At Least One Probability vs Complement

At least one and none are complements: every result has either zero occurrences or one or more, with nothing between them and no overlap. So $P(\text{at least one}) = 1 - P(\text{none})$, and the right-hand side stays a single term however many trials there are, while the direct route sums exactly one, exactly two, exactly three, and onward.

Roll a fair six-sided die four times. One roll misses a 6 with probability $5/6$. If the rolls are independent and each carries the same probability, four misses in a row is $(5/6)^4 = 625/1296 \approx 0.4823$, so $P(\text{at least one } 6) = 1 - 0.4823 = 0.5177$. Enumerating all $6^4 = 1296$ ordered results confirms it: 671 of them contain a 6, and $671/1296 \approx 0.5177$.

Independence and a constant success probability are conditions, not decoration, and the power step is what fails without them. Deal 4 cards from a standard deck and the chance of at least one ace is $1 - \binom{48}{4} / \binom{52}{4} = 1 - 194580/270725 \approx 0.2813$. Treating the draws as independent gives $1 - (48/52)^4 \approx 0.2740$, wrong because each card dealt changes the deck for the next.

Two wrong sentences turn up constantly. At least one is the opposite of exactly one, so subtract $P(X = 1)$: that gives $1 - 500/1296 \approx 0.6142$ for the dice above against the true 0.5177, because it throws away an outcome you wanted and keeps the one you did not. And four rolls at $1/6$ each is $4/6 \approx 0.667$: adding probabilities across trials fails its own sanity check, since six rolls would then reach 1, a certainty, while the true value is $1 - (5/6)^6 \approx 0.6651$.

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.