Mutually exclusive events

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.