MUTUALLY EXCLUSIVE EVENTS

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Mutually Exclusive Events

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◇Overview

Two events $A$ and $B$ are Mutually Exclusive (or Disjoint) if they share no common outcomes in the sample space: $A \cap B = \emptyset$. Consequently, they cannot occur simultaneously on a single trial of an experiment, so $P(A \cap B) = 0$.

◇Key Definitions

• Disjoint Set Condition: $A \cap B = \emptyset \implies P(A \cap B) = 0$.
• Special Additive Rule: For mutually exclusive events $A$ and $B$, $P(A \cup B) = P(A) + P(B)$.
• Pairwise Mutually Exclusive Collection: Events $E_1, E_2, \dots, E_k$ are mutually exclusive if $E_i \cap E_j = \emptyset$ for all $i \ne j$, giving $P\left(\bigcup_{i=1}^k E_i\right) = \sum_{i=1}^k P(E_i)$.
• Mutually Exclusive vs. Independent: Mutually exclusive is a set-theoretic property ($A \cap B = \emptyset$); independence is a probabilistic property ($P(A \cap B) = P(A)P(B)$). For $P(A), P(B) > 0$, they are opposites!

◇Worked Example

Problem: A single card is drawn from a standard $52$-card deck. Let $A$ be drawing a King ($4$ cards), $B$ be drawing a Queen ($4$ cards), and $C$ be drawing a Heart ($13$ cards).

• $A$ and $B$ are mutually exclusive ($A \cap B = \emptyset$), so $P(A \cup B) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$.
• $A$ and $C$ are not mutually exclusive because $A \cap C = \{\text{King of Hearts}\} \ne \emptyset$.

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• Related Notes: ◇intersection, ◇Fundamental Principle of Counting