Introduction
On the basis of the Bernoulli test , probabilities are used to evaluate our “chance” that an event appears in our computational context. The principle of the test is to split our population into two phases: Pile / Face, Good / Not Good, Male / Female, 1/6 against 5/6 which is our chance to get a 1 for a roll of die. .. She was born in 1713 with the publication of “Ars conjectandi“, the posthumous work of the Swiss Bâlois Jacob Bernoulli.
It is called Bernoulli’s scheme to repeat the same proof indefinitely. Below, an example of a Bernoulli scheme for n = 4.
Le principe
The principle of calculating a probability is simple. It consists in calculating the ratio between our number of winning paths and our number of possible paths :
Calculation method
The method consists of calculating our number of winning paths and the number of possible paths. This calculation depends on the case in which we are.

With order 
Without order 
with put back 
Case 1 
Case 2 
Without put back 
Case 3 
Case 4 
 With order : This indicates that we take into account the order of occurrence of events. Thus 123 is different from 321. And conversely.
 With put back : This indicates that at each draw, the previous event may appear again. So, we can have 111. And conversely.
Some définitions
 n : The number of draws we make. Example, at the Euromillions, we have a first draw of 5 balls, then a draw of 2 balls. We will have n = 5 for the first case and n = 2 for the second case.
 k : The number of successes we want to have among our n draws. At euromillions, we dream of having the 5 winning balls and the 2 complementary ones too. We will have k = 5 for the first case and 2 for the next case.
 N : The number of possibilities we have for the first draw. For example, for a roll of a die 6, we have 6 possibilities in the first draw.
 M : The number of winning items in which we draw. For example, if we want to evaluate our probability of drawing a black ball in an urn containing 5 black balls and 10 white balls, then M will be 5.
Case 1 : With order and put back
For each draw, all alternatives are possible (we put the ball in each draw) and the order of obtaining the figures is important. That means that 123 and 231 are not the same and we can have 113. We call this a plist.
umber of possibles pListes : N^{n}
The number of winning paths is equal to the number of times we make an attempt to succeed. Example :
We want to identify the number of possible combinations for your phone’s PIN. This code is 4 digits, and for each digit we can have 10 numbers. So we are in a discounted case, then the 10 numbers can be used for each digit. The number of code possibilities is therefore 10^{4} = 10 000, 1 chance out of 10 000. If we consider that we have 3 attempts before blocking our phone, we have 3/10000 chances.
Case 2 : without order and put back
In this case, the order of appearance does not matter and all possibilities are open for each draw. We can have 111,222 … and have 112. This is the case where we are going to use the Binomiale law. This is called pSuites and we use the Binomial law to perform the calculation.
Nb of winning psuites
Nb of possible psuites
We roll 4 rolls of a wellbalanced die. we want to know what is our probability of obtaining 3 times the number 6. We obtain the following result :
Probability = 4 / 259, that is to say 1.54% chance of getting 3 times our number of 6.
Case 3 : With order and without put back
If the first digit is a 1, the second digit can be a 2 or a 3, which is the number of possibilities of the start – 1 since the same number can not be chosen twice. This case is called an Arrangement.
Nb of possible arrangement
Take the example of the tiercé. The challenge is to calculate our probability of having the good ranking 3 horses out of 24. The order is necessarily important and the horse which is in first position can not be in second at the same time. We want to maximize our chances of winning and we take 3 “ grids “. We are getting : 12144 possibles arrangements, that is to say 3/12144 chances to win
Case 4 : without order and with our put back
We are in the case of most games of chance, like lotto for example. In this case, we take again the number of possibilities of the case 2 which one divides by the number of draw. This case is called a combination , and we use the Hypergeométric law to perform the calculation. The winning number of combinations is :
Nb of winning combinations
Nb of possibles combinations
In lotto, we have 5 digits. Knowing that we have the choice with 49 possibilities for the 5 digits without discount, we have 1 906 884 possible combinations.
If we take the lucky number, 1 number having 10 possibilities, we will have 19,068,840 combinations.
Probability calculation in specific cases
Rules of operation
Description  Formula  Example  

Complement  We want to calculate the non occurrence of event A  P(no A) = 1 – P(A)  In a 52card game, the probability of not having a heart is : 1 – ¼ = 75% 
Addition  We want to calculate the probability of occurrence of event A OR B.  Our events are incompatible : P (A or B) = P(A) + P(B)  In a 52card game, the probability of having a heart or clover is : 13/52 + 13/52 = 50% 
Our events are compatible : P (A or B) = P(A) + P(B) – P(A and B)  In a 52card game, the probability of having an ace or a heart is : 4/52 + 13/52 – (4/52 * 13/52) = 30,8% 

Multiplication  We want to calculate the probability of occurrence of event A AND B  Our events are independent P(A and B) = P(A) * P(B)  By shooting 2 cards one after the other and with put back In a 52card game, the probability of having 2 hearts is : (13/52) * (13/52) = 6,25% 
Our events are dependent P(A and B) = P(A) * P(B I A) P(B I A) = P(A et B)/P(A)  By shooting 2 cards one after the other and without put back In a 52card game, the probability of having 2 hearts is : (13/52) * (12/51) = 5,88% 