Disjoint events are as dependent as events get

Events $A$ and $B$ are mutually exclusive (or disjoint) when they cannot both occur on the same repetition of the random process. In sample-space terms they share no outcomes, so

$$P(A \text{ and } B) = 0.$$

Drawing one card, "the card is a heart" and "the card is a club" are mutually exclusive: no card is both. Rolling one die, "the roll is even" and "the roll is 3" are mutually exclusive. But "the card is a heart" and "the card is a face card" are not: the jack, queen, and king of hearts sit in both events.

The test is mechanical: name an outcome that satisfies both. If one exists, the events are not mutually exclusive, however unrelated they sound. Being disjoint is a property of the outcomes, not of the topics the events describe.

For any two events,

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).$$

The subtraction is bookkeeping: outcomes in both events get counted in $P(A)$ and again in $P(B)$, so one copy is removed. In a single card draw, $P(\text{heart or face card}) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} \approx 0.423$, where the 3 are the face cards of hearts. Skipping the subtraction gives $\frac{25}{52}$, and the three cards counted twice are exactly the difference.

When $A$ and $B$ are mutually exclusive, $P(A \text{ and } B) = 0$ and the rule collapses to $P(A \text{ or } B) = P(A) + P(B)$. That is not a separate rule; it is the general rule with a zero in it. Two habits follow. Subtract only when there is an overlap to subtract, since removing a nonexistent intersection from disjoint events understates the answer. And read "or" inclusively: in probability, "$A$ or $B$" means at least one of them, including both.

The 0-to-1 rule catches this error for free. Adding overlapping percentages in a survey routinely produces a total above 1, which is the signal that an intersection was counted twice.

The most costly confusion in this unit is treating mutually exclusive and independent as the same idea. They are close to opposites.

Independence means learning that $B$ occurred does not change the probability of $A$: $P(A \mid B) = P(A)$. Mutual exclusivity means learning that $B$ occurred tells you $A$ definitely did not: $P(A \mid B) = 0$. If $P(A) > 0$, those two statements contradict each other, so:

1. Two events with positive probability that are mutually exclusive are necessarily dependent.
2. Two independent events with positive probability necessarily overlap, since $P(A \text{ and } B) = P(A)P(B) > 0$.

A concrete pair: rolling a die once, "the roll is a 1" and "the roll is a 6" are disjoint, and knowing the roll was a 1 changes $P(6)$ from $\frac{1}{6}$ to 0. That is dependence at full strength. Meanwhile rolling a die twice, "first roll is 1" and "second roll is 6" are independent and can perfectly well both happen.

The source of the confusion is the word: "exclusive" sounds like "separate", and separate sounds like unrelated. Nothing about the picture is unrelated. On a Venn diagram, disjoint circles do not touch, and that gap is precisely what makes one event's occurrence informative about the other.

An event and its complement are the standard mutually exclusive pair: $A$ and $A^c$ cannot both occur, and one of them must, so $P(A) + P(A^c) = 1$.

More generally, a partition of the sample space is a set of mutually exclusive events that together cover every outcome, so their probabilities sum to 1. The categories of any single categorical variable form a partition: a student's grade is exactly one of freshman, sophomore, junior, or senior, so those four probabilities add to 1 and any two of them are disjoint.

This is where two-way tables and Venn diagrams meet. Each interior cell of a two-way table is an intersection, the cells are mutually exclusive, and they partition the sample space, which is why the joint proportions across the whole table sum to 1. Verifying that a set of events partitions the space, disjoint and exhaustive, is often faster than computing the probability asked for directly.