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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.

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.

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.

Put $a=P(A)$, $b=P(B)$ and $j=P(A\cap B)$. Because $a,b>0$, the conditional probability inequality $P(A\mid B)>P(A)$ is equivalent to $j>ab$. But that same inequality gives $j/a>b$, or $\boxed{P(B\mid A)>P(B)}$. Thus positive association of two events is symmetric.

For the complement of an event, $P(B^c)=1-b>0$, andTherefore $B^c$ strictly repels $A$; equivalently $P(B^c\mid A)=1-j/a<1-b=P(B^c)$. Both directions are stated because the wording switches the grammatical roles of attraction and repulsion. The substantive conclusion is the same: the conditional probabilities involving the complement are strictly smaller than their unconditional counterparts.

For independent fair dice, the probability of avoiding one on each die is $5/6$. ThereforeThe complement of an event gives

The complement of an event here is that all three dice avoid six, with probability $(5/6{)}^3$. Therefore

Take the complement of an event: fewer than two successes means zero or one. These mutually exclusive events have probabilities $(1-y{)}^5$ and $5y(1-y{)}^4$. Hence

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.

Model two fair independent coin tosses, recording their order. Here $H$ denotes heads and $T$ tails, andThe power set $\mathcal{P}(\Omega )$ includes every question expressible as a subset of these four outcomes. The mass $1/4=(1/2)(1/2)$ comes from the fairness and independence assumptions. These are exact model probabilities, rather than measured toss frequencies.

Let $A$ be the event that the first toss is heads and $B$ the event that the second toss is heads. ThenThe union means “at least one head,” while the intersection means “both heads.” Their probabilities are $3/4$ and $1/4$. Counting the union through the inclusion-exclusion principle gives $1/2+1/2-1/4=3/4$; alternatively its complement of an event is $\{TT\}$, of probability $1/4$.

On the same sample space, assigning masses $3/8,1/8,1/8,3/8$ to $HH,HT,TH,TT$ would still make each toss fair: $\mathbb{P}(A)=\mathbb{P}(B)=1/2$. But $\mathbb{P}(A\cap B)=3/8\ne 1/4$, so the tosses would be dependent. The event of at least one head would have probability $5/8$. Changing the probability measure changes the answer even when the sample space and the single-toss probabilities stay the same.