Events

A random experiment produces one outcome from a fixed collection called the sample space $S$. Any subset $A \subseteq S$ is an event; it "occurs" whenever the actual outcome happens to lie inside $A$.

Sizes of events. A simple (elementary) event contains exactly one outcome, such as $\{H\}$ for a coin toss. A compound event contains more than one outcome, such as $A = \{2,4,6\}$ for "an even number" on a die. Two boundary cases matter: the sure event is $S$ itself (it always occurs, since the outcome is always somewhere in $S$), and the impossible event is $\varnothing$ (it never occurs, since no outcome lies in an empty set).

Combining events as sets. Because events are sets, everyday phrases translate directly into set algebra:

• 'Not $A$' is the complement $A' = S - A$ — the event that $A$ fails to occur.
• '$A$ or $B$' is the union $A \cup B$ — the event that at least one of $A$, $B$ occurs.
• '$A$ and $B$' is the intersection $A \cap B$ — the event that both occur together.

where: $S$ is the sample space, $A$ and $B$ are events (subsets of $S$), and $x$ denotes a generic outcome. No new machinery is needed — everything you already know about sets carries over unchanged to events.

Two relationships between events. $A$ and $B$ are mutually exclusive if $A \cap B = \varnothing$: they share no outcome, so they cannot both happen on the same trial. A family $A_1, A_2, \dots, A_n$ is exhaustive if $A_1 \cup A_2 \cup \cdots \cup A_n = S$: between them, they cover every possible outcome, so something in the family must occur. When a family is both mutually exclusive and exhaustive, it partitions $S$ into non-overlapping pieces that together account for the whole sample space — a structure that becomes essential once probabilities are assigned, because each outcome then belongs to exactly one piece.

There is no separate "logic of chance" to learn here — the whole point of writing events as sets is that the logic students already own for sets transfers over without change. "Not $A$", "$A$ or $B$", and "$A$ and $B$" are not new ideas dressed up in new notation; they are complement, union and intersection, and every identity you know from set theory — $A \cup A' = S$, $A \cap A' = \varnothing$, De Morgan's laws — automatically becomes a true statement about events. This is why the language is worth learning carefully before any probability numbers appear: get the set description of an event wrong, and every later computation built on it inherits the mistake. The reason mutually exclusive and exhaustive matter so much together is that a mutually exclusive and exhaustive family is a partition of $S$: no outcome is left out (exhaustive), and no outcome is double-counted across two different events in the family (mutually exclusive). Picture $S$ as a sheet of paper. An exhaustive family covers the whole sheet with no gaps; a mutually exclusive family cuts the sheet with no pieces overlapping. Only when both hold at once do you get a clean tiling — every single outcome sitting in exactly one piece. That property is precisely what later lets probabilities of a partition add up correctly to $1$, and it is why exam questions repeatedly ask you to check both conditions rather than just one: exhaustive without mutually exclusive still leaves overlaps (an outcome counted twice), and mutually exclusive without exhaustive still leaves gaps (an outcome counted nowhere).

A single die is thrown once. With $S = \{1,2,3,4,5,6\}$, let $A$ = "an even number appears" and $B$ = "a number greater than 4 appears". Find $A \cap B$, $A \cup B$ and $A'$, and check whether the pair $\{A, A'\}$ is mutually exclusive and exhaustive.

Step 1. Write down $A$ and $B$ as subsets of $S$. "Even" picks out multiples of 2, and "greater than 4" picks out numbers strictly above 4:

Step 2. Find $A \cap B$, the outcomes common to both descriptions — even and greater than 4:

Since $A \cap B = \{6\} \neq \varnothing$, $A$ and $B$ share an outcome, so they are not mutually exclusive.

Step 3. Find $A \cup B$, every outcome satisfying at least one description:

Step 4. Find $A'$, the complement of $A$ with respect to $S$ (not with respect to $B$ or any other set) — the odd numbers:

Step 5 — Competitive stretch. Check $\{A, A'\}$ against the partition conditions. Mutually exclusive: $A \cap A' = \{2,4,6\} \cap \{1,3,5\} = \varnothing$, so yes. Exhaustive: $A \cup A' = \{2,4,6\} \cup \{1,3,5\} = \{1,2,3,4,5,6\} = S$, so yes. Any event and its complement always form a partition of $S$ into exactly two pieces — a fact that holds for every event $A$, not just this one. Answer: $A \cap B = \{6\}$, $A \cup B = \{2,4,5,6\}$, $A' = \{1,3,5\}$, and $\{A, A'\}$ is mutually exclusive and exhaustive.

• Confusing "mutually exclusive" ($A \cap B = \varnothing$, they can't both happen) with "exhaustive" ($A \cup B = S$, together they cover everything) — neither property implies the other.
• Calling any small-looking subset a "simple" event; a simple event must contain exactly one outcome, while a compound event can contain any number greater than one.
• Taking the complement of $A$ against the wrong set — $A'$ is always $S - A$, measured against the full sample space, not against another event.
• Treating $S$ or $\varnothing$ as outcomes themselves; they are events (sets of outcomes), and $\varnothing$ in particular contains none at all.

Ask the tutor: Stuck on turning a tricky compound-event description — like "at least two heads" or "sum is a multiple of 3" — into a set, or unsure whether two events you've listed are mutually exclusive, exhaustive, both, or neither? Ask the AI tutor to walk through the sample space with you, outcome by outcome.