Sample Spaces and Events
Before we can put a number on the chance of something happening, we have to write down what “something” even is. That “something” is an event, and events live inside a sample space. The two definitions are absolutely fundamental: every later theorem references them, and most beginner mistakes come from a fuzzy notion of one or the other.
The Sample Space $S$
An experiment is anything whose outcome is not predictable in advance. The set of all possible outcomes is the sample space, written $S$ (Ross also uses $\Omega$).
The sample space depends on what we choose to record. The same physical experiment can have several different sample spaces depending on the question. A few canonical examples:
| Experiment | Sample space $S$ | $|S|$ |
|---|---|---|
| Toss a coin once | $\{H, T\}$ | $2$ |
| Toss a coin twice (record sequence) | $\{HH, HT, TH, TT\}$ | $4$ |
| Toss a coin twice (record number of heads) | $\{0, 1, 2\}$ | $3$ |
| Roll a six-sided die | $\{1, 2, 3, 4, 5, 6\}$ | $6$ |
| Roll two dice (ordered pair) | $\{(i, j) : 1 \le i, j \le 6\}$ | $36$ |
| Lifetime of a light bulb (hours) | $[0, \infty)$ | uncountable |
| Toss a coin until first head | $\{H, TH, TTH, TTTH, \ldots\}$ | countably infinite |
Events
An event is any subset of the sample space.
Two extreme events show up in every proof:
- The certain event $S$ itself: it always occurs because every outcome is in $S$.
- The impossible event $\emptyset$: it never occurs because no outcome is in $\emptyset$.
An event made of a single outcome, $\{s\}$, is called an elementary event. Bigger events are unions of elementary events.
Examples
Roll one die so $S = \{1,2,3,4,5,6\}$. Some events:
- “Even”: $E = \{2, 4, 6\}$.
- “Greater than 4”: $F = \{5, 6\}$.
- “Prime”: $G = \{2, 3, 5\}$.
- “Eight”: $H = \emptyset$ (impossible).
Combining Events: Set Operations
Events are sets, and we combine them with the standard set operations. The translation between English and set notation is the most important reflex to build in this chapter.
| Operation | English | Definition |
|---|---|---|
| $E \cup F$ | "$E$ or $F$" (or both) | $\{s \in S : s \in E \text{ or } s \in F\}$ |
| $E \cap F$ or $EF$ | "$E$ and $F$" | $\{s \in S : s \in E \text{ and } s \in F\}$ |
| $E^c$ or $\bar E$ | "not $E$" | $\{s \in S : s \notin E\}$ |
| $E \setminus F$ | "$E$ but not $F$" | $E \cap F^c$ |
| $E \subseteq F$ | "$E$ implies $F$" | every $s \in E$ satisfies $s \in F$ |
Ross often writes the intersection without the symbol — $EF$ instead of $E \cap F$ — because intersection corresponds to the word “and”, which carries no extra symbol in English. We will use both styles; pick whichever reads naturally.
Why does $\cup$ mean “or” and $\cap$ mean “and”?
Because event $E$ “occurs” means “the outcome lies in $E$”. So:
- “$E$ or $F$ occurs” means the outcome is in $E$, or in $F$, or in both — that is exactly the union $E \cup F$.
- “$E$ and $F$ both occur” means the outcome is in $E$ and in $F$ — the intersection $E \cap F$.
- “$E$ does not occur” means the outcome is in $S \setminus E = E^c$.
Generalised Unions and Intersections
For a family of events $E_1, E_2, \ldots, E_n$ (or even infinitely many) we write
$\displaystyle \bigcup_{i=1}^{n} E_i = E_1 \cup E_2 \cup \cdots \cup E_n, \qquad \bigcap_{i=1}^{n} E_i = E_1 \cap E_2 \cap \cdots \cap E_n.$
$\bigcup_i E_i$ occurs iff at least one $E_i$ occurs; $\bigcap_i E_i$ occurs iff every $E_i$ occurs. The same notation works for countable unions $\bigcup_{i=1}^{\infty} E_i$, which we will need when we state the third axiom.
Mutually Exclusive (Disjoint) Events
Two events $E$ and $F$ are mutually exclusive (also called disjoint) when
$E \cap F = \emptyset$,
that is, they cannot both occur. A family $\{E_i\}$ is pairwise mutually exclusive if $E_i \cap E_j = \emptyset$ for every $i \ne j$.
This idea is so central that the third axiom of probability is stated only for disjoint events: probabilities of disjoint events add.
Mutually exclusive vs complementary
Complementary events are always mutually exclusive, but the reverse is not true. $E$ and $E^c$ partition $S$ into exactly two pieces; mutually exclusive events $E, F$ might leave a third region $S \setminus (E \cup F)$ uncovered.
De Morgan's Laws
Negation flips $\cup$ and $\cap$:
$\displaystyle \left( \bigcup_{i=1}^{n} E_i \right)^c = \bigcap_{i=1}^{n} E_i^c, \qquad \left( \bigcap_{i=1}^{n} E_i \right)^c = \bigcup_{i=1}^{n} E_i^c.$
In English: "none of the events occur" = $\bigcap_i E_i^c$ = $\left(\bigcup_i E_i\right)^c$. "Not all of the events occur" = $\bigcup_i E_i^c$ = $\left(\bigcap_i E_i\right)^c$.
Why It Works
Pick any outcome $s \in S$. Then $s \in (\bigcup_i E_i)^c$ iff $s \notin \bigcup_i E_i$ iff $s$ is in no $E_i$ iff $s \in E_i^c$ for every $i$ iff $s \in \bigcap_i E_i^c$. The two sides agree element by element, so they're the same set. The other identity follows by complementing both sides of the first one and using $(E^c)^c = E$.
Visual: Venn Diagrams Step by Step
Step through the four basic regions of a two-event Venn diagram and see how each set operation lights up a different combination of regions.
▶ Visual: Building $E \cup F$, $E \cap F$, $E^c$, and $E \setminus F$
Each step toggles a different set operation. Watch how the shaded regions correspond to "or", "and", "not", and "but not".
Properties That Will Show Up Everywhere
The set algebra you'll use in proofs:
- Commutative. $E \cup F = F \cup E$, $E \cap F = F \cap E$.
- Associative. $(E \cup F) \cup G = E \cup (F \cup G)$, similarly for $\cap$.
- Distributive. $E \cap (F \cup G) = (E \cap F) \cup (E \cap G)$ and $E \cup (F \cap G) = (E \cup F) \cap (E \cup G)$.
- Identity. $E \cup \emptyset = E$, $E \cap S = E$.
- Complement. $E \cup E^c = S$, $E \cap E^c = \emptyset$, $(E^c)^c = E$.
Worked Example
Roll a single die. Let $E$ = "even", $F$ = "greater than $3$".
- $S = \{1,2,3,4,5,6\}$, $E = \{2,4,6\}$, $F = \{4,5,6\}$.
- $E \cup F = \{2,4,5,6\}$ ("even or larger than 3").
- $E \cap F = \{4,6\}$ ("even and larger than 3").
- $E^c = \{1,3,5\}$ ("odd").
- $E \setminus F = \{2\}$ ("even but not larger than 3").
- $E^c \cap F^c = \{1, 3\}$. By De Morgan this equals $(E \cup F)^c = S \setminus \{2,4,5,6\} = \{1,3\}$. ✓
Common Mistakes
- Confusing “mutually exclusive” with “independent”. They're unrelated until we define independence in the next chapter; in fact, two events with positive probability cannot be both mutually exclusive and independent.
- Picking the wrong sample space. If outcomes you treat as equally likely are not actually equally likely, every probability you compute will be wrong.
- Forgetting that $E$ and $F$ may overlap. $|E \cup F|$ is not $|E| + |F|$ in general — that's a Part 4 inclusion–exclusion lesson.
- Reading "$E$ but not $F$" as $E^c \cap F$ instead of $E \cap F^c$.
Practice
- Toss a coin three times. Write out $S$. Let $E$ = "exactly two heads" and $F$ = "first toss is a head". Find $E$, $F$, $E \cup F$, $E \cap F$, $E^c$.
- Two cards are drawn without replacement from a 52-card deck. Describe in words: $E \cap F$, $E^c$, $E \setminus F$ where $E$ = "first card is an ace", $F$ = "second card is a heart".
- Three students sit an exam. Express “at least one fails” both with $\cup$ and using De Morgan with $\cap$.
- Roll two dice. Show that "sum is even" and "first die is odd" are not mutually exclusive but neither contains the other.
- Prove that for any events $A, B$: $A = (A \cap B) \cup (A \cap B^c)$, and that the two pieces are disjoint. (This identity will be the engine of the addition rule in Part 4.)