Chapter 02

Easy

Probability Axioms

00 · Symbol Glossary

$\mathbb{P}(A)$P of A — probability of event A

A number in $[0,1]$ the model assigns to the event $A$ . Which assignments are allowed is fixed by the Kolmogorov axioms of this chapter, whatever interpretation (frequency, belief, market-implied) sits behind the numbers.

$(\Omega,\mathbb{P})$Probability model

A sample space together with a probability on its events. In this chapter $\Omega$ is finite or countable and every subset of $\Omega$ is an event.

$p(\omega)$p of omega — outcome weight

The probability of the single outcome $\omega$ , $p(\omega)=\mathbb{P}(\{\omega\})$ . On a finite or countable space the weights determine every event's probability by summation.

$\sum_{\omega\in A}$Sum over outcomes in A

Add up a quantity over every outcome in $A$ . The basic computation of this chapter is $\mathbb{P}(A)=\sum_{\omega\in A}p(\omega)$ .

$|A|$Cardinality of A

The number of outcomes in a finite event $A$ . Used in the uniform model $\mathbb{P}(A)=|A|/|\Omega|$ .

$\mathbb{P}(A\cap B)$Joint probability — both occur

Probability that $A$ and $B$ occur together. This is the overlap term that inclusion–exclusion subtracts, and the quantity naive risk aggregation forgets.

$A_n\uparrow A$Nested events increasing to A

$A_1\subseteq A_2\subseteq\cdots$ with union $A$ . Countable additivity forces the probabilities to settle: $\mathbb{P}(A_n)\to\mathbb{P}(A)$ . A cumulative default curve is this kind of nested sequence.

01 · From Sets to Numbers

Chapter 01 built the language: a sample space $\Omega$ , and events as subsets of it. A probability model adds one more ingredient, a rule $\mathbb{P}$ that assigns each event a number.

There are several ways to read those numbers. A frequency reading says $\mathbb{P}(A)$ is the long-run fraction of runs on which $A$ occurs. A belief reading says it is the price at which you would bet on $A$ . A market-implied reading backs probabilities out of traded prices, such as default probabilities implied by credit spreads. These readings disagree about where the numbers come from, but they all agree on the rules the numbers must obey. Those rules are Kolmogorov's axioms, and every later result in this subject — Bayes' rule, expectation, the limit theorems — is derived from them.

Rules, not sources

The axioms do not tell you what $\mathbb{P}(\text{default})$ is for a given firm. Estimating that is a statistics problem, taken up from Chapter 13 onward. The axioms only tell you which collections of numbers are internally consistent — and that alone rules out a surprising number of real-world risk reports.

One set of rules, more than one measure

A long-run default frequency and a probability backed out of a credit spread can be different numbers for the same event. Both still have to be non-negative, total to $1$ , and add on disjoint events. The axioms police consistency; they do not choose between a historical reading and a market-implied one. Which reading a price uses is a modeling decision made outside this chapter.

02 · The Kolmogorov Axioms

Definition — Probability (finite additivity)

Let $\Omega$ be a sample space. A function $\mathbb{P}$ that assigns a real number to every event $A\subseteq\Omega$ is a probability if:

Non-negativity: $\mathbb{P}(A)\ge 0$ for every event $A$ .
Normalization: $\mathbb{P}(\Omega)=1$ .
Additivity: if $A\cap B=\emptyset$ , then $\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)$ .

Plain language: probabilities are never negative, something certainly happens, and for events that cannot happen together, the chance that one of them happens is the sum of their chances. Probability behaves like a fixed total amount of mass, $1$ , spread over the outcomes; the mass sitting on two non-overlapping regions is the sum of the two masses.

Applying additivity repeatedly extends it to any finite number of events. If $A_1,\ldots,A_n$ are pairwise disjoint ( $A_i\cap A_j=\emptyset$ for $i\neq j$ ), then

\mathbb{P}\Big(\bigcup_{i=1}^n A_i\Big)=\sum_{i=1}^n \mathbb{P}(A_i)

In particular, for a partition $\{B_1,\ldots,B_n\}$ of $\Omega$ the cell probabilities sum to $1$ .

Countable additivity

The full axiom requires the same identity for a countably infinite sequence of pairwise disjoint events. For a finite $\Omega$ the two versions say the same thing. On a countable space such as "days until the first down day," $\Omega=\{1,2,3,\ldots\}$ , the countable version is what guarantees that $\mathbb{P}(A)$ is the sum of the outcome weights in $A$ even when $A$ is infinite. Continuous spaces (Chapter 06) need it as well, along with care about which sets count as events; nothing in this chapter depends on those details.

03 · First Consequences

Every rule below is derived from the three axioms alone. Prove each once and use it freely afterwards.

Step-by-step — Four rules from three axioms

Impossible event.

\Omega

and

\emptyset

are disjoint and

\Omega\cup\emptyset=\Omega

, so additivity gives

\mathbb{P}(\Omega)=\mathbb{P}(\Omega)+\mathbb{P}(\emptyset)

. Hence

\mathbb{P}(\emptyset)=0

Complement rule.

A

and

A^c

are disjoint with union

\Omega

, so

1=\mathbb{P}(A)+\mathbb{P}(A^c)

. Hence

\mathbb{P}(A^c)=1-\mathbb{P}(A)

Difference and monotonicity. If

A\subseteq B

, then

B

is the disjoint union of

A

and

B\setminus A

, so

\mathbb{P}(B)=\mathbb{P}(A)+\mathbb{P}(B\setminus A)

. Hence

\mathbb{P}(B\setminus A)=\mathbb{P}(B)-\mathbb{P}(A)

, and because that difference is non-negative,

\mathbb{P}(A)\le\mathbb{P}(B)

Upper bound. Every event satisfies

A\subseteq\Omega

, so monotonicity gives

\mathbb{P}(A)\le 1

. The upper bound was never an axiom; it is a consequence.

\mathbb{P}(\emptyset)=0, \qquad \mathbb{P}(A^c)=1-\mathbb{P}(A), \qquad A\subseteq B \implies \mathbb{P}(A)\le\mathbb{P}(B), \qquad 0\le\mathbb{P}(A)\le 1

Example — Survival and cumulative default

A bond has a one-year default probability of $2\%$ and a five-year cumulative default probability of $9\%$ . Let $A$ be "defaults within one year" and $B$ be "defaults within five years."

Survival through year one: $\mathbb{P}(A^c)=1-0.02=0.98$ .
Defaulting within one year implies defaulting within five, so $A\subseteq B$ and the numbers are consistent with monotonicity: $0.02\le 0.09$ .
Defaulting in years two to five is $B\setminus A$ , with $\mathbb{P}(B\setminus A)=0.09-0.02=0.07$ .

❌ A default curve that falls with the horizon

A model reports a five-year cumulative default probability of $8\%$ and a seven-year cumulative default probability of $7\%$ for the same issuer.

Why it breaks: "default within five years" is a subset of "default within seven years," so monotonicity forces $\mathbb{P}(\text{within } 5)\le\mathbb{P}(\text{within } 7)$ . Read through the difference rule, the curve implies $\mathbb{P}(\text{default in years 6 to 7})=0.07-0.08=-0.01$ , a negative probability.

Consequence: any price built from this curve — a seven-year credit default swap, say — is computed from an inconsistent model. Cumulative default and cumulative loss curves must be non-decreasing in the horizon; checking that is the cheapest sanity test a curve can get.

04 · Building a Probability From Outcome Weights

On a finite or countable space you never need to list $\mathbb{P}(A)$ for every one of the possibly many events. It is enough to give one weight per outcome.

Definition — Probability From Outcome Weights

Let $\Omega$ be finite or countable. Choose weights $p(\omega)$ with

p(\omega)\ge 0 \text{ for every } \omega\in\Omega, \qquad \sum_{\omega\in\Omega}p(\omega)=1

and define, for every event $A$ ,

\mathbb{P}(A)=\sum_{\omega\in A}p(\omega)

Then $\mathbb{P}$ satisfies the axioms: sums of non-negative numbers are non-negative, the total weight is $1$ , and if $A$ and $B$ are disjoint, no outcome is counted in both sums. Conversely, every probability on such a space arises this way, with $p(\omega)=\mathbb{P}(\{\omega\})$ .

Plain language: put a non-negative weight on each outcome so the weights total one. The probability of an event is the total weight of the outcomes inside it.

Example — One-day return buckets

An analyst models tomorrow's index return with five buckets: big down (below $-2\%$ ), down, flat, up, big up (above $+2\%$ ), with weights

p(\text{big down})=0.05,\quad p(\text{down})=0.42,\quad p(\text{flat})=0.03,\quad p(\text{up})=0.45,\quad p(\text{big up})=0.05

The weights are non-negative and sum to $1$ , so this is a valid model. Then $\mathbb{P}(\text{some rise})=0.45+0.05=0.50$ , $\mathbb{P}(\text{move larger than } 2\% \text{ either way})=0.05+0.05=0.10$ , and $\mathbb{P}(\text{not flat})=1-0.03=0.97$ by the complement rule. Nothing forces the five weights to be equal, and in a model of real returns they should not be.

Definition — Uniform Probability on a Finite Space

If $\Omega$ is finite and every outcome receives the same weight $p(\omega)=1/|\Omega|$ , then

\mathbb{P}(A)=\frac{|A|}{|\Omega|}

Plain language: when all outcomes are equally likely, a probability is a count of favorable outcomes divided by a count of possible outcomes. This is where combinatorics enters probability — but it is one special choice of weights, not the definition of probability.

Example — Fair up/down days

Model three trading days by $\Omega=\{U,D\}^3$ with the uniform weight $1/8$ on each outcome. "At least two up days" is $\{UUD,UDU,DUU,UUU\}$ , so its probability is $4/8=1/2$ . "Up on day 2" has $4$ outcomes, also probability $1/2$ . "Up all three days" is the single outcome $UUU$ , with probability $1/8$ .

❌ Applying the uniform model on the wrong space

Someone reasons: "Over two fair trading days the number of up days is $0$ , $1$ , or $2$ , so each count has probability $1/3$ ."

Why it breaks: uniformity was a property of the ordered space $\{UU,UD,DU,DD\}$ , where each outcome has weight $1/4$ . The count "exactly one up day" collects two of those outcomes, $UD$ and $DU$ , while the counts $0$ and $2$ collect one each. The correct weights on the coarser space are $1/4$ , $1/2$ , $1/4$ . Merging outcomes into buckets of different sizes destroys uniformity.

Consequence: a binary option paying on "exactly one up day" would be priced at $1/3$ instead of $1/2$ — off by a third of the payoff. The same error appears in the shortcut "either the firm defaults or it does not, so the probability is 50%": two outcomes do not make two equally likely outcomes. Uniformity is a modeling claim that needs a reason, usually symmetry on the finest space.

05 · Inclusion–Exclusion

Additivity only applies to disjoint events. When events overlap, adding their probabilities counts the overlap twice.

Step-by-step — Deriving the two-event formula

Split the union into disjoint pieces.

A\cup B

is the disjoint union of

A\setminus B

A\cap B

, and

B\setminus A

Apply additivity three times.

\mathbb{P}(A\cup B)=\mathbb{P}(A\setminus B)+\mathbb{P}(A\cap B)+\mathbb{P}(B\setminus A)

, while

\mathbb{P}(A)=\mathbb{P}(A\setminus B)+\mathbb{P}(A\cap B)

and

\mathbb{P}(B)=\mathbb{P}(B\setminus A)+\mathbb{P}(A\cap B)

Compare. Adding the last two equations gives

\mathbb{P}(A)+\mathbb{P}(B)=\mathbb{P}(A\cup B)+\mathbb{P}(A\cap B)

. The overlap appears once too often on the left.

Definition — Inclusion–Exclusion

For any events $A$ and $B$ ,

\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)

For three events,

\mathbb{P}(A\cup B\cup C)=\mathbb{P}(A)+\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(A\cap B)-\mathbb{P}(A\cap C)-\mathbb{P}(B\cap C)+\mathbb{P}(A\cap B\cap C)

Plain language: add the single events, subtract every pairwise overlap because it was counted twice, then add back the triple overlap because subtracting three pairs removed it once too often. The pattern continues with alternating signs for more events.

Example — Two default indicators

Let $D_1$ be "firm 1 defaults" and $D_2$ be "firm 2 defaults," with $\mathbb{P}(D_1)=0.04$ , $\mathbb{P}(D_2)=0.03$ , and $\mathbb{P}(D_1\cap D_2)=0.01$ . Then

\mathbb{P}(\text{at least one defaults})=0.04+0.03-0.01=0.06

not $0.07$ . By the complement rule, both survive with probability $0.94$ . The overlap $0.01$ is large relative to the marginals: firms in the same sector tend to default together.

Since $\mathbb{P}(A\cap B)\ge 0$ , inclusion–exclusion gives two bounds that hold with no knowledge of the overlap at all:

\mathbb{P}\Big(\bigcup_{i=1}^n A_i\Big)\le\sum_{i=1}^n\mathbb{P}(A_i), \qquad \mathbb{P}(A\cap B)\ge\mathbb{P}(A)+\mathbb{P}(B)-1

The first is the union bound, obtained by dropping the subtracted overlap and repeating over $n$ events. The second rearranges the two-event formula using $\mathbb{P}(A\cup B)\le 1$ . Together with monotonicity, $\mathbb{P}(A\cap B)\le\min(\mathbb{P}(A),\mathbb{P}(B))$ , they bracket what the joint behavior can be when only the marginals are known. For two events these upper and lower bounds on the intersection are the Fréchet–Hoeffding bounds. Chapter 10 returns to the same bracket once dependence is modeled directly, instead of being left as an unknown overlap.

❌ Summing overlapping risks

A quarterly risk report lists three stress scenarios: rates shock with probability $0.15$ , equity drawdown $0.20$ , credit-spread blowout $0.10$ . It states the probability of "at least one stress event" as $0.15+0.20+0.10=0.45$ .

Why it breaks: the sum equals $\mathbb{P}(A\cup B\cup C)$ only when the scenarios are pairwise disjoint. Stress scenarios are the opposite of disjoint — a spread blowout usually arrives with an equity drawdown. Every overlap is counted more than once.

Consequence: $0.45$ is only an upper bound. Monotonicity gives the lower bound $0.20$ , since the union contains the equity drawdown. If the other two scenarios occur only alongside an equity drawdown, the true answer is $0.20$ . With enough listed risks, the naive sum exceeds $1$ — a reliable sign that the aggregation, not the world, is wrong. Combining risks needs the joint probabilities, not only the marginals.

Practice — Bracketing a basket

Five bonds each have default probability $0.03$ over the year, and nothing is known about how their defaults are related. Use monotonicity and the union bound to show that $\mathbb{P}(\text{at least one default})$ lies between $0.03$ and $0.15$ . Describe a joint arrangement of defaults that attains each end of the range.

06 · Countable Spaces and Running Totals

A finite list of buckets is not always enough. "Trading day of the first down move" has one outcome per positive integer. On that space, finite additivity does not by itself control an infinite union. The countable form of the axiom does.

Example — A valid weight on each waiting time

Let $\Omega=\{1,2,3,\ldots\}$ and put weight $p(k)=(1/2)^k$ on day $k$ . These weights are positive, and the geometric series sums to one:

\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k=\frac{1/2}{1-1/2}=1

so the list is a probability. The first down move falls within three days with probability

\mathbb{P}(\{1,2,3\})=\frac12+\frac14+\frac18=\frac78

and falls after day 3 with probability $1-7/8=1/8$ . There is no outcome called "never": that description is the empty set, and $\mathbb{P}(\emptyset)=0$ . Survival through $n$ days has probability $(1/2)^n$ , which can be made arbitrarily small and is still not the probability of an extra outcome you forgot to list. Chapter 04 derives weights of this shape from a constant daily down-move probability; here they are simply a list that obeys the axioms.

Definition — Continuity Along Nested Events

If $A_1\subseteq A_2\subseteq\cdots$ and $A=\bigcup_{n=1}^{\infty} A_n$ , countable additivity implies

\mathbb{P}(A_n)\to\mathbb{P}(A)

The increments $C_1=A_1$ and $C_n=A_n\setminus A_{n-1}$ are disjoint and union to $A$ , so $\mathbb{P}(A_n)$ is a partial sum of their probabilities and the infinite sum is $\mathbb{P}(A)$ . In the waiting-time model, $A_n=\{1,2,\ldots,n\}$ increases to $\Omega$ and $\mathbb{P}(A_n)=1-(1/2)^n$ increases to $1$ .

Plain language: as you lengthen the horizon, the probability of a nested event settles to the probability of everything it eventually covers. A cumulative default curve is exactly such a sequence. The number at the longest printed tenor is $\mathbb{P}(A_n)$ for that $n$ , not automatically $\mathbb{P}(A)$ .

❌ Adding cumulative probabilities across tenors

A credit memo lists cumulative default probabilities of $2\%$ , $5\%$ , and $9\%$ at one, three, and five years, then a "total default risk" of $2\%+5\%+9\%=16\%$ .

Why it breaks: "default by year 1" is a subset of "default by year 3", which is a subset of "default by year 5". The $5\%$ already contains the $2\%$ . Adding the three numbers counts the early defaults once for every later tenor. Additivity applies to the increments, which are disjoint: $2\%$ in the first year, $5\%-2\%=3\%$ over the next two years, and $9\%-5\%=4\%$ over the two years after that.

Consequence: the five-year cumulative probability is $9\%$ , the last number on the curve, and the increments sum to that same $9\%$ . The $16\%$ figure is not a probability of anything in the model. Chapter 03 shows a related trap with conditional (given-survival) default probabilities, which are not increments either and can sum to more than $1$ .

07 · Exercises

EXERCISE 2.1

Write $B$ as the disjoint union of $A$ and $B\setminus A$ , apply additivity, and use non-negativity on the leftover piece. For the bound, take $B=\Omega$ .

Since $A\subseteq B$ , the sets $A$ and $B\setminus A$ are disjoint and their union is $B$ . Additivity gives $\mathbb{P}(B)=\mathbb{P}(A)+\mathbb{P}(B\setminus A)$ , so $\mathbb{P}(B\setminus A)=\mathbb{P}(B)-\mathbb{P}(A)$ . Non-negativity gives $\mathbb{P}(B\setminus A)\ge 0$ , so $\mathbb{P}(A)\le\mathbb{P}(B)$ . Taking $B=\Omega$ and using normalization, $\mathbb{P}(A)\le\mathbb{P}(\Omega)=1$ . Only the three axioms were used; the upper bound $1$ never had to be assumed.

Using only the three axioms, prove that $A\subseteq B$ implies $\mathbb{P}(B\setminus A)=\mathbb{P}(B)-\mathbb{P}(A)$ and $\mathbb{P}(A)\le\mathbb{P}(B)$ . Deduce that $\mathbb{P}(A)\le 1$ for every event.

EXERCISE 2.2

The intersection is capped above by the smaller marginal (monotonicity) and below by $\mathbb{P}(A)+\mathbb{P}(B)-1$ . Then feed each extreme into inclusion–exclusion to get the range for the union.

Monotonicity gives $\mathbb{P}(A\cap B)\le\min(0.7,0.8)=0.7$ . The lower bound gives $\mathbb{P}(A\cap B)\ge 0.7+0.8-1=0.5$ . So $\mathbb{P}(A\cap B)\in[0.5,0.7]$ . Inclusion–exclusion gives $\mathbb{P}(A\cup B)=1.5-\mathbb{P}(A\cap B)\in[0.8,1.0]$ . The upper end $\mathbb{P}(A\cap B)=0.7$ means $\mathbb{P}(A\setminus B)=0$ : the index never rises in a month when spreads fail to tighten. The lower end $0.5$ means $A\cup B$ is certain: in every month at least one of the two happens.

Let $A$ be "the index rises this month" and $B$ be "credit spreads tighten this month," with $\mathbb{P}(A)=0.7$ and $\mathbb{P}(B)=0.8$ . Which values of $\mathbb{P}(A\cap B)$ are consistent with the axioms? What range does that imply for $\mathbb{P}(A\cup B)$ , and what does each extreme say about how the two events relate?

EXERCISE 2.3

Apply the three-event formula directly. For "exactly one," peel the overlaps off each marginal: the probability that only firm 1 defaults is $\mathbb{P}(D_1)$ minus the parts of $D_1$ shared with $D_2$ or $D_3$ .

(a) The marginals sum to $0.12$ , the pairwise overlaps to $0.024$ , so

$\mathbb{P}(D_1\cup D_2\cup D_3)=0.12-0.024+0.002=0.098.$

(b) Only $D_1$ and $D_2$ : $0.010-0.002=0.008$ . Only $D_1$ and $D_3$ : $0.008-0.002=0.006$ . Only $D_2$ and $D_3$ : $0.006-0.002=0.004$ . Then only firm 1: $0.05-0.008-0.006-0.002=0.034$ ; only firm 2: $0.04-0.008-0.004-0.002=0.026$ ; only firm 3: $0.03-0.006-0.004-0.002=0.018$ . Exactly one default has probability $0.034+0.026+0.018=0.078$ . All seven pieces are non-negative and add to $0.098$ , which confirms both the answer in (a) and that the inputs are consistent.

(c) The naive sum $0.12$ overstates the true $0.098$ by about $22\%$ . The no-default probability is $1-0.098=0.902$ , not $0.88$ .

A basket holds three bonds with one-year default probabilities $\mathbb{P}(D_1)=0.05$ , $\mathbb{P}(D_2)=0.04$ , $\mathbb{P}(D_3)=0.03$ , pairwise joint default probabilities $\mathbb{P}(D_1\cap D_2)=0.010$ , $\mathbb{P}(D_1\cap D_3)=0.008$ , $\mathbb{P}(D_2\cap D_3)=0.006$ , and $\mathbb{P}(D_1\cap D_2\cap D_3)=0.002$ . (a) Find the probability of at least one default. (b) Find the probability of exactly one default. (c) By how much does simply adding the three marginals overstate the answer to (a)?

EXERCISE 2.4

A geometric series $\sum_{k=0}^{\infty} ar^k$ with ratio $r$ between $0$ and $1$ sums to $a/(1-r)$ . Here $a=1/3$ and $r=2/3$ . "Never" is not an element of $\Omega=\{1,2,3,\ldots\}$ .

(a) The sum is $\sum_{k=1}^{\infty}(1/3)(2/3)^{k-1}$ . The first term is $1/3$ and the ratio is $2/3$ , so the sum equals $(1/3)/(1-2/3)=1$ . The weights are a probability. (b) $\mathbb{P}(\{1,2\})=(1/3)+(1/3)(2/3)=1/3+2/9=5/9$ . (c) The tail $\mathbb{P}(\{3,4,\ldots\})=1-5/9=4/9$ , which is also $(1/3)(2/3)^2/(1-2/3)=(4/27)/(1/3)=4/9$ . (d) No outcome in $\Omega$ is "the down move never arrives", so that description is $\emptyset$ and has probability $0$ . Survival past day $n$ is the tail $(2/3)^n$ , which tends to $0$ and never needs its own outcome.

Let $\Omega=\{1,2,3,\ldots\}$ be the day of the first down move, with weights $p(k)=\tfrac13\left(\tfrac23\right)^{k-1}$ . (a) Show that the weights sum to $1$ . (b) Compute the probability that the first down move falls on day 1 or day 2. (c) Compute the probability that it falls after day 2. (d) Why is the probability that a down move never arrives equal to $0$ in this model, even though every finite horizon still has positive probability of all up days?

08 · Chapter Summary

Concept	Meaning
Axioms	$\mathbb{P}(A)\ge 0$ , $\mathbb{P}(\Omega)=1$ , additive on disjoint events
Complement rule	$\mathbb{P}(A^c)=1-\mathbb{P}(A)$
Monotonicity	$A\subseteq B$ implies $\mathbb{P}(A)\le\mathbb{P}(B)$ ; cumulative curves never fall
Outcome weights	non-negative $p(\omega)$ summing to $1$ ; $\mathbb{P}(A)$ is the weight inside $A$
Uniform model	equal weights; favorable count over total count, valid only on the right space
Inclusion–exclusion	add events, subtract pairwise overlaps, add back the triple overlap
Union bound	summing marginals gives an upper bound, not the probability of the union
Fréchet–Hoeffding	$\mathbb{P}(A)+\mathbb{P}(B)-1 \le \mathbb{P}(A\cap B) \le \min(\mathbb{P}(A),\mathbb{P}(B))$
Nested cumulatives	add the increments $\mathbb{P}(A_n)-\mathbb{P}(A_{n-1})$ , never the cumulative levels themselves
Countable additivity	infinite disjoint unions add; waiting-time weights must sum to $1$

Next: Chapter 03 — Conditional Probability & Bayes, which updates probabilities once some information about the outcome has been observed.