A random experiment is one whose outcome cannot be predicted with certainty in advance, even though every possible result is known beforehand. Two features must hold: the experiment has more than one possible result, and the set of all possible results is known before the trial. Tossing a coin, rolling a die, or drawing a card from a well-shuffled pack are all random experiments — you know the menu of possibilities, but not which one will turn up on a given trial. By contrast, "heating water to $100^{\circ}$C at sea level" is deterministic: the outcome is fixed, so it is not a random experiment.
Each individual result of the experiment is an outcome (or sample point). The set of all possible outcomes is the sample space, written $S$ (some books use $U$ or $\Omega$). For a single die, $S = \{1, 2, 3, 4, 5, 6\}$; for one toss of a coin, $S = \{H, T\}$. The total number of sample points is $n(S)$.
Listing the sample space correctly is the single most important skill in this chapter — a wrong $S$ poisons every probability that follows. A few standard cases:
| Experiment | Sample space $S$ | $n(S)$ |
|---|---|---|
| One coin | $\{H, T\}$ | $2$ |
| Two coins (or one coin twice) | $\{HH, HT, TH, TT\}$ | $4$ |
| Three coins | $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$ | $8$ |
| One die | $\{1, 2, 3, 4, 5, 6\}$ | $6$ |
| Two dice | $\{(1,1), (1,2), \dots, (6,6)\}$ (ordered pairs) | $36$ |
| Drawing one card | the $52$ cards (or $\{$spade, heart, diamond, club$\}$ for suit) | $52$ |
For multi-stage (sequential) trials — toss a coin, then roll a die; or draw cards one after another — pair each result of the first stage with every result of the next. A tree diagram makes this systematic: each branch is one outcome. For a coin then a die, $S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}$, so $n(S) = 2 \times 6 = 12$. The same multiplication principle gives $2^n$ outcomes for $n$ coins and $6^n$ for $n$ dice.
A standard pack of 52 playing cards recurs throughout the chapter, so memorise its structure: four suits — spades $\spadesuit$, hearts $\heartsuit$, diamonds $\diamondsuit$, clubs $\clubsuit$ — of $13$ cards each. Hearts and diamonds are red ($26$ cards); spades and clubs are black ($26$ cards). Each suit runs Ace, $2$–$10$, Jack, Queen, King; the face cards are Jack, Queen, King ($3$ per suit, so $12$ in all).
An event is any subset of the sample space — a collection of outcomes we choose to single out. "Getting an even number on a die" is the event $E = \{2, 4, 6\}$, a subset of $S$. This is the bridge to the chapter on Sets: events are sets, and the sample space is the universal set. We say an event $E$ occurs if the actual outcome of the trial belongs to $E$.
Because events are sets, the set operations carry direct meanings. The complement $\overline{A}$ (also written $A'$) is the event "$A$ does not happen" and equals $S - A$. The union $A \cup B$ is "$A$ or $B$ (or both)"; the intersection $A \cap B$ is "$A$ and $B$ together"; the difference $A - B = A \cap \overline{B}$ is "$A$ but not $B$".
| Type of event | Meaning | Example (single die) |
|---|---|---|
| Simple (elementary) | Exactly one outcome | $\{4\}$ — getting a 4 |
| Compound | Two or more outcomes | $\{2, 4, 6\}$ — an even number |
| Sure (certain) | The whole sample space $S$ | $\{1,2,3,4,5,6\}$ — a number $\le 6$ |
| Impossible | The empty set $\varnothing$ | getting a 7 |
| Mutually exclusive | $A \cap B = \varnothing$ (cannot both occur) | $\{1,3,5\}$ and $\{2,4,6\}$ |
| Exhaustive | Union covers all of $S$ | $\{1,2,3\}$ and $\{3,4,5,6\}$ |
Two events $A$ and $B$ are mutually exclusive (disjoint) when they share no outcome, so they can never happen on the same trial. A collection of events is exhaustive when their union is the entire sample space, so at least one of them must occur. When events are both mutually exclusive and exhaustive they partition $S$ into non-overlapping pieces that together account for everything — for example $A$ and its complement $\overline{A}$ always partition $S$.
Two relations are worth stating cleanly. Events $A$ and $B$ are mutually exclusive exactly when $A \cap B = \varnothing$; they are exhaustive exactly when $A \cup B = S$. These two ideas are independent of each other: events can be mutually exclusive without being exhaustive (two faces of a die), exhaustive without being mutually exclusive (overlapping sets whose union is $S$), both, or neither.
Deeper Insight — probability is set theory with a measuring tape: The reason the chapter on Sets sits in the same syllabus as this one is no accident. A sample space is just a universal set, an event is just a subset, and "or", "and", "not" translate exactly into union, intersection, and complement. This is why the vocabulary feels familiar: "mutually exclusive" is the everyday name for disjoint sets, and "exhaustive" simply restates that a family of subsets covers the universal set. Once you accept that an event is a set of outcomes, the laws you already know — De Morgan's laws, the distributive laws — transfer wholesale, which is why "neither $A$ nor $B$" equals $\overline{A} \cap \overline{B} = \overline{A \cup B}$. Probability then adds one new ingredient on top of this set machinery: a way to measure how large an event is, relative to the whole sample space. Get the set picture firm first — especially the habit of writing out $S$ in full — and the numerical rules in the next two topics will feel like bookkeeping rather than new ideas.