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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 $q>0$.

Events $A$ and $B$ are independent whenFor a possible sum $2\le t\le 12$, let $N_t$ be the number of ordered die pairs with sum $t$. ThenIf $t-i\in \{1,\dots ,6\}$, then $\mathbb{P}(A_t\cap B_i)=1/36$; otherwise it is zero. Independence for a possible sum therefore requiresso $N_t=6$. This occurs only for $t=7$, and then $7-i$ is valid for every $i=1,\dots ,6$. ThusIf impossible sums are admitted as events of probability zero, they are trivially independent as well.

Divide the keystrokes into disjoint blocks of five, and let $A_n$ be the event that block $n$ is exactly HELLO. The events are independent events, andConsequentlyPart (c) shows that infinitely many of these block events occur almost surely. Each occurrence is an occurrence of HELLO in the full typed sequence, so

WriteThe Russo-Seymour-Welsh theorem and the Harris-FKG inequality give the standard one-arm extension estimate: there is $c>0$ such thatuniformly in $r$. Indeed, on the one-arm event to scale $r$, a fixed finite collection of open rectangle crossings in the annulus ${\Lambda }_{2r}\setminus {\Lambda }_r$, each having probability bounded below by RSW, joins that arm to $\partial {\Lambda }_{2r}$; FKG multiplies the lower bounds. Iteration shows that $\pi (ar)$ and $\pi (r)$ are comparable for every fixed $a>0$. This proves the estimate suggested in the hint.

Fix $x\in \partial {\Lambda }_n$ and choose $r=\lfloor n/4\rfloor$. If $0⟷x$, there is an open arm from $0$ to distance $r$ and another from $x$ to distance $r$. These are independent events because they use disjoint edge sets, so the extension estimate gives

For the reverse inequality, take one-arm events from $0$ and $x$ at scale comparable with $n$, in disjoint boxes. A fixed collection of open crossings of rectangles of bounded aspect ratio joins the two arms. The Russo-Seymour-Welsh theorem bounds the probability of every added crossing below uniformly in $n$ and in the position of $x$ along the four sides; the Harris-FKG inequality and the arm-extension estimate therefore giveThis is the usual RSW gluing lemma for two one-arm events. Enlarging the constants handles the finitely many small $n$, proving the claim with positive constants $c_1,c_2$.

There are three even faces, so $\mathbb{P}(A)=1/2$. The sum is even precisely when both faces have the same parity, giving $9+9=18$ outcomes and $\mathbb{P}(B)=1/2$. Both $A$ and $B$ occur precisely when both faces are even, giving nine outcomes. ThereforeThus $A$ and $B$ are independent events.

No: pairwise independence does not imply that the three are independent events. Take $C=\{Y\text{ is even}\}$. Then $\mathbb{P}(C)=1/2$, and $\mathbb{P}(A\cap C)=9/36=1/4$ by independence of the two rolls. Also $B\cap C$ is precisely the event that both faces are even, so $\mathbb{P}(B\cap C)=1/4$. Thus all three pairs of events are independent.

But $A\cap B\cap C$ also consists of the nine even-even outcomes, soThis violates the triple-intersection requirement for independent events.

Under independent daily tosses, attending every lecture is the intersection of $24$ independent events of probability $3/4$. Hence

For independent events from fair dice, the probability of a six on each roll is $1/6$. Multiplication gives

Assume that wrong-note events are independent events and that each note has the same error probability $p$, with $0\le p\le 1$. Then their count $N$ has a binomial distribution. Choose the $n$ mistaken positions in $\left(\genfrac{}{}{0ex}{}{100}{n}\right)$ ways and multiply the probabilities of the errors and correct notes:The endpoint cases $p=0$ or $p=1$ are understood by continuity, giving deterministic counts. More than half means at least $51$ mistakes, soThe single-note error probability alone would not determine this count distribution if mistakes were correlated; independence is the additional standard modeling assumption.

The three tosses are independent events, each with head probability $1/2$. Consequently

For $B_n=\{X_n>a_n\}$, the definition of the common cumulative distribution function givesThe strict exceedance matters here: an atom of a measure at $a_n$ belongs to $\{X_n\le a_n\}$, so it is already included in $F(a_n)$. The events $B_n$ are mutually independent events, since each depends on a separate member of the independent random variables $X_n$.

If ${\sum }_n(1-F(a_n))<\infty$, countable subadditivity gives, for every $N$,Continuity of probability then gives $\mathbb{P}({\mathrm{limsup}}_n{B}_n)=0$. This is the First Borel-Cantelli lemma and does not need independence.

If the series diverges, independence and $1-u\le e^{-u}$ give, for every fixed $N$,Taking $M\to \infty$ shows that the probability of no exceedance after $N$ is zero. A countable union over $N$ shows that finitely many exceedances has probability zero. This is the Second Borel-Cantelli lemma. Together with the preceding event equality, it proves

Suppose the events $(B_n)$ are mutually independent events. Then $B_k$ is independent of ${\mathcal{F}}_{k-1}$, first by factorization against finite intersections of previous events and then by part (a). ThereforeIf ${\sum }_k\mathbb{P}(B_k)=\infty$, the sum in part (ii) is infinite almost surely, and consequentlyThis is the Second Borel-Cantelli lemma. In fact part (ii) also yields probability zero in the convergent case for these independent events.

For $0<p<1$, independence of the remaining indicators givesBecause $0<k<n$, the latter probability is positive for interior $p$. The two events are independent events exactly when their joint probability equals $p\mathbb{P}(S_n=k)$, which is equivalent to $p=k/n$.

At $p=0$ or $p=1$, the event $\{S_n=k\}$ is null, and a null event is independent of every event. Therefore the full answer, including the degenerate endpoints, is

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.

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.

For finite sets, the number of ordered choices of one element from each is the product of the individual numbers of choices. More generally, if stage $j$ has exactly $n_j$ possible choices after every permitted history of earlier stages, the total number of full choice sequences is ${\prod }_j{n}_j$, by mathematical induction on the stages. This counts $b^a$ functions between sets of sizes $a,b$, and counts the injective functions by the falling factorial $b(b-1)\cdots (b-a+1)$ when $a\le b$.

With five people, choose a captain in five ways and a different deputy in four ways: $5\cdot 4=20$ ordered assignments. The second choice depends on the first because the captain is excluded, but its number of permitted options is always four. The rule of product therefore applies without any assumption about independent events. To count two-person teams instead, each subset has two possible role assignments, so division by $2!=2$ gives ten teams. This division works because every team was counted equally often.

Take independent random variables $U_n$, each with the continuous uniform distribution on $[0,1]$, and set $A_n=\{U_n\le p_n\}$. Then $\mathbb{P}(A_n)=p_n$, and the events are mutually independent events. The product probability space supplies a model for the entire infinite sequence; a finite list of simulated values is only an illustration.

For $p_n=1/(n+1{)}^2$, the p-series is summable:The Basel problem supplies the total sum, and the integral test for convergence gives the displayed tail bound. The First Borel-Cantelli lemma makes the limsup of events have probability zero. In fact, the tail bound implies that the probability of even one alarm at or after $m$ is at most $1/m$; for $m=100$, it is at most $0.01$. This bound remains valid for dependent events with the same individual probabilities.

For $p_n=1/(n+1)$, the harmonic series diverges, so the Second Borel-Cantelli lemma makes the limsup of events have probability one. There is also an exact finite-horizon calculation using independent events:For any fixed $m$, this finite product of numbers tends to zero as $M\to \infty$. Every infinite tail therefore contains an alarm almost surely, even though the individual probabilities tend to zero.