Complement vs Mutually Exclusive Events

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.

Two events $A$ and $B$ are mutually exclusive, or disjoint, when no outcome belongs to both, so $P(A \cap B) = 0$ ($\cap$ is the intersection sign, read "and"). Said with conditional probability, $P(A \mid B) = 0$: once $B$ has happened, $A$ is off the table for that trial. The addition rule then loses its overlap term and becomes $P(A \cup B) = P(A) + P(B)$, and the same plain sum extends to any collection of events that are disjoint in every pair.

Roll one fair die, with $A = \{1, 2\}$ and $B = \{5, 6\}$. Each has probability $2/6 \approx 0.3333$, no outcome is in both, so $P(A \cap B) = 0$ and $P(A \cup B) = 0.3333 + 0.3333 = 0.6667$.

"A and B cannot happen together, so neither one affects the other, so they are independent." Put numbers on that same pair. Independence demands $P(A \cap B) = P(A)P(B)$, and here $P(A)P(B) = (1/3)(1/3) = 1/9 \approx 0.1111$ while $P(A \cap B) = 0$. The two are not equal, so the events are dependent, and about as dependent as events get: learning that $B$ occurred drops the probability of $A$ from $1/3$ to 0. In general, if $P(A) > 0$ and $P(B) > 0$ and the events are mutually exclusive, then $P(A \cap B) = 0 \ne P(A)P(B)$, so they cannot be independent. Disjoint is the opposite of unrelated, not a version of it.

The condition that both probabilities exceed 0 is doing real work. Let $B$ be "the die shows a 7", so $P(B) = 0$. Then $P(A \cap B) = 0$ and $P(A)P(B) = 0$ as well, which makes $A$ and $B$ disjoint and independent at the same time. That degenerate case is the only escape, which is why the rule is always stated with both probabilities above zero.

Justifying that two events are mutually exclusive from their joint probability is the whole of topic 2.5.