Introduction to Probability and Event Operations

Note on Inclusion: If we say $A \subset B$, it means that if event $A$ occurs, then event $B$ must also occur. For example, if $A$ is rolling a 2 on a die and $B$ is rolling an even number, then $A \subset B$.

Mathematical Operations on Events

Just as we perform arithmetic on numbers, we can perform logical operations on events to create new ones.

Union and Intersection

• Union ($A \cup B$): The event that $A$ occurs, or $B$ occurs, or both occur.
• Intersection ($A \cap B$): The event that both $A$ and $B$ occur simultaneously.

Difference and Disjointness

• Difference ($A - B$): The set of outcomes that belong to $A$ but do not belong to $B$.
• Disjoint (Mutually Exclusive) Events: Two events are disjoint if they have no outcomes in common, meaning their intersection is empty ($A \cap B = \emptyset$).
• Pairwise Disjoint: If you have a collection of multiple events where any two selected events are disjoint, the entire collection is said to be pairwise disjoint.

Algebraic Properties

To solve complex probability problems, you must master the following algebraic laws:

1. De Morgan’s Laws: These describe how complements interact with unions and intersections: $(A \cup B)^c = A^c \cap B^c$ $(A \cap B)^c = A^c \cup B^c$
2. Distributive Properties: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
3. Associative Properties: The grouping of operations does not change the result when performing the same operation repeatedly: $(A \cup B) \cup C = A \cup (B \cup C)$

Worked Example: Tossing a Coin Three Times

Let's apply these concepts to a practical scenario. Suppose a coin is tossed three times. The sample space $S$ contains eight outcomes: $S = {HHH, THH, HTH, HHT, HTT, THT, TTH, TTT}$

Define the following events:

• $A$: At least one head is obtained.
• $B$: A head is obtained on the second toss.
• $C$: A tail is obtained on the third toss.
• $D$: No heads are obtained.

Problem 1: Find $A^c$ Since $A$ is "at least one head," the complement $A^c$ must be "no heads." Looking at our sample space, the only outcome with no heads is ${TTT}$. Thus, $A^c = D = {TTT}$.

Problem 2: Find $B \cap C$ We look for outcomes that satisfy both $B$ (head on 2nd toss) and $C$ (tail on 3rd toss):

• $B = {HHH, THH, HHT, THT}$
• $C = {HHT, HTT, THT, TTT}$ The common elements are ${HHT, THT}$. Therefore, $B \cap C = \{HHT, THT\}$.

Exam Tip: When dealing with "at least one" problems, it is often much faster to calculate the probability of the complement ("none") and subtract it from 1.