Algebra of Events

The algebra of events is the set of rules and regulations that shows how the event has occurred. These operations on the events show us how the values are related to each other. Fundamental operations are union, intersection, and complement. These operations can help analysts to predict the likelihood of an event.

Algebra of Events

The set of outcomes from an experiment is known as an Event.

When a die is thrown, sample space $S = \{1, 2, 3, 4, 5, 6\}.$

Let $A = \{2, 3, 5\},$ $B = \{1, 3, 5\},$ $C = \{2, 4, 6\} $

Here, $A$ is the event of the occurrence of prime numbers, $B$ is the event of the occurrence of odd numbers and $C$ is the event of the occurrence of even numbers.

Also, observe that $A, B,$ and $C$ are subsets of $S$.

Now, What is the Occurrence of an event?

From the above example, the experiment of throwing a die. Let $E$ denote the event “ a number less than $4$ appears”. If any of $' 1’$ or $ '2'$ or $'3'$ had appeared on the die then we say that event $E$ has occurred.

Thus, the event $E$ of a sample space $S$ is said to have occurred if the outcome $ω$ of the experiment is such that $ω ∈ E$. If the outcome $ω$ is such that $ω ∉ E$, we say that the event $E$ has not occurred.

Complimentary Event

For every event $A$, there corresponds to another event $A$' which contains all outcomes in sample space that are not covered in $A$. Such event is called the complementary event to $A$. It is also called the event ‘not $A$’.

For example, take the experiment ‘of tossing two coins’. The sample space is

$S = \{HH, HT, TH, TT\}$

Let $A=\{HT, TH\}$ be the event ‘only one tail appears’

Thus the complementary event ‘not $ A$’ to the event $A$ is

$A’ = \{HH, TT\}$

or $ A’ = \{ω : ω ∈ S$ and $ω ∉A\} = S - A$

The Event $'A$ or $B’$

As we have studied in the first chapter, ‘Sets’, the union of two sets $A$ and $B$ denoted by $A ∪ B$ contains all those elements which are either in $A$ or in $B$ or in both.

When the sets $A$ and $B$ are two events associated with a sample space, then $A ∪ B$ is the event ‘either $A$ or $B$ or both’. This event $A ∪ B$ is also called $'A$ or $B’.$

Therefore

Event
$
\begin{aligned}
\mathbf{A} \text { or } \mathbf{B} & =\mathbf{A} \cup \mathbf{B} \\
& =\{\omega: \omega \in \mathbf{A} \text { or } \omega \in \mathbf{B}\}
\end{aligned}
$

The Event $'A$ and $B’$

The intersection of two sets $A ∩ B$ is the set of those elements which are common to both $A$ and $B$. i.e., which belong to both $'A$ and $B’.$

If $A$ and $B$ are two events, then the set $A ∩ B$ denotes the event $'A$ and $B’.$

The Event $'A$ but not $B’$

The $A – B$ is the set of all those elements which are in $A$ but not in $B$. Therefore, the set $A – B$ may denote the event $'A$ but not $B’.$

Also, $A - B = A ∩ B´$ or $A - ( A ∩ B).$

Equally Likely Events

Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face $(1, 2, 3, 4, 5, or 6)$ is as likely to occur as any other face. If you toss a fair coin, a Head $(H)$ and a Tail $(T)$ are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.

Exhaustive events

Consider the experiment of rolling a die. The associated sample space is

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

Let us define the following events

$ A:$ ‘a prime number number less than $6$ appears’,

$B:$ ‘a number less than $2$ appears’

and $C:$ ‘a number greater than $3$ appears’.

Then $A = \{2, 3, 5\}, B = \{1\}$ and $C = \{4, 5, 6\}.$

Observe that, $A ∪ B ∪ C = \{2, 3, 5\} ∪ \{1\} ∪ \{4, 5, 6\} = \{1, 2, 3, 4, 5, 6\} = S$

So if union of given events equals sample space, then these events are called a system of exhaustive events. Thus events A, B and C are called exhaustive events in this case.

In general, $E_1, E_2, \ldots, E_n$ are $n$ events of a sample space $S$ and if

$\mathrm{E}_1 \cup \mathrm{E}_2 \cup \mathrm{E}_3 \cup \ldots \cup \mathrm{E}_n=\bigcup_{i=1}^n \mathrm{E}_i=\mathrm{S}$

then $E_1, E_2, \ldots, E_n$ are called exhaustive events.

Mutually exclusive events

Consider the experiment of rolling a die. The associated sample space is

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

Consider events, $A$ ‘an odd number appears’ and $B$ ‘an even number appears’

$A = \{1, 3, 5\}$ and $B = \{2, 4, 6\}$

Clearly $A ∩ B = φ$, i.e., $A$ and $B$ are disjoint sets.

In general, two events $A$ and $B$ are called mutually exclusive events if the occurrence of any one of them excludes the occurrence of the other event, i.e., if they can not occur simultaneously. In this case the sets $A$ and $B$ are disjoint.

Solved Examples Based on Algebra of Events:

Example 1: Which of the following events are exhaustive events in the case of a deck of $52$ cards?

1) $P$ = Drawing a King

$Q$ = Drawing a Black card.

2) $P$ = Drawing a numbered card

$Q$ = Drawing a Black card.

3) $P$ = Drawing a numbered card

$Q$ = Drawing a Red card

$R$ = Drawing a Face card.

4) $P$ = Drawing Black cards

$Q$ = Drawing face cards.

Solution

Since numbered cards are $\{A, 2, \cdots \ldots \ldots \ldots, 10\}$
Red cards are $\{R 1, R 2, \cdots \cdots \cdots \cdots \cdots, R J, R Q, R K, R A\}$and face cards are $\{J, Q, K\}$.

The union of these three events gives us the sample space.

So, these three are exhaustive events.

Hence, the answer is the option (3).

Example 2: Which of the following types of numbers are mutually exclusive?

1) $\mathbb{R}, \mathbb{Z}$
2) $R^{+}, R$
3) $\mathbb{Z}, \mathbb{N}$
4) $W, R^{-}$

Solution

Here $R$= Real Number

$R^ +$= Positive real Number

$\mathbb{Z}=$ Integers, $W=$ Whole Number, $N=$ Natural Number,
Since $W>0$ and $R^{-}<0$

So $W$ and $\mathrm{R}^{-}$do not have any common element. So these are mutually exclusive events.

Hence, the answer is the option (4).

Example 3: A die is thrown. Let $A$ be the event that the number obtained is greater than $3$. Let $B$ be the event that the number obtained is less than $5$. Then $P(A \cup B)$ is
1) $\frac{2}{5}$
2) $\frac{3}{5}$
3) $0$
4) $1$

Solution

Algebra of events

$A \cup B \rightarrow$ at least one event.
$A \cap B \rightarrow$ both occur simultaneously.
$A \cap \bar{B}=A-B$, the occurrence of event A but not B .
$\bar{A} \cap \bar{B} \rightarrow$ neither A or B.]

where $A$ & $B$ are any two events.

Set $A=\{4,5,6\}$
$\operatorname{Set} B=\{1,2,3,4\}$
Set $A \cup B=\{1,2,3,4,5,6\}$
Which is a set of all possible outcomes. Hence

$P(A \cup B)=1$

Hence, the answer is the option 4.

Example 4: Let $A$ and $B$ be two events such that $P(\overline{A \cup B})=\frac{1}{6}, \quad P(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{1}{4}$, where $\bar{A}$ stands for the complement of event $A$. Then events $A$ and $B$ are

1) equally likely but not independent

2) equally likely and mutually exclusive

3) mutually exclusive and independent

4) independent but not equally likely

Solution

We have,

$
\begin{aligned}
& P(A \cup B)=1-P(\overline{A \cup B})=\frac{5}{6} \\
& P(A \cap B)=\frac{1}{4} \\
& P(A)=1-P(\bar{A})=\frac{3}{4}
\end{aligned}
$

Also,

$
\begin{aligned}
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \frac{5}{6}=\frac{3}{4}+P(B)-\frac{1}{4} \\
& P(B)=\frac{1}{3}
\end{aligned}
$

Now,

$
P(A \cap B)=\frac{1}{4} \text { and } P(A) \cdot P(B)=\frac{3}{4} \times \frac{1}{3}=\frac{1}{4}
$

so, $P(A) \cdot P(B)=P(A \cap B)$
Therefore, Events A \& B are independent
But as $P(A) \neq P(B)$, so they are not equally likely.
Also as $P(A \cap B)$ does not equal $0$ , so they are not mutually exclusive

Hence, the answer is the option 4.

Example 5: Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are $0.3$ and $0.2,$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is (assume that different planes hitting the target are independent events)

1) $0.2$

2) $0.7$

3) $0.06$

4) $0.14$

Solution

Let $ A$ be the event that the first plane hits the target, and

$B$ be the event that the second plane hits the target

So we need $P(A' ∩ B)$

As $A$ and $B$ will hit the target independently, A and B are independent events, and thus $A'$ and $B$ are also independent events

So $P(A' ∩ B) = P(A').P(B) = (1 - P(A)).(P(B)) = (0.7)*(0.2) = 0.14$

Hence, the answer is the option 4.

Summary
The algebra of events provides a framework for analyzing the relationship between different events in probability. These methods are widely used in real-life applications providing insights and solutions to complex problems. Mastery of these concepts can help in solving gaining deeper insights and contributing meaningfully to real-life problems.