20$ and simultaneous success has probability $19/20$. This second model shows how overlap can leave the guarantee conservative.

For $n$ requirements, an equal allocation ${\delta }_i=\delta /n$ is convenient; unequal allocations also work when some requirements are harder to satisfy. For infinitely many requirements indexed by $i\ge 1$, the allocation ${\delta }_i=\delta /2^i$ uses total budget $\delta$ by the geometric series. A fixed positive allowance for every requirement instead gives an infinite sum and no positive simultaneous-success guarantee from the union bound. The individual probability bounds must cover all the failure events being combined: omitting possible failure modes leaves the resulting guarantee about the listed requirements only. The same addition of error allowances underlies the Bonferroni correction in simultaneous hypothesis tests.

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.

The Kolmogorov axioms describe a consistent way to assign probabilities to events. Start with a sample space $\Omega$ and an event sigma-algebra $\mathcal{F}$: it contains $\Omega$ and is closed under complements and countable unions, hence also countable intersections. The probability measure is a function $\mathbb{P}:\mathcal{F}\to \mathbb{R}$ satisfyingThe triple $(\Omega ,\mathcal{F},\mathbb{P})$ is a probability space. The third axiom is countable additivity; the infinite series means the limit of a sequence of nonnegative partial sums. It applies to pairwise disjoint events, not to overlapping events and not to arbitrary uncountable unions.^[ref]

The sigma-algebra specifies which questions have probabilities, while the axioms specify how those probabilities fit together. For a finite sample space one can take the entire power set; a continuous model commonly uses a Borel sigma-algebra. Nonnegativity and normalization do not require equally likely outcomes or independence: those are additional modelling choices.

For an exact six-outcome model, take $\Omega =\{1,2,3,4,5,6\}$, $\mathcal{F}=\mathcal{P}(\Omega )$, and assign $\mathbb{P}(\{k\})=k/21$. The total is one because $1+2+3+4+5+6=21$. Define the probability of an event by summing the weights of its outcomes. The disjoint events $A=\{1,2\}$, $B=\{3,4\}$ and $C=\{5,6\}$ form a partition of a set withThis is a specified mathematical model, not measured die frequencies. The figure makes probability literal area: each region has the same height and a width proportional to its assigned mass. Regrouping disjoint events preserves their total area.

Several familiar rules are consequences, not further axioms. Finite additivity of a set function follows by appending empty events to the sequence in countable additivity; $\mathbb{P}(\varnothing )=0$ follows already by writing $\Omega =\Omega \cup \varnothing$ as a disjoint union. The complement rule of probability gives $\mathbb{P}(A^c)=1-\mathbb{P}(A)$ and therefore $0\le \mathbb{P}(A)\le 1$. Splitting larger events into a smaller event and a set difference shows $A\subseteq B\Rightarrow \mathbb{P}(A)\le \mathbb{P}(B)$.

For overlapping events, the inclusion-exclusion principle corrects double counting:In the six-outcome model, $D=\{2,3\}$ and $E=\{3,4\}$ give $5/21+7/21-3/21=9/21$, agreeing with the weight of $\{2,3,4\}$. This decomposition is useful because a complicated event can be assembled from simpler questions with known probabilities.

On a finite sample space, a family of pairwise disjoint events has only finitely many nonempty members, so finite additivity of a set function already handles every countable disjoint union. On an infinite sample space, the countable axiom additionally controls what is left beyond every finite stage. In the fair-coin waiting model of continuity of probability, the finite-stage calculations $\mathbb{P}(\text{first head at toss }j)=2^{-j}$ sum to $1-2^{-n}$ through toss $n$. Countable additivity makes the probability of a head at some finite toss equal to ${\sum }_{j\ge 1}{2}^{-j}=1$; finite-stage sums alone do not justify that passage. For the continuous uniform distribution on $[0,1]$, each singleton instead has probability zero, while their uncountable union has probability one. Countable, rather than unrestricted, additivity accommodates both models.

Historically, Kolmogorov's 1933 book Grundbegriffe der Wahrscheinlichkeitsrechnung established the influential axiomatic framework connecting probability theory with measure theory. The three conditions above are the modern presentation on a sigma-algebra. Shafer and Vovk's historical study explains that the original presentation began with finite additivity on an algebra of sets and added continuity for decreasing events with empty intersection, which supplies the countable extension.^[ref]

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.

Colour each edge of the complete graph $K_n$ independently red or blue, with probability $1/2$ for each colour. Every four-element subset $A$ of vertices determines exactly one copy of $K_4$, containing six edges. These six edges are all red with probability $2^{-6}$ and all blue with probability $2^{-6}$. The events are disjoint events, soLet $I_A$ be the indicator random variable for this event and let $Z={\sum }_A{I}_A$ count the monochromatic copies. Linearity of expectation givesThe indicator random variables for overlapping copies need not be independent. Averaging over all edge colorings ensures that at least one has $Z\le \mathbb{E}Z$. Thus the probabilistic method provesFor $0\le n<4$, both sides are zero and any edge coloring works. Copies are counted by their four-vertex subsets, rather than by their $4!$ possible labellings.

Because the events are pairwise disjoint events, their count can only be zero or one. The event that the count is zero is the complement of an event of their union. Finite additivity therefore gives

Now use $N\ge 1$ for the fixed number of independently located ships. Normalize the radius to one and write their positions as unit vectors $a_1,\dots ,a_N$. A point $x$ on the sphere is reached by ship $i$ precisely when $a_i\cdot x>0$, so each ship reaches an open hemisphere. Coverage fails exactly when there is a unit vector $x$ with $a_i\cdot x\le 0$ for every $i$. Equivalently, all ship positions lie in some closed hemisphere.

With probability one no pair of ship vectors is antipodal and no three are linearly dependent: the exceptional choices lie on sets of zero surface area. For such a configuration, containment in a closed hemisphere implies containment in an open hemisphere. To see this, choose its inward normal $u$, so $u\cdot a_i\ge 0$. At most two vectors lie on its boundary. If there are two, their sum has positive dot product with each, since they are not antipodal; if there is one, use that vector itself. Perturb $u$ slightly in this direction. The boundary dot products become positive while all already positive dot products remain positive. Thus the strict and non-strict containment conditions have the same probability here.

Condition on the unoriented axes $\{a_i,-a_i\}$. Choose representatives $b_i$ by any fixed sign convention. By uniform sampling and independence, the actual positions are $a_i={\epsilon }_i{b}_i$ with $N$ independent fair signs ${\epsilon }_i\in \{-1,1\}$. Conditional on these axes, each of the $2^N$ sign choices has probability $2^{-N}$.

The great circles $b_i\cdot u=0$ partition the sphere into $R_N=N(N-1)+2$ open regions. In each region the sign vector $\sigma (u)=(\mathrm{sgn}(b_i\cdot u){)}_i$ is constant. No sign vector labels two different regions: its strict inequalities define a convex cone, and normalized line segments within that cone connect any two points of its spherical section. The regions of a general-position great-circle arrangement therefore yield exactly $R_N$ realized sign vectors.

The chosen positions lie in an open hemisphere if and only if some $u$ has $u\cdot a_i>0$ for every $i$, that is, ${\epsilon }_i=\mathrm{sgn}(b_i\cdot u)$ for every $i$. Failure of coverage thus corresponds exactly to one of the $R_N$ realized sign choices. These choices are mutually exclusive events, each of conditional probability $2^{-N}$, so the first part applies:The result is independent of the axes. Averaging gives the coverage of a sphere by random open hemispheres:For $N=1,2,3$ this is zero; for $N=4$ it is $1/8$. If no ship lands, coverage has probability zero separately. The counting uses equiprobable orientation signs, not an assumption that the spherical regions have equal area.

For a family $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$, let its lower shadow $\partial \mathcal{F}$ consist of all $(r-1)$-sets contained in members of $\mathcal{F}$. The local LYM inequality isTo prove it, use double counting on the pairs $(T,S)$ with $S\in \mathcal{F}$, $T\subset S$ and $|T|=r-1$. Each $S$ contributes $r$ pairs; each $T\in \partial \mathcal{F}$ contributes at most $n-r+1$. Hence $r|\mathcal{F}|\le (n-r+1)|\partial \mathcal{F}|$, exactly the displayed Local LYM inequality after using $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=(n-r+1)\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$.

To deduce the LYM inequality, start with an antichain $\mathcal{A}$ in the Boolean lattice and assign it weight ${\sum }_{S\in \mathcal{A}}{\left(\genfrac{}{}{0ex}{}{n}{|S|}\right)}^{-1}$. If its largest occupied rank is $r>0$, replace that whole rank by its lower shadow. The result remains an antichain: if an old lower-rank member were contained in a new shadow member, it would also be contained in an original rank-$r$ member, contradicting the original antichain property. Nor can a new member be contained in an old one, since every old member has size at most $r-1$ and equality would give the same contradiction. The Local LYM inequality shows that this replacement does not decrease the weight. Repeating eventually leaves an antichain on rank zero, whose weight is at most one. ThereforeFor the empty family the conclusion is immediate; the same argument includes the singleton family containing the empty set.

The antichains of maximum size are exactly the full middle levels. More precisely, for even $n$ only $\left(\genfrac{}{}{0ex}{}{[n]}{n/2}\right)$ attains the bound in Sperner theorem; for odd $n$ either $\left(\genfrac{}{}{0ex}{}{[n]}{(n-1)/2}\right)$ or $\left(\genfrac{}{}{0ex}{}{[n]}{(n+1)/2}\right)$ does. To prove that there are no others, put $M=\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$. If $|\mathcal{A}|=M$, the LYM inequality and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\le M$ imply equality in every term's bound $1/\left(\genfrac{}{}{0ex}{}{n}{r}\right)\ge 1/M$. Thus every member lies on a middle level. For even $n$, there is only one such level, so it must be full.

For odd $n=2r+1$, let $X$ be the rank-$r$ members and $Y$ the rank-$(r+1)$ members. Consider the bipartite graph of inclusions between those levels. Both parts have size $M$, and the degree of a vertex is $r+1$. If $N(X)$ is the neighbourhood of $X$, double counting gives $|N(X)|\ge |X|$. Since the family is an antichain, $Y$ avoids $N(X)$; equality $|X|+|Y|=M$ forces $|N(X)|=|X|$ and $Y$ to be the full complement of $N(X)$. Equality in the degree count means that every edge into $N(X)$ comes from $X$, so there are no edges between $X\cup N(X)$ and its complement.

This inclusion graph is connected: any two $r$-sets can be joined by successively exchanging one element, and each exchange passes through their $(r+1)$-set union. Every upper vertex has a lower neighbour. Thus $X$ must be empty or the whole lower level, proving the classification. For $n=1$ the graph is just one edge, and for $n=0$ the unique maximum family is $\{\varnothing \}$. This establishes equality in Sperner theorem without assuming that equality in the LYM inequality alone always forces a middle level.

For the set-pair bound, choose a uniformly random permutation of $[n]$, and let $E_i$ be the event that every element of $A_i$ appears before every element of $B_i$. Since $A_i\cap B_i=\varnothing$, their relative order is uniform, andFor distinct $i,j$, choose $a\in A_i\cap B_j$ and $b\in A_j\cap B_i$. These are distinct because $A_i\cap B_i$ is empty. On $E_i$, $a$ precedes $b$; on $E_j$, $b$ precedes $a$. Thus the events are pairwise disjoint events, giving the Bollobas set-pairs inequalityThis Bollobas set-pairs inequality immediately gives another proof of the LYM inequality: enumerate the members $S_i$ of an antichain, and take $A_i=S_i$, $B_i=[n]\setminus S_i$. For $i\ne j$, incomparability means $S_i\setminus S_j\ne \varnothing$, while $A_i\cap B_i=\varnothing$. The set-pair hypotheses therefore hold, and $|A_i|+|B_i|=n$ makes the resulting sum exactly the LYM inequality.

The partition theorem is the law of total probability. If $B_1,B_2,\dots$ form a finite or countable partition of a sample space into pairwise disjoint events with positive probabilities, then for any event $A$,Indeed, $A$ is the disjoint union of the events $A\cap B_j$, so countable additivity gives the formula after applying the definition of conditional probability. Zero-probability members can be omitted.

There are $6^3=216$ unrestricted placements, since one colour has $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ occupancy vectors. For any specified empty box, the two remaining boxes admit $3^3=27$ placements. For any specified pair of empty boxes, the one remaining box admits one placement. All three boxes cannot be empty.

The inclusion-exclusion principle therefore givesThe empty-box events are not disjoint events, so subtracting their counts without restoring their pairwise overlaps would be incorrect.

Starting at $X$, the two possible two-step returns are $X\to X\to X$ and $X\to Y\to X$. They are disjoint events, with probabilities $1/4$ and $3/16$. ThereforeEach path probability uses the specified Markov property to multiply its successive transition probabilities.

A probability measure is a nonnegative, countably additive measure with total mass one. Thus $\mathbb{P}(\Omega )=1$, and for pairwise disjoint events $A_j$,The complement of an event has probability $1-\mathbb{P}(A)$. The probability of a union of overlapping events needs a correction: $\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)$. Counting the overlap once is the simplest instance of the inclusion-exclusion principle.

A zero-probability event need not be impossible as a member of the sample space. Under the continuous uniform distribution on $[0,1]$, every individual point has probability zero, while the whole interval has probability one. Countable additivity applies to countable unions, not arbitrary uncountable ones; there is no contradiction.

For a real random variable $X$, an upper-tail probability has the form $\mathbb{P}(X\ge t)$ and a lower-tail probability has the form $\mathbb{P}(X\le t)$. Each is the probability of an event lying beyond a chosen threshold in one direction. A tail probability need not be small: the threshold determines how much of the probability distribution it includes. A concentration inequality bounds such probabilities, often for deviations from an expected value.

Endpoint conventions matter. Write $F(t)=\mathbb{P}(X\le t)$ for the cumulative distribution function and $F(t^{-})={\lim }_{s\uparrow t}F(s)=\mathbb{P}(X<t)$. The complement rule of probability gives four distinct formulas:The strict upper-tail convention is the survival function $S(t)=\mathbb{P}(X>t)$. Its value differs from the inclusive upper tail probability by exactly the probability mass at the threshold:A singleton with positive probability mass is an atom of a measure. Thus $1-F(t)$ excludes an atom of a measure at $t$, whereas $1-F(t^{-})$ includes it. These endpoint identities follow directly from the cumulative distribution function properties in Kyle Siegrist, Distribution and Quantile Functions: endpoint probabilities and right tails.

As the threshold increases, both upper tail probabilities are nonincreasing functions. The strict survival function is a right-continuous function; the inclusive upper tail probability takes the upper endpoint value at a downward jump, equal to the left limit of the survival function. Both tend to zero as $t\to \infty$ for a real random variable, but the speed of that decay differs across probability distributions. The cumulative distribution function properties and their proof from continuity of probability measures are given in MIT 6.436J, Lecture 4: random variables and cumulative distribution function properties.

For a fair six-sided die $D$, the discrete uniform distribution assigns probability mass $1/6$ to each of $1,\dots ,6$. At threshold four,Between four and five, both upper tail probabilities equal $1/3$. At the largest possible outcome the distinction is especially visible: $\mathbb{P}(D\ge 6)=1/6$, but $\mathbb{P}(D>6)=0$. The diagram shows exact probabilities for this mathematical model; its dots specify which endpoints belong to each step function.

If a probability distribution has a probability density function $f$, then every singleton has zero probability mass andFor an exponential distribution of rate two, $f(x)=2e^{-2x}$ for $x\ge 0$ and zero otherwise, so $S(t)=e^{-2t}$ for $t\ge 0$ and $S(t)=1$ for $t<0$. Hence $\mathbb{P}(X\ge 1)=e^{-2}\approx 0.135335$. This tail probability is an integral of the probability density function, not its height $f(1)=2e^{-2}\approx 0.270671$. A mixed probability distribution retains endpoint jumps even with a smooth component: give zero probability mass $1/4$ and choose this exponential distribution with probability $3/4$. Then $\mathbb{P}(X\ge 0)=1$, $\mathbb{P}(X>0)=3/4$, and $S(t)=(3/4)e^{-2t}$ for $t\ge 0$.

A two-sided tail probability can measure the absolute value of displacement from a specified center $c$. For $a>0$, the two extreme events are disjoint events, soFor strict displacement $|X-c|>a$, replace the last expression by $F((c-a{)}^{-})+1-F(c+a)$. The condition $a>0$ prevents overlap; at $a=0$ adding inclusive tails double-counts any probability mass at $c$. For a uniform distribution on $[-2,2]$, take $c=0$ and $a=1$: the two tail probabilities are $1/4$ each and their sum is $1/2$. Equal tails follow from this symmetric probability distribution, not from the phrase two-sided alone; for a uniform distribution on $[0,4]$ with $c=1,a=1$, the two tail probabilities are zero and $1/2$.

The tail-sum formula for expectation turns a whole curve of tail probabilities into an expected value. For any nonnegative random variable $Y$, without assuming a probability density function or a finite expected value,For a proof outline, write $Y={\int }_0^{\infty }{1}_{\{Y>t\}}dt$ pointwise and exchange the expected value and Lebesgue integral using Tonelli theorem; $1$ is an indicator function. The inclusive version has the same Lebesgue integral: positive probability masses occur at only a countable set of thresholds, which has Lebesgue measure zero. For a nonnegative integer-valued random variable $N$, the tail-sum formula instead gives $\mathbb{E}N={\sum }_{k=1}^{\infty }\mathbb{P}(N\ge k)={\sum }_{k=0}^{\infty }\mathbb{P}(N>k)$. The integer hypothesis is essential for these sums. The integral identity, its indicator function proof, and the integer version appear in Kyle Siegrist, Additional Properties of Expected Value: integration and summation of right tails.

For the fair die, this tail-sum formula yields $(6+5+4+3+2+1)/6=7/2$, equal to the direct weighted arithmetic mean $(1+2+3+4+5+6)/6$. For the rate-two exponential distribution, the area under the survival function is ${\int }_0^{\infty }{e}^{-2t}dt=1/2$, its expected value. The first panel below shades an integral of the probability density function for a single threshold; the second shades an integral of the survival function across all nonnegative thresholds. These are analytical illustrations, with the displayed curves truncated at three; their exact total areas use an infinite integration range.

Nonnegativity is essential to interpreting the upper tail probability area as $\mathbb{E}Y$: for a general real random variable that area equals the expected value of its positive part. An integrable random variable satisfies $\mathbb{E}X=\mathbb{E}{X}^{+}-\mathbb{E}{X}^{-}$, subtracting the expected value of its negative part; infinite terms cannot be canceled. Even when $Y$ is finite almost surely, its expected value can be infinite. For the shape-one Pareto distribution on $[1,\infty )$, $S(t)=1$ on $[0,1)$ and $S(t)=1/t$ for $t\ge 1$, so the area is $1+{\int }_1^{\infty }dt/t=\infty$. The finite versus infinite expected value distinction is illustrated in Kyle Siegrist, Definitions and Basic Properties: exponential and Pareto expected values.

Why are tail probabilities useful? They answer threshold questions directly, and a concentration inequality can bound them without knowing the full probability distribution. For nonnegative $Y$ with finite expected value, Markov inequality gives $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$ for $t>0$: a large threshold relative to the expected value forces rarity. The quantile function reverses the question by choosing a threshold for a prescribed cumulative level. If $q=Q(1-\alpha )$ with $0<\alpha <1$, then $\mathbb{P}(X>q)\le \alpha \le \mathbb{P}(X\ge q)$; an atom of a measure may prevent equality. For the die, $Q(0.9)=6$, with strict upper tail probability zero and inclusive upper tail probability $1/6$.

A tail probability describes rarity under a specified probability distribution; it is not automatically a conditional probability after extra information. For $u\ge t$ with $\mathbb{P}(X\ge t)>0$,because the threshold events are nested. Thus the die has $\mathbb{P}(D\ge 5)=1/3$ but $\mathbb{P}(D\ge 5\mid D\ge 4)=2/3$. Conditioning changes the sample space and renormalizes its probability measure; the same extreme outcomes can become common in that restricted population.