Complement vs Event

The complement of an event $A$ is written $A^c$ (read "A complement"), and it holds exactly the outcomes of the sample space that $A$ does not. Two conditions define it together, and both matter: no outcome belongs to both sets, and no outcome is left out of the pair. Complementing twice returns you to where you started, since $(A^c)^c = A$, and the same pair of conditions is what forces $P(A) + P(A^c) = 1$, stated as the complement rule.

Roll a fair die and let $A$ be "at most 2", the set $\{1, 2\}$, so $P(A) = 2/6 \approx 0.3333$. Then $A^c$ is $\{3, 4, 5, 6\}$ with $P(A^c) = 4/6 \approx 0.6667$, and the two add to exactly 1.

"A and B cannot both happen, so B is the complement of A." Try $B = \{3, 4\}$ on that same die. It is mutually exclusive with $A$, since no outcome sits in both, yet $P(A) + P(B) = 2/3$ rather than 1, because 5 and 6 belong to neither. Complementary events are always mutually exclusive; mutually exclusive events are complementary only when they also leave nothing out.

Complementing a compound event flips the connector, and getting that backwards is the expensive error. Take $C = \{1, 3, 5\}$, the odd rolls. Then $A \cap C = \{1\}$, so the complement of "A and C" is $\{2, 3, 4, 5, 6\}$ at $5/6 \approx 0.8333$. That set is $A^c \cup C^c$, the union of the two complements, not $A^c \cap C^c$, which is $\{4, 6\}$ at $2/6 \approx 0.3333$. The rule runs the other way too: the complement of a union is the intersection of the complements, which is exactly why the complement of "at least one" is "none".

A complement is always taken relative to a stated sample space. Change what counts as possible and the same event acquires a different complement, so $S$ has to be settled before the word means anything.

An event is a subset of the sample space, named with a capital letter such as $A$ or $B$. An event holding a single outcome is called simple, one holding several is compound. Add up the probabilities of every outcome it contains and you have $P(A)$, so when those outcomes are equally likely the adding collapses into counting.

Roll one fair die. Let $A$ be "the roll is even", the set $\{2, 4, 6\}$, and let $B$ be "the roll is greater than 3", the set $\{4, 5, 6\}$. Then $P(A) = 3/6 = 0.5$ and $P(B) = 3/6 = 0.5$. Events combine into new events: "A and B" is $\{4, 6\}$ with probability $2/6 \approx 0.3333$, and "A or B" is $\{2, 4, 5, 6\}$ with probability $4/6 \approx 0.6667$.

"A or B means one of them but not both." Read that way, "A or B" would be $\{2, 5\}$ with probability $2/6 \approx 0.3333$, which is wrong. In probability "or" is inclusive: the event holds whenever at least one of $A$ and $B$ holds, so 4 and 6 stay in. The mirror-image slip is to add, $0.5 + 0.5 = 1$, which claims every roll satisfies "A or B" when a 1 or a 3 satisfies neither. The addition rule exists to strip out the outcomes counted twice.

Not every event can be handled by listing. "The mean of 40 measurements exceeds 12" is an event on a sample space far too large to write down, and its probability comes from a model rather than a count. At the two extremes, the empty set and the whole sample space are both events, with probabilities 0 and 1.

Events are what the rest of Unit 2 operates on. The complement of an event, the union of two events, and their intersection are all events themselves, which is what lets the rules be chained together.