## What is Probability?

It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.

**P(A)=number of successful trials ÷ Total number of trials**

- The value is always expressed between zero and one, both inclusive.
- It has been introduced in math and statistics to predict how likely events are to happen.

## Types of Probability

There are four types of probability.

**Theoretical probability:**The probability that is determined based on reasoning is called theoretical probability. According to theoretical probability, if*n*outcomes occur an equal number of times, then there is an equal chance of each of them happening. So, each outcome is assigned the same probability.**Experimental probability:**In this, we determine the probability of any event by doing actual experiments. The outcomes of each event are recorded and the likelihood of occurrence of the event is calculated from the experimental data.**Subjective probability:**It is based on personal observations. In this, people assign values to events based on the likelihood of an event occurring. However, the estimates may not be accurate as they may be biased, and individuals may assign higher values to events that benefit them.**Axiomatic probability:**Axiomatic probability states some axioms based on which all the probabilities are calculated. It states the following axioms:

- The probability of any event E is greater than or equal to 0, i.e., $P(E)\ge 0$.

- The probability of sample space is exactly equal to one.
- The probability of either of two mutually exclusive events happening is the sum of their individual probabilities.

## Probability of Multiple Events

Now, let's go onto the case where there will be two events. When we require the probability of both the events occurring simultaneously or the probability of one of them or both of the two events occurring, we will require the probability laws to execute the calculations.

There are two such laws.

- Addition law or rule
- Multiplication rule

The multiplication law of probability is applied to find the probability of two or more events taking place simultaneously. The multiplication law helps to find the probability of event A and event B occurring together. Here, we will be primarily focusing on the “addition rule of probability”.

## Addition Rule (Law) of Probability

The addition rule states that the probability of occurrence of event A or event B is the difference of the sum of the individual probabilities of A and B and the probability of A and B occurring together (i.e., probability of overlap). Subtracting the probability of both events is necessary to avoid the problem of double-counting, where A and B are the subsets of the universal set U or from the same sample space. Sample space is defined as the collection of all possible outcomes of an event.

The rule is outlined as $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$.

Where,

$P\left(A\right)$– Probability of event A

$P\left(B\right)$– Probability of event B

$P\left(A\cup B\right)$– Probability that either of A or B happens

$P\left(A\cap B\right)\text{}$– Probability of A and B occurring together

The Venn diagram for the addition rule is depicted below:

### Example of Addition Rule of Probability

We have to draw one card out of a well-shuffled deck of 52 cards. So what are the chances of getting a diamond or a face card?

Solution:

Let A denote drawing a diamond and B denote drawing a face card. As there are 13 diamonds and a total of 12 face cards (3 of each suit) but only 3 face cards of diamonds, we get:

$P\left(A\right)\text{}=\frac{13}{52}$

$P\left(B\right)\text{}=\frac{12}{52}$

$P\left(A\cap B\right)=\frac{3}{52}$

Using the addition rule, we calculate:

$\begin{array}{l}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\text{}\\ =\frac{13}{52}+\frac{12}{52}-\frac{3}{52}\\ =\frac{22}{52}\\ =\frac{11}{26}\end{array}$

## Types of Events

There are predominantly two types of events:

- Mutually exclusive events or disjoint events
- Independent events

### Mutually Exclusive Events

Two events *A* and *B*, are said to be mutually exclusive if they do not intersect. These are the events that cannot happen simultaneously (at the same time) or cannot occur at the same time and therefore do not overlap each other. For example, while testing a switch, "fail" and "operate" are mutually exclusive.

The probability is represented as $P\left(A\cap B\right)=0$, where A and B are two mutually exclusive events of the same sample space, or A and B are disjoint events.

The Venn diagram for two mutually exclusive events is:

So, if two events A and B are mutually exclusive, then the addition rule of the probability formula reduces to $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$as $P\left(A\cap B\right)=0$.

The Venn diagram for non-mutually exclusive events is:

Here A and B are two non-mutually exclusive events that do not overlap each other, and their intersection is shown in the figure.

Let us consider one more example to understand the addition rule of probability for mutually exclusive events.

**Example**: Let us assume a deck of 52 playing cards, and a card at random is drawn. What are the chances that the drawn card is king or queen?

**Solution**:

Let A represent the event that the queen is drawn and B as a king is drawn.

The events A and B are disjoint as no king is equal to the queen.

So, from the addition rule:

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$

$P\left(A\right)\text{}=\frac{4}{52}$

$P\left(B\right)\text{}=\frac{4}{52}$

So,

$\begin{array}{l}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{52}+\frac{4}{52}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2}{13}\end{array}$

## Formulas

- The addition rule of probability is given as:

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\text{}$

- The addition rule can be extended to three events A, B and C as follows:

$P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(B\right)-P\left(A\cap B\right)-P\left(A\cap C\right)\text{}-P\left(B\cap C\right)+P\left(A\cap B\cap C\right)$

Here,

A, B and C correspond to the same sample space.

$P\left(A\right)$– Probability of event A.

$P\left(B\right)$– Probability of event B.

$P\left(C\right)$– Probability of event C.

$P\left(A\cup B\cup C\right)$- Probability that either A, B, or C happens.

$P\left(A\cap B\right)$– Probability of A and B happening together.

$P\left(B\cap C\right)$– Probability of B and C happening together.

$P\left(A\cap C\right)$– Probability of A and C happening together.

$P\left(A\cap B\cap C\right)$– Probability of A, B, and C happening together.

## Context and Application

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

- B.Sc. in Math
- B.Sc. in Statistics
- M.Sc. in Math
- M.Sc. in Statistics

## Practice Problem

**Problem 1:** Calculate the probability of drawing an ace or a card of spade from a pack of cards?

**Solution:** Consider A as the event of drawing an ace.

The probability of drawing an ace can be denoted as $P\left(A\right)\text{}=\frac{4}{52}$.

There are a total of 52 cards and four aces cards in a pack, so the probability of drawing an ace can be denoted as $P\left(A\right)\text{}=\frac{4}{52}$.

Consider S as the event of drawing a spade.

The probability of drawing a card of spades can be denoted as $P\left(S\right)\text{}=\frac{13}{52}$.

The total number of cards of spades in a deck of 52 cards is 13. So, the probability of drawing a card of spades can be denoted as $P\left(S\right)\text{}=\frac{13}{52}$.

Now, the probability of drawing an ace of spade is $\frac{1}{52}$.

This means that $P\text{}\left(A\text{}\cap \text{}S\right)\text{}=\frac{1}{52}$.

So, to calculate the required probability, we have:

$\begin{array}{c}P\left(A\cup S\right)=P\left(A\right)+P\left(S\right)-P\left(A\cap S\right)\text{}\\ \text{=}\frac{4}{52}+\frac{13}{52}-\frac{1}{52}\\ =\frac{16}{52}\\ =\frac{4}{13}\end{array}$

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