# Event Relations and Operations

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In the last section, we have defined events as subsets of the sample space ${\displaystyle S}$. In interpreting events as sets, we can apply the same operations and relations to events, that we know from basic set theory. We shall now recapitulate some of the most important concepts of set theory.

## Subsets and Complements

${\displaystyle A}$ is subset of ${\displaystyle B}$ is denoted by ${\displaystyle \ A\subset B.}$   Thus if event ${\displaystyle A}$ occurs, ${\displaystyle B}$ occurs as well. ${\displaystyle A}$ and ${\displaystyle B}$ are equivalent events if and only if (abbreviated as ’iff’) ${\displaystyle \ A\subset B}$ and ${\displaystyle B\subset A}$. If ${\displaystyle A\subset B}$ then we define the complement of ${\displaystyle A}$ ${\displaystyle \ }$ denoted by ${\displaystyle \ {\overline {A}}}$ to be the set of points in ${\displaystyle B}$ that are not in ${\displaystyle A}$,

## Union of Sets

The set of points belonging to either the set ${\displaystyle A}$ or the set ${\displaystyle B}$ is called the union of sets ${\displaystyle A}$ and ${\displaystyle B}$, and is denoted by ${\displaystyle A\cup B}$.  Thus if the event ’${\displaystyle A\,or\,B}$’ has occurred, then a basic outcome in the set ${\displaystyle A\cup B}$ has taken place.

Set union can be extended to ${\displaystyle n}$ sets and hence ${\displaystyle n}$ events ${\displaystyle A_{1},A_{2},\ldots ,A_{n}}$: in which case we have${\displaystyle A_{1}\cup A_{2}\cup \ldots \cup A_{n}=\cup _{i=1}^{n}A_{i}}$ Example: Rolling a die once Define ${\displaystyle A=\{1,2\}}$ and ${\displaystyle B=\{2,4,6\}}$ Then ${\displaystyle A\cup B=\{1,2,4,6\}}$ General results:${\displaystyle A\cup A=A}$${\displaystyle A\cup S=S}$ where ${\displaystyle S}$ is the sample space.${\displaystyle A\cup \emptyset =A}$ where ${\displaystyle \emptyset }$ is the null set, the set with no elements in it.${\displaystyle A\cup {\overline {A}}=S}$

## Intersection of Sets

The set of points common to the sets ${\displaystyle A}$ AND ${\displaystyle B}$ is known as intersection of ${\displaystyle A}$ and ${\displaystyle B}$, ${\displaystyle A\cap B}$.  Thus if the event ’${\displaystyle A\,and\,B}$’ has occurred, then a basic outcome in the set ${\displaystyle A\cap B}$ has taken place.

Set intersection can be extended to ${\displaystyle n}$ sets and hence to ${\displaystyle n}$ events ${\displaystyle A_{1},A_{2},\ldots ,A_{n}}$:${\displaystyle A_{1}\cap A_{2}\cap \ldots \cap A_{n}=\cap _{i=1}^{n}A_{i}}$ Example: Rolling a die once Define ${\displaystyle A=\{1,2\}}$ and ${\displaystyle B=\{2,4,6\}}$ Then ${\displaystyle A\cap B=\{2\}}$ General results:${\displaystyle A\cap A=A}$${\displaystyle A\cap S=A}$${\displaystyle A\cap \emptyset =\emptyset }$${\displaystyle A\cap {\overline {A}}=\emptyset }$ ${\displaystyle \emptyset \cap S=\emptyset }$ Disjoint events: Two sets or events are to said to be disjoint (or mutually exclusive) if their intersection is the empty set: ${\displaystyle A\cap B=\emptyset }$. Interpretation: events ${\displaystyle A}$ and ${\displaystyle B}$ cannot occur simultaneously. By definition, ${\displaystyle A}$ and ${\displaystyle {\overline {A}}}$ are mutually exclusive. The reverse doesn’t hold, i.e. disjoint events aren’t necessarily complements of each other. Example: Rolling a die once Define ${\displaystyle A=\{1,3,5\}}$ and ${\displaystyle B=\{2,4,6\}}$ Then ${\displaystyle B={\overline {A}}}$ and ${\displaystyle A={\overline {B}}}$ ${\displaystyle \Rightarrow A\cap B=A\cap {\overline {A}}=\emptyset }$ Interpretation: events (sets) ${\displaystyle A}$ and ${\displaystyle B}$ are disjoint and complementary events. Define ${\displaystyle C=\{1,3\}}$ and ${\displaystyle B=\{2,4\}}$ ${\displaystyle \Rightarrow C\cap D=\emptyset }$ Interpretation: events ${\displaystyle C}$ and ${\displaystyle D}$ are disjoint but not complementary.

## Logical Difference of Sets or Events

The set or event ${\displaystyle C}$ is the logical difference of events ${\displaystyle A}$ and ${\displaystyle B}$ if it represents the event: ’${\displaystyle A}$ has occurred but ${\displaystyle B}$ has not occurred’ i.e. it is the outcomes in ${\displaystyle A}$, that are not in ${\displaystyle B}$: ${\displaystyle A\backslash B=C\equiv A\cap {\overline {B}}}$

Example: Rolling a six-sided die once Define ${\displaystyle A=\{1,2,3\}}$ and ${\displaystyle B=\{3,4\}}$ Then${\displaystyle A\backslash B=C=\{1,2\}}$ and ${\displaystyle B\backslash A=\{4\}}$

## Disjoint Decomposition of the Sample Space

A set of events ${\displaystyle A_{1},A_{2},\ldots ,A_{n}}$ is called disjoint decomposition of ${\displaystyle S}$, if the following conditions hold:

• ${\displaystyle A_{i}\neq \emptyset \quad \left(i=1,2,\ldots ,n\right)}$
• ${\displaystyle A_{i}\cap A_{k}=\emptyset \quad \left(i\neq k;i,k=1,2,\ldots ,n\right)}$
• ${\displaystyle A_{1}\cup A_{2}\cup \ldots \cup A_{n}=S}$

One can think of such a decomposition as a partition of the sample space where each basic outcome falls into exactly one set or event. Sharing a birthday cake results in a disjoint decomposition or partition of the cake. Example: Rolling a six-sided dice Sample space: ${\displaystyle S=\{1,2,3,4,5,6\}}$ Define ${\displaystyle A_{1}=\{1\},A_{2}=\{3,4\},A_{3}=\{1,3,4\},A_{4}=\{5,6\},A_{5}=\{2,5\},A_{6}=\{6\}}$. Claim: one possible disjoint decomposition is given by ${\displaystyle A_{1},A_{2},A_{5},A_{6}}$. Proof: ${\displaystyle A_{1}\cap A_{2}=\emptyset }$,${\displaystyle A_{1}\cap A_{5}=\emptyset }$,${\displaystyle A_{1}\cap A_{6}=\emptyset }$,${\displaystyle A_{2}\cap A_{5}=\emptyset }$,${\displaystyle A_{2}\cap A_{6}=\emptyset }$,${\displaystyle A_{5}\cap A_{6}=\emptyset }$,${\displaystyle A_{1}\cup A_{2}\cup A_{5}\cup A_{6}=S}$.

## Some Set Theoretic Laws

De Morgan’s laws${\displaystyle {\overline {A\cap B}}={\overline {A}}\cup {\overline {B}}}$${\displaystyle {\overline {A\cup B}}={\overline {A}}\cap {\overline {B}}}$ Associative laws${\displaystyle \left(A\cap B\right)\cap C=A\cap \left(B\cap C\right)}$${\displaystyle \left(A\cup B\right)\cup C=A\cup \left(B\cup C\right)}$ Commutative laws${\displaystyle A\cap B=B\cap A}$${\displaystyle A\cup B=B\cup A}$ Distributive laws${\displaystyle A\cap \left(B\cup C\right)=\left(A\cap B\right)\cup \left(A\cap C\right)}$${\displaystyle A\cup \left(B\cap C\right)=\left(A\cup B\right)\cap \left(A\cup C\right)}$

## Summary

Verbal Technical Algebraic
If ${\displaystyle A}$ occurs, then ${\displaystyle B}$ occurs also ${\displaystyle B}$ is subset of ${\displaystyle A}$ ${\displaystyle A\subset B}$
${\displaystyle B}$ and ${\displaystyle A}$ always occur together ${\displaystyle A}$ and ${\displaystyle B}$ are equivalent events ${\displaystyle A\equiv B}$
${\displaystyle A}$ and ${\displaystyle B}$ cannot occur together ${\displaystyle A}$ and ${\displaystyle B}$ are disjoint events ${\displaystyle A\cap B=\emptyset }$
${\displaystyle A}$ occurs ${\displaystyle \ }$ if and only ${\displaystyle B}$ does not occur ${\displaystyle A}$ and ${\displaystyle B}$ are complementary events ${\displaystyle B={\overline {A}}}$
${\displaystyle A}$ occurs if and only if at least one ${\displaystyle A_{i}}$ occurs ${\displaystyle A}$ is union of ${\displaystyle A_{i}}$ ${\displaystyle A=\cup _{i}A_{i}}$
${\displaystyle A}$ occurs if and only if all ${\displaystyle A_{i}}$ occur ${\displaystyle A}$ is intersection of all ${\displaystyle A_{i}}$ ${\displaystyle A=\cap _{i}A_{i}}$