The axioms of probability They are mathematical propositions referring to the theory of probability, which do not merit demonstration. The axioms were established in 1933 by the Russian mathematician Andrei Kolmogorov (1903-1987) in his work Fundamentals of Probability Theory and laid the foundations for the mathematical study of probability.
When carrying out a certain random experiment ξ, the sample space E is the set of all the possible results of the experiment, also called events. Any event is denoted as A and P (A) is the probability of its occurrence. Then Kolmogorov established that:
-Axiom 1 (not negativity): the probability of any event A occurring is always positive or zero, P (A) ≥0. When the probability of an event is 0, it is called impossible event.
-Axiom 2 (certainty): whenever some event that belongs to E, its probability of occurrence is 1, which we can express as P (E) = 1. It is what is known as a sure event, since when conducting an experiment, there is certainly a result.
-Axiom 3 (addition): in the case of two or more incompatible events two by two, called A1, TOtwo, TO3…, The probability that event A occurs1 plus the Atwo plus the A3 and so on, it is the sum of the probabilities that each one happens separately.
This is expressed as: P (A1 U Atwo U A3 U…) = P (A1) + P (Atwo) + P (A3) + ...
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The axioms of probability are widely used in a multitude of applications. For example:
A thumbtack or tack is thrown into the air, and when it falls to the floor there is the option of landing with the point up (U) or with the point down (D) (we will not consider other possibilities). The sample space of this experiment consists of these events, then E = U, D.
By applying the axioms we have:
P (E) = 1 (Axiom 2)
But P (E) = P (U) + P (D) (Axiom 3), because these events are mutually incompatible, or disjoint. The thumbtack does not fall with the tip up or down at the same time, it is one or the other, but not both, as other possibilities are not being considered. Then:
P (U) + P (D) = 1
P (U) = 1 - P (D)
If it is equally likely that it will land with the tip up or down, P (U) = P (D) = ½ (Axiom 1). However, it may be that the construction and design of the thumbtack is more likely to fall one way or another. For example, it may be that P (U) = ¾ while P (D) = ¼ (Axiom 1).
Note that in both cases, the sum of the probabilities gives 1. However, the axioms do not indicate how to assign the probabilities, at least not completely. But they do affirm that they are numbers between 0 and 1 and that, as in this case, the sum of all is 1.
The axioms of probability are not a method of assigning the value of probability. For this there are three options that are compatible with the axioms:
Each event is assigned the same probability of happening, then the probability of occurrence is defined as:
P (A) = number of cases favorable to event A / number of possible cases
For example, what is the probability of drawing an ace from a deck of French cards? The deck has 52 cards, 13 of each suit and there are 4 suits. Each suit has 1 aces, so in total there are 4 aces:
P (as) = 4/52 = 1/13
Laplace's rule is limited to finite sample spaces, where each event is equally probable.
Here the experiment has to be repeatable, since the method is based on performing a large number of repetitions.
Let's make i repetitions of the experiment ξ, of which we find that n is the number of times that certain event A occurs, then the probability that this event occurs is:
P (A) = limi → ∞ (neither)
Where n / i is the relative frequency of an event.
Defining P (A) in this way satisfies Kolmogorov's axioms, but has the drawback that many tests have to be performed for the probability to be appropriate..
A person or a group of people can agree to assign the probability to an event, through their own judgment. This method has the disadvantage that different people can assign different probabilities to the same event..
In the experiment of simultaneously tossing 3 honest coins, obtain the probabilities of the events described:
a) 2 heads and a tail.
b) 1 head and two tails
c) 3 crosses.
d) At least 1 face.
Heads are denoted by C and tails by X. But there are several ways to get two heads and a tail. For example, the first two coins can land heads and the third can land tails. Or the first can fall heads, the second tails and the third heads. And finally the first can be tails and the remaining heads.
To answer the questions it is necessary to know all the possibilities, which are described in a tool called tree diagram or tree of probabilities:
The probability that any coin will come up heads is ½, the same is true for tails, since the coin is honest. In the right column are listed all the possibilities that the toss has, that is, the sample space.
From the sample space, the combinations that respond to the requested event are chosen, since the order in which the faces appear is not important. There are three favorable events: CCX, CXC and XCC. The probability of each event happening is:
P (CCX) = ½. ½. ½ = 1/8
The same happens for the CXC and XCC events, each one has a 1/8 probability of happening. Therefore the probability of getting exactly 2 heads is the sum of the probabilities of all favorable events:
P (2-sided) = 1/8 + 1/8 + 1/8 = 3/8 = 0.375
Finding the probability that exactly two crosses occur is a problem analogous to the previous one, there are also three favorable events taken from the sample space: CXX, XCX and XXC. Therefore:
P (2 crosses) = 3/8 = 0.375
Intuitively we know that the probability of getting 3 tails (or 3 heads) is lower. In this case the event sought is XXX, at the end of the right column, whose probability is:
P (XXX) = ½. ½. ½ = 1/8 = 0.125.
It is requested to obtain at least 1 face, this means that 3 faces, 2 faces or 1 face can come out. The only event incompatible with this is the one in which 3 tails come out, whose probability is 0.125. Therefore the probability sought is:
P (at least 1 head) = 1 - 0.125 = 0.875.
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