Probability refers to a numerical measure of the likelihood that some event occurs. A probability can take any value between $0$ and $1$, or equivalently between $0 \%$ and $100 \%$, with $0$ representing an impossibility and $1$ representing absolute certainty.

There are four main types of approaches to computing probability. These are Theoretical, Experimental, Subjective, or Axiomatic approaches.

Let’s understand what is theoretical probability with examples.

## What is Theoretical Probability?

Theoretical probability is the statistical concept that measures the likelihood (probability) of something happening. In a classic sense, it means that every statistical experiment will contain elements that are equally likely to happen (equal chances of occurrence of something). Therefore, it is also known as **classical probability** and its concept is the simplest form of probability that has equal odds of something happening.

The theoretical or classical approach of probability is well suited to describing problems involving the outcomes of coin flips, dice rolls, or cards drawn from a well-shuffled deck. The idea of a ”fair” coin or die is quite natural, with heads or tails, or each number, being equally likely to appear. Likewise, a well-shuffled deck of cards would be completely randomized, so any particular card has an equal chance of being drawn.

### Theoretical Probability Examples

**Ex 1:** The typical example of theoretical(or classical) probability would be rolling a fair die because it is equally probable that the top face of the die will be any of the 6 numbers on the die: 1, 2, 3, 4, 5, or 6.

**Ex 2:** Another example of (or classical) probability would be tossing an unbiased coin. There is an equal probability that your toss will yield either head(H) or tail(T).

**Ex 3:** Picking a card from a well-shuffled deck of playing cards is another example of theoretical/classical probability. Each of the 52 cards has an equal likelihood of being selected.

**Ex 4:** Guessing a multiple choice quiz (MCQs) test with (say) four possible answers A, B, C, or D is one more example of theoretical/classical probability. Each option (choice) has the same odds (equal chances) of being picked (assuming you pick randomly and do not follow any pattern).

## How to Find Theoretical Probability? (Theoretical Probability Formula)** **

The theoretical approach to probability is based on assuming that an experiment has $n$ equally likely possible outcomes. The probability of each outcome is then equal to the ratio $\frac{1}{n}$.

The probability of a simple event happening is the number of times the event can happen(favourable events), divided by the number of possible events (outcomes).

Mathematically, it can be represented as $\text{P} (\text{E}) = \frac{m}{n}$,

where $\text{P} (\text{E})$ means “probability of event $\text{E}$” (event $\text{E}$ is whatever event you are looking for, like getting a head when a coin is tossed, picking out an Ace from a well-shuffled deck of cards),

$m$ is the frequency or number of possible times the event could happen(also known as the number of favourable events) and

$n$ is the number of times the event could happen(also known as the number of sample points in the sample space).

For example, the probability of getting $5$ on a fair die is $\frac{1}{6}$. In other words, one possible outcome (there is only one way to roll a $5$ on a fair die) is divided by the number of possible outcomes (i.e., $6$).

## Real-Life Examples of Theoretical Probability

Theoretical probability or classical probability can only be applied when there are a finite number of choices that have equal probability. As such, it’s difficult to find many classical probability examples in real life because most things in life do not have equal probability. But it doesn’t mean it’s not important to learn. The reason behind learning theoretical or classical probability is that it’s a building block for other areas of probability such as the counting rule. It also extends to more complex situations, like quantum theory, which shares many important properties with classical probability theory.

Some of the real-world situations where theoretical or classical probability is used are

- Flipping a coin to make the two teams choose their openings during a cricket match.
- Guessing on a multiple-choice test. Let’s say you have four choices: A, B, C, and D. Each of these options has an equal chance of being correct, which means you have a 25% chance of getting the question right.
- Lottery machines like those used for Mega Millions or Powerball contain numbered ping pong balls that are mixed with spinning paddles or jets of air. Each of the ping pong balls has an equal probability of being chosen.

## Theoretical Probability Problems** **

.Let’s solve some problems on theoretical probability to understand the concept.

### Theoretical Probability Problems on Tossing of Coin(s)

#### Case 1: When a Single Coin is Tossed

A coin has two faces commonly referred to as ‘Head’(denoted by H), and ‘Tail’(denoted by T). The sample space when a coin is tossed is $\text{S} = \{\text{H}, \text{T} \}$, and therefore, the total number of possible outcomes is $n = 2$.

**Example 1:** Find the probability of the coin landing on the head when it is tossed.

Let $\text{E}$ be the event of the coin landing on the head.

The total number of possible outcomes $n = 2$.

The number of favourable events $m = 1$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{1}{2} = 0.5$

#### Case 2: When Two Coins are Tossed(or a Single Coin is Tossed Twice)

The sample space when two coins are tossed(or a single coin is tossed twice) is $\text{S} = \{\text{HH}, \text{HT}, \text{TH}, \text{TT} \}$, and therefore, the total number of possible outcomes is $n = 4$.

**Example 2:** Find the probability of getting two heads when a coin is tossed twice.

Let $\text{E}$ be the event of getting two heads when a coin is tossed twice.

The total number of possible outcomes $n = 4$.

The favourable event = $\{\text{HH} \}$

Therefore, the number of favourable events $m = 1$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{1}{4} = 0.25$

**Example 3:** Find the probability of getting one head and one tail when two coins are tossed simultaneously.

Let $\text{E}$ be the event of getting one head and one tail when two coins are tossed simultaneously.

The total number of possible outcomes $n = 4$.

The favourable events = $\{\text{HT}, \text{TH} \}$

Therefore, the number of favourable events $m = 2$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{2}{4} = \frac{1}{2} = 0.5$

#### Case 3: When Three Coins are Tossed(or a Single Coin is Tossed Thrice)

The sample space when three coins are tossed(or a single coin is tossed thrice) is $\text{S} = \{\text{HHH}, \text{HHT}, \text{HTH}, \text{HTT}, \text{THH}, \text{THT}, \text{TTH}, \text{TTT} \}$, and therefore, the total number of possible outcomes is $n = 8$.

**Example 4:** Find the probability of the coin landing on the tail all three times when it is tossed thrice.

Let $\text{E}$ be the event of getting tail all three times when it is tossed thrice.

The total number of possible outcomes $n = 8$.

The favourable event = $\{\text{TTT} \}$

Therefore, the number of favourable events $m = 1$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{1}{8} = 0.125$

**Example 5:** Find the probability of getting one tail when three coins are tossed simultaneously.

Let $\text{E}$ be the event of getting one tail when three coins are tossed simultaneously.

The total number of possible outcomes $n = 8$.

The favourable event = $\{\text{HHT}, \text{HTH}, \text{THH} \}$

Therefore, the number of favourable events $m = 3$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{3}{8} = 0.375$

**Example 6:** Find the probability of getting one head when three coins are tossed simultaneously.

Let $\text{E}$ be the event of getting one head when three coins are tossed simultaneously.

The total number of possible outcomes $n = 8$.

The favourable event = $\{\text{HTT}, \text{THT}, \text{TTH} \}$

Therefore, the number of favourable events $m = 3$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{3}{8} = 0.375$

**Note:** Total number of possible outcomes or number of sample points in a sample space when $n$ coins are tossed simultaneously(or a coin is tossed $n$ times) is $2^n$.

### Theoretical Probability Problems on Rolling of Dice

#### Case 1: When a Single Die is Rolled

A die has six faces numbered $1$, $2$, $3$, $4$, $5$, and $6$. The sample space when a die is rolled is $\text{S} = \{1, 2, 3, 4, 5, 6 \}$, and therefore, the total number of possible outcomes is $n = 6$.

**Example 1:** Find the probability of getting a number $4$, when a die is rolled.

Let $\text{E}$ be the event of getting a number $4$ when a die is rolled.

The total number of possible outcomes $n = 6$.

The favourable event = $\{4 \}$

Therefore, the number of favourable events $m = 1$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{1}{6} = 0.167$

**Example 2:** Find the probability of getting an odd number, when a die is rolled.

Let $\text{E}$ be the event of getting an odd number when a die is rolled.

The total number of possible outcomes $n = 6$.

The favourable event = $\{1, 3, 5 \}$

Therefore, the number of favourable events $m = 3$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{3}{6} = \frac{1}{2} = 0.5$

**Example 3:** Find the probability of getting a composite number, when a die is rolled.

Let $\text{E}$ be the event of getting a composite number when a die is rolled.

The total number of possible outcomes $n = 6$.

The favourable event = $\{4, 6 \}$

Therefore, the number of favourable events $m = 2$.

Therefore, $\text{P}( \text{E}) = \frac{m}{n} = \frac{2}{6} = \frac{1}{3} = 0.333$

#### Case 2: When a Pair of Dice is Rolled or a Single Die is Rolled Twice

The sample space when a pair of dice is rolled or a single die is rolled twice is $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$

$ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), $

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), $

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), $

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), $

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$.

**Example 4:** Find the probability of getting a doublet when a pair of dice is rolled.

The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), $

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), $

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), $

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), $

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), $

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$.

The events with doublet = $\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 1)\}$, and therefore, the total number of favourable events is $m = 6$.

Therefore, the probability of dice showing up a doublet $\text{P}(\text{E}) = \frac{m}{n} = \frac{6}{36} = \frac{1}{6}$.

**Example 5:** Find the probability of getting a sum of $9$ when a pair of dice is rolled.

The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$

$ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), $

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), $

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), $

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), $

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$.

The events with a sum of $9$ = $\{(3, 6), (4, 5), (5, 4), (6, 3)\}$, and therefore, the total number of favourable events is $m = 4$.

Therefore, the probability of dice showing up a sum of $9$ $\text{P}(\text{E}) = \frac{m}{n} = \frac{4}{36} = \frac{1}{9}$.

### Theoretical Probability Problems on Playing Cards

A deck of playing cards has $52$ cards divided into $4$ suits. The suits are Diamond, Heart, Spade, and Club. Each of the suits has $13$ cards – $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $\text{J}$, $text{Q}$, $text{K}$, and $text{A}$. The cards from $2$ to $10$ are called number cards and the cards from $\text{J}$ to $text{K}$ are called face cards. The face cards are referred to as $\text{J} – \text{Jack}$, $\text{Q} – \text{Queen}$, $\text{K} – \text{King}$, and $\text{A} – \text{Ace}$. Further, the colour of the cards in the diamonds and hearts suits is red and the colour of the cards in the spades and clubs suits is black.

**Note:**

- The cards $\text{J} – \text{Jack}$, $\text{Q} – \text{Queen}$, $\text{K} – \text{King}$ are called face cards.
- There are $12$ face cards in a deck of $52$ playing cards, $3$ in each of the $4$ suits.

#### Theoretical Probability Problems on Playing Cards

**Example 1:** A card is picked from a well-shuffled deck of playing cards. Find the probability that the card selected is an Ace.

The number of cards in a deck $n = 52$.

The number of cards in a deck that is Ace $m = 4$.

Therefore, the probability of picking an Ace card = $\frac {4}{52} = \frac{1}{13}$.

**Example 2:** A card is picked from a well-shuffled deck of playing cards. Find the probability that the card selected is a Black card.

The number of cards in a deck $n = 52$.

The number of black cards in a deck $m = 26$ ($13$ spades and $13$ clubs).

Therefore, the probability of picking a Black card = $\frac {26}{52} = \frac{1}{2}$.

**Example 3:** A card is picked from a well-shuffled deck of playing cards. Find the probability that the card selected is a face card.

The number of cards in a deck $n = 52$.

The number of face cards in a deck $m = 12$ ($3$ cards, viz, Jack, Queen, and King in each of the four suits).

Therefore, the probability of picking a face card = $\frac {12}{52} = \frac{3}{13}$

## Difference Between Theoretical Probability and Experimental Probability

The following are the main differences between theoretical probability and experimental probability.

Theoretical Probability | Experimental Probability |

Theoretical probability is based on what is expected to happen in an experiment, without actually conducting it | Empirical probability is based on the data which is obtained after an experiment is carried out |

It is the result of $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$ | It is the result of $\frac{\text{Number of occurrences of an event}}{\text{Total number of trials}}$ |

Example: A coin is tossed. Sample Space $\text{S } = \{\text{H}, \text{T} \}$ and $\text{P}( \text{Head}) = \frac{1}{2} = 0.5$ and $\text{P}( \text{Tail}) = \frac{1}{2} = 0.5$ | Example: A coin is tossed $10$ times. It is recorded that the coin landed on the head $7$ times and on the tail $3$ times, then $\text{P} ( \text{Head})= \frac{7}{10} = 0.7$ and $\text{P} ( \text{Tail}) = \frac{3}{10} = 0.3$ |

## Practice Problems

- Define theoretical probability.
- Find the probability of getting two tails when a pair of coins are tossed simultaneously.
- Find the probability of getting one head and one tail when a pair of coins are tossed simultaneously.
- Find the probability of getting three heads when three coins are tossed simultaneously.
- Find the probability of getting one head when three coins are tossed simultaneously.
- Find the probability of getting two tails when three coins are tossed simultaneously.
- Find the probability of getting a prime number when a die is rolled.
- Find the probability of getting an odd number when a die is rolled.
- Find the probability of getting a doublet when a pair of dice is rolled.
- Find the probability of getting a sum of $11$ when a pair of dice is rolled.
- Find the probability of picking a black jack from a well-shuffled deck of cards.
- Find the probability of picking a card whose value is less than $5$ from a well-shuffled deck of cards.
- Find the probability of picking a number card from a well-shuffled deck of cards.

## FAQs

### What is the theoretical probability?

A theoretical approach to probability assumes that every statistical experiment will contain elements that are equally likely to happen (equal chances of occurrence of something).

### What is the formula for theoretical probability?

This formula for theoretical probability is expressed as $\text{Theoretical Probability } = \frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$.

### What are the 4 types of probability?

The four major types of probabilities are

a) Theoretical Probability(or Classical Probability)

b) Experimental Probability(or Empirical Probability)

c) Subjective Probability

d) Axiomatic Probability

### What’s the difference between theoretical and experimental probability?

Theoretical probability assumes that every statistical experiment will contain elements that are equally likely to happen (equal chances of occurrence of something). A coin is equally likely to land heads or tails, so the theoretical probability of getting heads is $\frac{1}{2}$. Experimental probability describes how frequently an event actually occurred when an experiment is performed.

## Conclusion

Theoretical probability is one of the four approaches used to find the probability of an event. The approach of theoretical probability assumes that every statistical experiment will contain elements that are equally likely to happen. The reason behind learning theoretical or classical probability is that it’s a building block for other areas of probability.