UNION

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Union

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◇Overview

The Union of two events $A$ and $B$, denoted $A \cup B$, is the event consisting of all outcomes $\omega \in S$ that belong to $A$, or to $B$, or to both $A$ and $B$ (inclusive OR). In words, $A \cup B$ occurs if and only if at least one of the events $A$ or $B$ occurs.

◇Key Definitions

• Set Definition: $A \cup B = \{\omega \in S : \omega \in A \text{ or } \omega \in B\}$.
• Countable Union ($\bigcup_{i=1}^n A_i$): Occurs if and only if $\omega \in A_i$ for at least one index $i \in \{1, \dots, n\}$.
• Monotonicity Property: $A \subseteq A \cup B$ and $B \subseteq A \cup B$, so $\max(P(A), P(B)) \le P(A \cup B) \le \min(1, P(A) + P(B))$.
• Boole's Inequality (Union Bound): For any finite or countable sequence of events $A_1, A_2, \dots, A_n$:
  $P\left(\bigcup_{i=1}^n A_i\right) \le \sum_{i=1}^n P(A_i)$

◇Worked Example

Problem: On a single roll of a fair six-sided die ($S = \{1, 2, 3, 4, 5, 6\}$), let $A = \{1, 2, 3\}$ and $B = \{2, 4, 6\}$. Find $A \cup B$ and $P(A \cup B)$.

Solution:

• Combining unique elements from both sets gives $A \cup B = \{1, 2, 3, 4, 6\}$.
• $P(A \cup B) = \frac{|A \cup B|}{|S|} = \frac{5}{6}$ (which matches $P(A) + P(B) - P(A \cap B) = \frac{3}{6} + \frac{3}{6} - \frac{1}{6} = \frac{5}{6}$ since $A \cap B = \{2\}$).

◇References & Navigation

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• Related Notes: ◇Complement, ◇Event