Mutually Exclusive Events

What does "mutually exclusive" mean?

Two events are mutually exclusive (or disjoint) when they cannot happen simultaneously. In set language, their intersection is empty: $A \cap B = \varnothing$, so $P(A \cap B) = 0$.

A single card cannot be both a King and a Queen at the same time, so "drawing a King" and "drawing a Queen" are mutually exclusive.

The rule (for mutually exclusive events only)

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

Extended to many events: if $A_1, A_2, \ldots, A_n$ are pairwise mutually exclusive, $P(A_1 \cup A_2 \cup \cdots \cup A_n) = P(A_1) + P(A_2) + \cdots + P(A_n)$

The general addition rule (for any two events) subtracts the overlap: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. When they are mutually exclusive, that overlap is zero and you get the simpler rule above.

Step-by-step solution

Step 1 — Find $P(\text{King})$. There are 4 Kings in 52 cards: $P(K) = \dfrac{4}{52} = \dfrac{1}{13}$.

Step 2 — Find $P(\text{Queen})$. Similarly, $P(Q) = \dfrac{4}{52} = \dfrac{1}{13}$.

Step 3 — Check mutual exclusivity. A card is either a King or a Queen (or neither); it cannot be both. ✓

Step 4 — Add: $P(K \text{ or } Q) = \tfrac{4}{52} + \tfrac{4}{52} = \tfrac{8}{52} = \boxed{\tfrac{2}{13}}$

That's about 15.4%.

Verification by direct count

There are $4 + 4 = 8$ favorable cards (4 Kings + 4 Queens), out of 52. $P = 8/52 = 2/13$. ✓

Contrast: events that are NOT mutually exclusive

$P(\text{King OR Heart})$ requires subtracting the overlap (the King of Hearts), because that single card is both. See the Addition Rule problem for the full treatment.

Visual intuition

On a Venn diagram, mutually exclusive events appear as two non-overlapping circles. The probability of their union is just the sum of the two separate areas, with nothing counted twice.

Common mistakes

• Confusing "mutually exclusive" with "independent." They are not the same:
  • Mutually exclusive: cannot both happen → $P(A \cap B) = 0$.
  • Independent: knowledge of one does not change the other → $P(A \cap B) = P(A) P(B)$. In fact, if $P(A), P(B) > 0$ and $A, B$ are mutually exclusive, they are not independent (knowing one happened rules the other out).
• Adding overlapping events directly. If events can co-occur, pure addition over-counts the overlap.
• Forgetting the rule extends to more than two events — but only if they are pairwise mutually exclusive.

Try it in the visualization

Toggle the "events are mutually exclusive" switch and see the Venn-diagram circles slide apart (no overlap) vs. cross (overlap). The addition formula updates to match, showing where the correction term appears.