The **classical probability** is a particular case of calculating the probability of an event. It is defined as the quotient between the events favorable to said event and the total number of possible events, provided that each of these events are all equally likely. Classical probability is also known as a priori probability or theoretical probability.

The desire to anticipate things is part of human nature at all times: we all wonder if it will rain the next day or if a certain football team will play in the first division or not next season. There is archaeological evidence that people played games of chance around 40,000 years ago.

However, the first book on probabilities is due to the Dutch astronomer Christian Huygens who called it *Reasonings related to the game of dice*. As we can see, classical probability has its origins in games of chance.

The die has a long history, it is a cubic piece whose faces are numbered with dots from one to six. When rolling an honest die just once: what is the probability that you will get, say, a five?

It is very simple: there is only one face among the 6 marked with five points, therefore the probability P is:

P = 1/6

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**Calculation in classical probability**

This way of calculating the probability of an event is an application of Laplace’s rule, first stated in 1812 by the French mathematician Pierre de Laplace (1749-1827).

Let A be an event for which we want to know its probability of occurrence P(A), then:

*P(A) = number of cases favorable to event A / number of possible cases*

The result of this operation is always a positive number between 0 and 1. If an event has a probability of 0, it means that it will not happen.

On the other hand, if the probability of occurrence is equal to 1, it means that it will happen anyway and in any case, the probability that an event will occur, added to the probability that it will not occur, is equal to 1:

Here we have denoted the probability that event A does not occur by a bar above the letters.

Obviously, on a legal die, any of the 6 heads has the same probability of coming up, therefore the probability of getting a head with 5 must be 1/6.

An important detail is the following: to apply Laplace’s rule, the number of possible cases must be finite, that is, we must be able to count them and obtain a natural number.

In the dice example there are 6 possible cases and only one favorable event. The set of possible cases is called *sample space*.

When applying Laplace’s rule, it is convenient to carefully analyze the sample space, including all possible events, that is, it must be complete and ordered, so that no event escapes being accounted for.

**The sample space and events**

The sample space is usually denoted by the letter S or the Greek letter Ω (capital omega) and was a concept introduced by Galileo.

A dice player asked the wise man why it is more difficult to get a 9 by rolling three dice than a 10, so Galileo calculated the possible ways to get a 9, and then he did the same with 10. Finally he calculated the respective probabilities, finding that, indeed, P(9) < P(10).

**Sample space with few elements**

If the sample space consists of few elements, they are listed as a set. For example, suppose you want to find the probability that in a family with two children, both are of the same sex.

We can apply classical probability by correctly determining the sample space. If M = female and H = male, the sample space of children is:

S = {(M,M), (H,H), (M,H), (H,M)}

Each element of the sample space is an event, for example, the event (M,M) means that the two children of this family are women.

Given the sample space, calculating the requested probability is very simple, since there are only 2 favorable cases out of 4, for both children to be of the same sex: (M,M) and (H,H), therefore:

P (both children of the same sex) = 2/4 = 0.5

**sample space with many elements**

When the sample space consists of many elements, it is better to give a general rule for finding it. For example, if t is the lifetime of a piece of equipment, the sample space is:

*S *= {*you*∕*you *≥0}

Which reads like this: “all values of t such that t is greater than or equal to 0”. An event in this space could be that the device has a useful life of t = 2 years.

**Examples of classical probability**

The classical probability applies whenever the two premises indicated above are met, that is:

-All events are equally likely.

-The sample space is finite.

Therefore, there are situations in which classical probability cannot be applied, such as when one wants to anticipate whether a new treatment will cure a certain disease, or the probability that a machine will produce defective items.

Instead, it can be applied successfully in the following cases:

**throw of a dice**

As we have seen, the probability that a given head will come up is equal to 1/6.

**Draw a card from a deck**

We have a deck of 52 cards from a French deck, consisting of four suits: hearts, clubs, diamonds and spades. Then the probability of drawing a heart, knowing that there are 13 cards of each suit is:

P (heart) = 13/52

**toss of a coin**

This is a typical example of classical probability, since when tossing a coin there is always a probability equal to ½ of getting heads or tails.

**Draw colored marbles from a bag **

Inside a bag there can be N colored marbles, for example there are R red marbles, A blue marbles and V green marbles. The probability of drawing a red is:

P(R) = R / N

**solved exercises**

**– Exercise 1**

An honest die is rolled once. Calculate the following probabilities:

a) Draw an odd number.

b) Get a 2 or a 5.

c) Get a value less than 4.

d) Obtain a value less than or equal to 4.

e) Take a value different from 3

**Solution to**

The sample space is S = {1, 2, 3, 4, 5, 6}, the odd values are 1, 3 and 5, therefore out of 6 possible cases, there are three favorable cases:

P(odd) = 3/6 =1/2 = 0.5

**solution b**

We want to extract a 2 or a 5, that is, any of these cases is favorable, therefore:

P(2 or 5) = 2/6 = 1/3 = 0.33

**solution c**

In this case there are 3 favorable events: draw 1, 2 or 3:

P (less than 4) = 3/6 = ½ = 0.5

**solution d**

Here is an additional favorable event, because we are asked for values less than or equal to 4, so:

** **P (value less than or equal to 4) = 4/6 = 2/3 = 0.67

**solution and**

A roll other than 3 means any of the other values rolled:

**– Exercise 2**

In a box there is a blue, a green, a red, a yellow and a black ball. What is the probability that when you take a ball out of the box with your eyes closed, it will be yellow?

**Solution**

Event «E» is to take a ball out of the box with your eyes closed (if it is done with your eyes open the probability is 1) and that it is yellow.

There is only one favorable case, since there is only one yellow ball. The possible cases are 5, since there are 5 balls in the box.

Therefore, the probability of the event «E» is equal to P(E) = 1 / 5.

As can be seen, if the event is to draw a blue, green, red or black ball, the probability will also be equal to 1/5. Therefore, this is an example of classical probability.

**Observation**

If there had been 2 yellow balls in the box then P(E) = 2/6 = 1/3, while the probability of drawing a blue, green, red or black ball would have been equal to 1/6.

Since not all events have the same probability, then this is not an example of classical probability.

**– Exercise 3**

What is the probability that, when rolling a dice, the result obtained is equal to 5?

**Solution**

A dice has 6 faces, each with a different number (1,2,3,4,5,6). Therefore, there are 6 possible cases and only one case is favorable.

So, the probability that a 5 is rolled is equal to 1/6.

Again, the probability of getting any other result on the die is also equal to 1/6.

**– Exercise 4**

In a classroom there are 8 boys and 8 girls. If the teacher randomly selects a student from her class, what is the probability that the student chosen is a girl?

**Solution**

Event “E” is to choose a student at random. In total there are 16 students, but since you want to choose a girl, then there are 8 favorable cases. Therefore P(E) = 8/16 = 1/2.

Also in this example, the probability of choosing a boy is 8/16=1/2.

In other words, the chosen student is as likely to be a girl as it is to be a boy.

**References**

August, A. Probability. University of Puerto Rico. Retrieved from: docs.uprb.edu.

Galindo, E. 2011. Statistics: methods and applications. Prociencia Editors.

Jiménez, R. 2010. Mathematics II. 2nd. Edition. Prentice Hall.

Triola, M. 2012. Elementary Statistics. 11th. Edition. Addison Wesley.

Sangaku Maths. Laplace’s rule. Recovered from: sangakoo.com.