Complement

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.