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$\Omega$ is the sample space; its elements are elementary outcomes, also called sample points. An event is a measurable subset of $\Omega$, belonging to the sigma-algebra $\mathcal{F}$, rather than necessarily a single outcome.

Let $A_i$ be the event that candidate $i$ succeeds. The countable additivity axiom adds probabilities only for mutually disjoint events. Successes of different candidates are not mutually exclusive, so replacing the probability of their union by ${\sum }_i\mathbb{P}(A_i)=1$ misuses that axiom. Without disjointness one obtains only the union bound, not certainty. The same sum is the expected value of the number of successes, which also need not imply at least one.

If the ten outcomes are mutually independent events, with each success probability $0.1$, all fail with probability $0.9^{10}$. HenceThe independence assumption is essential. The marginal probabilities alone allow any union probability from $0.1$ up to $1$; perfectly overlapping success events give the lower endpoint and disjoint successes give the upper endpoint.

Conditioning on $Z=z$ has positive probability exactly when $z=2r+1$ for an integer $r\ge 0$. This is the same event as $X+Y=r$. The addition of independent Poisson random variables gives a Poisson distribution with parameter $\lambda +\mu$ for the sum. Therefore, for $0\le t\le r$,Thus $X\mid \{Z=2r+1\}\sim \mathrm{Binomial}(r,\lambda /(\lambda +\mu ))$. The case $r=0$ is a constant zero random variable. For even, nonpositive, or noninteger $z$, the conditioning event has zero probability, so an event-conditional probability distribution is not specified by the quotient formula. This is a two-component case of Poisson-multinomial conditioning.

Insert the Poisson distribution probability generating function into the preceding bound:For fixed $x>\lambda$, differentiate the exponent with respect to $s$. Its derivative is $\lambda -x/s$, and its second derivative is $x/s^2>0$, so its unique minimum occurs at $s=x/\lambda >1$. At that value,ConsequentlyThis Chernoff bound remains valid for noninteger positive thresholds $x>\lambda$; the event is then the same as exceeding or equalling its integer ceiling.

Let $R_1,R_2$ be the two red-card events. Sampling without replacement givesThis differs from the product $m^2/N^2$, sinceThus the events are not independent, for every positive $m,n$. A first red card reduces the remaining red fraction, a basic instance of how sampling without replacement need not be independent.

The expected value of an indicator random variable is the probability of its event. For a simple symmetric random walk absorbed at $0,k$, the gambler's ruin success probability from $i$ is $i/k$. HenceThe boundary values are $0$ at the left exit and $1$ at chamber $k$.

For $1\le i\le k$, the success probability is $i/k>0$, and the definition of conditional expectation on an event givesThereforeAt $i=k$ the value is zero because the trial has already ended. At the left absorbing state $i=0$, conditioning on success would be undefined because it has zero probability. The formula concerns the successful trajectories, so it need not equal the unconditional mean $i(k-i)$.

The random variables are independent random variables if the events determined by one are independent of the events determined by the other. For integer-valued variables, this is equivalent to the factorizationIndeed, independence gives the factorization by taking singleton events. Conversely, summing it over $x\in A$ and $y\in B$, for any subsets $A,B\subseteq \mathbb{Z}$, gives $\mathbb{P}(X\in A,Y\in B)=\mathbb{P}(X\in A)\mathbb{P}(Y\in B)$. Thus the singleton condition implies independence of all events in question.

For each integer $x$, the events $\{X=x,Y=y\}$, indexed by the integers $y$, are disjoint and their union is $\{X=x\}$. Therefore countable additivity gives the marginal probability mass functionThe nonnegative terms make this sum well-defined, even when the support is infinite.

Let $(B_j)$ be a finite or countable partition of a set of the sample space into disjoint events, with $\mathbb{P}(B_j)>0$. For any event $A$, the law of total probability isTo prove it, the intersections $A\cap B_j$ are disjoint and have union $A$. Countable additivity and the definition of conditional probability therefore giveIt is enough that the union of the partition events has probability one. Parts of probability zero contribute zero in the intersection formula and can be omitted before writing conditional probabilities; conditioning by division on a zero-probability part would be undefined.

A partition of a sample space is a finite or countable family of pairwise disjoint events $B_i$ whose union is the whole sample space. For an integrable random variable, the partition form of the law of total expectation isZero-probability cells contribute zero and are omitted because elementary event-conditioning on them is undefined. The formula also holds with extended nonnegative values when $X\ge 0$. Integrability makes the signed series legitimate in the finite-expectation case.

In independent Bernoulli trials with success probability $p$, $N$ counts failures before the first success. The event $N=n$ requires $n$ failures followed by one success, giving $(1-p{)}^{n}p$. This is the zero-based geometric distribution, rather than the count of trials including the success.

The events $\{X=k\}$, for $k\in \mathbb{Z}$, are pairwise disjoint. Their union is $\{X\in \mathbb{Z}\}$, and the integers form a countable set. By countable additivity,Countability is the necessary justification: one cannot apply this argument to an arbitrary uncountable union of zero-probability singletons. This is a special case of countable sets have zero probability under a density.

For $k\ge 1$, the event $\{\lfloor X\rfloor \ge k\}$ equals $\{X\ge k\}$. Integrating the Pareto distribution's density givesThe mass formula follows by subtracting consecutive tail probabilities, or by integrating over $[k,k+1)$. With the paper's convention $0\in \mathbb{N}$, include the separate endpoint values $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely. Substituting $k=0$ into the power expression would be invalid.

For the rooted infinite model defining percolation probability, the percolation critical probability isBelow $p_c$, the root has probability zero of belonging to an infinite percolation cluster. Above $p_c$, that probability is positive, by the definition and the monotone coupling of Bernoulli percolation. The definition alone does not determine behavior at $p_c$, or the numerical value of $p_c$: these require results specific to the graph and the model.

A finite graph with $M$ edges has $2^M$ open/closed configurations. The probability of any connectivity event is a finite sum of terms $p^k(1-p{)}^{M-k}$, hence a polynomial in $p$. The square's $2p^2-p^4$ is an example. These finite probabilities vary smoothly, although an infinite-system limit can distinguish a whole interval with $\theta (p)=0$ from one with $\theta (p)>0$. A steep finite connection curve and a mathematically proved percolation critical probability are different kinds of information.

For independent bond percolation on an infinite graph with a specified root $o$, let $C_o$ be its percolation cluster. The percolation probability isIt concerns the root, rather than the existence of an infinite percolation cluster somewhere else. Even above the percolation critical probability of the square lattice, an isolated root has probability $(1-p{)}^4$, because its four incident edges can all be closed. Thus $\theta (p)\le 1-(1-p{)}^4<1$ when $p<1$. Positive percolation probability does not mean every vertex is connected to infinity.

On the square lattice, the events that $o$ is joined by an open path to the boundary of $[-n,n{]}^2$ decrease as $n$ increases. Their probabilities converge to $\theta (p)$ by continuity from above of a measure. This relates an infinite question to finite observations, without equating any one finite boundary connection with survival forever. The monotone coupling of Bernoulli percolation shows that $\theta$ is a monotone function of $p$.

Percolation theory asks how many local connections must be available before a large connected structure appears. Start with a graph, randomly retain some edges or vertices, and study the resulting percolation clusters. In independent bond percolation, every edge is open with probability $p$, independently of the others. In independent site percolation, the random decisions concern vertices instead. A percolation cluster is a connected component of a graph formed by the available connections; an open path joins two vertices exactly when they belong to the same percolation cluster.

The attraction is that a local rule can produce a global change. One open edge says little about travel across a large graph. Many alternative graph paths can nevertheless make distant connections likely. The model separates the geometry of possible connections from the probability distribution of which connections are usable. This was the central perspective of Broadbent and Hammersley's 1957 paper, which discussed porous solids and disease transmission as interpretations of the same mathematical structure: ^[ref]. Independent bond percolation supplies a simplified model of those systems; the physical interpretation still requires justification of its independence assumptions.

For a completely calculable example, consider the four edges around a square and ask whether opposite corners $A$ and $C$ are connected. There are two graph paths, each using two edges. Each open path has probability $p^2$. Their events are independent events because the paths share no edges, even though they share endpoints. The inclusion-exclusion principle therefore givesAt $p=1/2$, the connection probability is $7/16$, larger than the $1/4$ for either chosen graph path. Counting all sixteen equally likely edge configurations gives the same answer: two configurations with exactly two open edges, four with three, and one with four connect $A$ to $C$. Connectivity benefits from alternative routes; adding their probabilities without subtracting their overlap would count the all-open configuration twice.

This finite calculation is already useful for studying redundant connections. The deeper question is what happens as the graph becomes arbitrarily large: does connectivity remain confined to finite percolation clusters, or can an infinite percolation cluster appear? The percolation probability and percolation critical probability make that question precise. A finite picture can illustrate the geometry, but cannot establish the existence of an infinite percolation cluster.

Let $X_1,\dots ,X_n$ be independent random variables, each choosing a label from $\{1,\dots ,n\}$. Put their probabilities in the matrix $P=(p_{ij})$, where $p_{ij}=\mathbb{P}(X_i=j)$. Each row sums to one. All choices are distinct precisely when their ordered list is one of the permutations of the $n$ labels. These are mutually exclusive events, and independence makes the probability of one list $\sigma$ equal to ${\prod }_i{p}_{i,\sigma (i)}$. Adding the list probabilities gives the permanent of a matrix:In particular, this permanent lies between zero and one.

For a two-person example,The distinct outcomes $(1,2)$ and $(2,1)$ have probabilities $3/5$ and $1/20$. ThereforeThe colliding outcomes $(1,1)$ and $(2,2)$ have probabilities $3/20$ and $1/5$, totaling $0.35$. All four outcomes are shown below; their probabilities sum to one.

If every choice has the uniform distribution on a finite set of $n$ labels, then $p_{ij}=1/n$ and the distinct-choice probability is $n!/n^n$. For three choices and three labels it is $6/27=2/9$. This counts a subset of all $n^n$ ordered outcomes, not independent pairwise collision events. Independence of the original choices is essential: without it, the individual rows of $P$ do not determine the joint probabilities.

For nontrivial events, strict attraction under conditional probability is equivalent to a positive covariance of their indicator variables. This condition is symmetric in the two events. Complementing exactly one of the events reverses the strict inequality. This two-event condition should be distinguished from association of an entire family of random variables.

The probability of an event is its value under a probability measure, a real number between zero and one. It compares the mass assigned to that event with the unit mass of the whole sample space. For a finite sample space with equally likely outcomes, $\mathbb{P}(A)=|A|/|\Omega |$. For unequal outcome masses, add their individual probabilities instead.

For example, selecting a coin with head probability $1/2$ with chance $4/5$, or a coin with head probability $3/4$ with chance $1/5$, gives head probability $(4/5)(1/2)+(1/5)(3/4)=11/20$. This is the law of total probability: averaging over the hidden selection is part of calculating the outcome's probability.