Algebra of Events

The Algebra of Events

In the chapter on sets, you learned how to combine sets using operations like union, intersection, difference, and complement. Since events are just subsets of a sample space $S$, we can combine them using the exact same set operations. This gives us a powerful language to describe complex outcomes from simple ones.

Let $A$, $B$, and $C$ be any events associated with an experiment whose sample space is $S$. Here is how we build new events from old ones.

---

1. The Complementary Event: ‘not A’

For every event $A$, there is a corresponding event called the complement of $A$, written as $A'$ or $A^c$. It is also called the event ‘not $A$’.

Definition.

$A' = \{\omega : \omega \in S \text{ and } \omega \notin A\} = S - A$.

In words: $A'$ contains every outcome of the experiment that is not in $A$.

Example. Toss three coins. The sample space is

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.

Let $A$ be the event ‘only one tail appears’, so $A = \{HTH, HHT, THH\}$.

Then the complementary event ‘not $A$’ is

$A' = \{HHH, HTT, THT, TTH, TTT\}$.

---

2. The Event ‘A or B’

Recall that the union of two sets $A$ and $B$, written $A \cup B$, contains all elements that are in $A$ or in $B$ (or in both). When $A$ and $B$ are events, $A \cup B$ is the event ‘$A$ or $B$’ — meaning the event that at least one of $A$ or $B$ occurs.

Definition.

Event ‘$A$ or $B$’ $= A \cup B = \{\omega : \omega \in A \text{ or } \omega \in B\}$.

Example. Roll a die once. Let $A = \{2, 3, 5\}$ (prime numbers) and $B = \{1, 3, 5\}$ (odd numbers). Then

‘$A$ or $B$’ $= A \cup B = \{1, 2, 3, 5\}$.

---

3. The Event ‘A and B’

The intersection of two sets $A$ and $B$, written $A \cap B$, contains only those elements that belong to both $A$ and $B$. As an event, $A \cap B$ is the event ‘$A$ and $B$’ — meaning both events occur together.

Definition.

Event ‘$A$ and $B$’ $= A \cap B = \{\omega : \omega \in A \text{ and } \omega \in B\}$.

Example. Throw a die twice. Let $A$ be ‘score on the first throw is six’ and $B$ be ‘sum of the two scores is at least 11’. Then

$A = \{(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$,

$B = \{(5,6), (6,5), (6,6)\}$.

So ‘$A$ and $B$’ $= A \cap B = \{(6,5), (6,6)\}$.

This pair of outcomes represents: “the first throw is six and the sum is at least 11.”

---

4. The Event ‘A but not B’

The set difference $A - B$ (also written $A \setminus B$) contains all elements that are in $A$ but not in $B$. As an event, $A - B$ is the event ‘$A$ but not $B$’ — meaning $A$ occurs while $B$ does not.

Definition.

Event ‘$A$ but not $B$’ $= A - B = \{\omega : \omega \in A \text{ and } \omega \notin B\}$.

A useful identity: $A - B = A \cap B'$.

Example. Using the same die-roll events $A = \{2, 3, 5\}$ and $B = \{1, 3, 5\}$,

‘$A$ but not $B$’ $= A - B = \{2\}$.

---

Worked Example (from the textbook)

Experiment: Roll a single die.

$S = \{1, 2, 3, 4, 5, 6\}$.

Let $A$ = ‘getting a prime number’ $= \{2, 3, 5\}$.

Let $B$ = ‘getting an odd number’ $= \{1, 3, 5\}$.

Write the sets for:

1. ‘$A$ or $B$’
2. ‘$A$ and $B$’
3. ‘$A$ but not $B$’
4. ‘not $A$’ Solution. …