It is unlikely that many people think about whether it is possible to calculate events that are more or less random. In simple terms, is it really possible to find out which side of the die in the dice will fall next time. It was this question that two great scientists asked, which laid the foundation for such a science as probability theory, in which the probability of an event in which is studied quite extensively.
Origin
If you try to define such a concept as probability theory, you get the following: this is one of the branches of mathematics that studies the constancy of random events. Of course, this concept does not really reveal the whole essence, therefore it is necessary to consider it in more detail.
I would like to start with the creators of the theory. As mentioned above, there were two of them, these were Pierre Fermat and Blaise Pascal. It was they who were one of the first to try, using formulas and mathematical calculations, to calculate the outcome of an event. On the whole, the rudiments of this science appeared in the Middle Ages. At that time, various thinkers and scientists tried to analyze gambling, such as roulette, dice, and so on, thereby establishing the pattern and percentage of the loss of a particular number. The foundation was laid in the seventeenth century precisely by the aforementioned scientists.
At first, their works could not be attributed to great achievements in this area, because all they did was simply empirical facts, and the experiments were made clearly, without the use of formulas. Over time, it turned out to achieve great results, which appeared as a result of observation of bone throwing. It was this tool that helped to derive the first intelligible formulas.
Like-minded people
One cannot but mention such a person as Christian Huygens in the process of studying a topic called the βtheory of probabilityβ (the probability of an event is covered in this particular science). This person is very interesting. He, like the scientists presented above, tried to derive the pattern of random events in the form of mathematical formulas. It is noteworthy that he did not do this together with Pascal and Fermat, that is, all his works did not intersect with these minds. Huygens derived the basic concepts of probability theory.
An interesting fact is that his work came out long before the results of the work of the discoverers, or rather, twenty years earlier. Among the designated concepts, the most famous are:
- the concept of probability as a measure of chance;
- mathematical expectation for discrete cases;
- probability multiplication and addition theorems.
Also, one cannot help but recall Jacob Bernoulli, who also made a significant contribution to the study of the problem. Carrying out his own trials independent of anyone, he managed to present a proof of the law of large numbers. In turn, the scientists Poisson and Laplace, who worked at the beginning of the nineteenth century, were able to prove the original theorems. From that moment on, for the analysis of errors in the course of observations, they began to use the theory of probability. Russian scientists, to be exact, Markov, Chebyshev and Diapunov, could not get around this science. Based on the work done by the great geniuses, they fixed this subject as a branch of mathematics. These figures worked already at the end of the nineteenth century, and thanks to their contribution, phenomena such as:
- the law of large numbers;
- theory of Markov chains;
- central limit theorem.
So, with the history of the origin of science and with the main persons who influenced it, everything is more or less clear. Now the time has come to concretize all the facts.
Basic concepts
Before touching on laws and theorems, it is worth exploring the basic concepts of probability theory. The event in it takes the leading role. This topic is quite voluminous, but without it it will not be possible to understand everything else.
An event in probability theory is any set of outcomes of the experience. There are not so few concepts of this phenomenon. So, the scientist Lotman, working in this area, said that in this case we are talking about what "happened, although it might not have happened."
Random events (probability theory pays special attention to them) is a concept that implies absolutely any phenomenon that has the ability to occur. Or, conversely, this scenario may not happen when many conditions are met. It is also worth knowing that it is random events that capture the entire volume of events that have occurred. Probability theory indicates that all conditions can be repeated continuously. Their conduct was called "experience" or "test".
A credible event is the phenomenon that will happen in this test one hundred percent. Accordingly, an impossible event is one that will not happen.
The combination of a pair of actions (conditionally case A and case B) is a phenomenon that occurs simultaneously. They are designated as AB.
The sum of the pairs of events A and B is C, in other words, if at least one of them happens (A or B), then we get C. The formula for the described phenomenon is written like this: C = A + B.
Incompatible events in probability theory imply that two cases are mutually exclusive. At the same time, they can never happen. Joint events in probability theory are their antipode. It is implied that if A happened, then it does not interfere with B.
Opposing events (probability theory considers them in great detail) are easy to understand. It is best to deal with them in comparison. They are almost the same as incompatible events in probability theory. But their difference lies in the fact that one of the many phenomena in any case should happen.
Equivalent events are those actions whose repeatability is equal. To make it clearer, one can imagine the throwing of a coin: the loss of one of its sides is equally probable of the loss of the other.
A favorable event is easier to consider with an example. Suppose there is episode B and episode A. The first is the roll of a dice with the appearance of an odd number, and the second is the appearance of the number five on the die. Then it turns out that A favors B.
Independent events in probability theory are projected on only two or more cases and imply the independence of an action from another. For example, A - tails out when tossing a coin, and B - getting a jack from the deck. They are independent events in probability theory. With this moment it became clearer.
Dependent events in probability theory are also admissible only for their set. They imply the dependence of one on the other, that is, the phenomenon of B can happen only if A has already happened or, conversely, has not happened, when this is the main condition for B.
The outcome of a random experiment consisting of one component is elementary events. Probability theory explains that this is a phenomenon that has occurred only once.
Basic formulas
So, the concepts of "event", "probability theory" were considered above, the definition of the basic terms of this science was also given. Now is the time to familiarize yourself directly with the important formulas. These expressions mathematically confirm all the main concepts in such a complex subject as probability theory. The likelihood of an event plays a huge role here.
Better to start with the basic formulas of combinatorics. And before embarking on them, it is worth considering what it is.
Combinatorics is primarily a branch of mathematics; it studies a huge number of integers, as well as various permutations of both the numbers themselves and their elements, various data, etc., leading to the appearance of a number of combinations. In addition to probability theory, this industry is important for statistics, computer science and cryptography.
So, now we can proceed to the presentation of the formulas themselves and their definition.
The first of them will be an expression for the number of permutations, it looks like this:
P_n = n β
(n - 1) β
(n - 2) ... 3 β
2 β
1 = n!
The equation is applied only if the elements differ only in the order of arrangement.
Now the placement formula will be considered, it looks like this:
A_n ^ m = n β
(n - 1) β
(n-2) β
... β
(n - m + 1) = n! : (n - m)!
This expression is applicable not only to the order of placement of the element, but also to its composition.
The third equation from combinatorics, and the last one, is called the formula for the number of combinations:
C_n ^ m = n! : ((n - m))! : m!
A combination refers to samples that are not ordered, respectively, and this rule applies to them.
It turned out to be easy to understand the combinatorics formulas, now we can proceed to the classical definition of probabilities. This expression looks like this:
P (A) = m: n.
In this formula, m is the number of conditions conducive to event A, and n is the number of absolutely all equally possible and elementary outcomes.
There are a large number of expressions, the article will not cover all, but the most important of them will be affected, such as, for example, the probability of the sum of events:
P (A + B) = P (A) + P (B) - this theorem is for adding up only incompatible events;
P (A + B) = P (A) + P (B) - P (AB) - and this for addition only compatible.
The probability of a product of events:
P (A β
B) = P (A) β
P (B) - this theorem is for independent events;
(P (A β
B) = P (A) β
P (B β£ A); P (A β
B) = P (A) β
P (A β£ B)) - and this one is for addicts.
The list of events will end the list. Probability theory tells us about the theorem Bayes that looks like this:
P (H_mβ£A) = (P (H_m) P (Aβ£H_m)): (β_ (k = 1) ^ n P (H_k) P (Aβ£H_k)), m = 1, ..., n
In this formula, H 1 , H 2 , ..., H n is the full group of hypotheses.
Let us dwell on this, further examples of the application of formulas for solving specific problems from practice will be considered.
Examples
If you carefully study any branch of mathematics, it cannot do without exercises and sample solutions. So is the theory of probability: events, examples here are an integral component, confirming scientific calculations.
Formula for the number of permutations
Suppose there are thirty cards in a card deck, starting with face value one. Next is the question. How many ways are there to fold a deck so that cards with a face value of one and two are not located side by side?
The task is set, now let's move on to its solution. First you need to determine the number of permutations of thirty elements, for this we take the above formula, it turns out P_30 = 30 !.
Based on this rule, we will find out how many options there are to fold the deck in different ways, but we need to subtract from them those in which the first and second cards will be next. To do this, start with the option when the first is above the second. It turns out that the first card can take twenty-nine places - from the first to twenty-ninth, and the second card from the second to thirtieth, it turns out only twenty-nine places for a pair of cards. In turn, the rest can take twenty-eight places, and in random order. That is, to swap twenty-eight cards, there are twenty-eight options P_28 = 28!
As a result, it turns out that if we consider the solution, when the first card is above the second, extra features will turn out 29 β
28! = 29!
Using the same method, you need to calculate the number of redundant options for the case when the first card is under the second. It also turns out 29 β
28! = 29!
It follows that the extra options are 2 β
29 !, while the necessary ways to collect the deck are 30! - 2 β
29 !. It remains only to count.
thirty! = 29! β
30; 30! - 2 β
29! = 29! β
(30 - 2) = 29! β
28
Now you need to multiply among themselves all the numbers from one to twenty nine, and then at the end multiply everything by 28. The answer is 2.4757335 β
γ10γ ^ 32
Solution example. Formula for the number of accommodation
In this problem, you need to find out how many ways there are to put fifteen volumes on one shelf, but provided that there are thirty volumes in total.
In this problem, the solution is a little simpler than in the previous one. Using the well-known formula, it is necessary to calculate the total number of arrangements of thirty volumes of fifteen.
A_30 ^ 15 = 30 β
29 β
28β
... β
(30 - 15 + 1) = 30 β
29 β
28 β
... β
16 = 202 843 204 931 727 360 000
The answer, respectively, will be equal to 202 843 204 931 727 360 000.
Now let's take the task a little harder. You need to find out how many ways you can arrange thirty books on two bookshelves, provided that only fifteen volumes can be on one shelf.
Before starting the solution, I would like to clarify that some tasks are solved in several ways, and there are two ways in this, but the same formula is used in both.
In this problem, you can take the answer from the previous one, because there we calculated how many times you can fill the shelf for fifteen books in different ways. It turned out A_30 ^ 15 = 30 β
29 β
28 β
... β
(30 - 15 + 1) = 30 β
29 β
28 β
... β
16.
We calculate the second shelf according to the rearrangement formula, because it contains fifteen books, while only fifteen remain. We use the formula P_15 = 15 !.
It turns out that in total there will be A_30 ^ 15 β
P_15 ways, but, in addition, the product of all numbers from thirty to sixteen will need to be multiplied by the product of numbers from one to fifteen, in the end we will get the product of all numbers from one to thirty, that is, the answer equal to 30!
But this problem can be solved in another way - easier. To do this, you can imagine that there is one shelf for thirty books. All of them are placed on this plane, but since the condition requires that there are two shelves, then we saw one long one in half, it turns out two by fifteen. From this it turns out that the options for the arrangement can be P_30 = 30 !.
Solution example. Formula for combination number
A variant of the third problem from combinatorics will now be considered. You need to know how many ways there are to arrange fifteen books, provided that you need to choose from thirty exactly the same.
For the solution, of course, the formula for the number of combinations will be applied. From the condition it becomes clear that the order of the same fifteen books is not important. Therefore, initially you need to find out the total number of combinations of thirty books of fifteen.
C_30 ^ 15 = 30! : ((30-15))! : fifteen ! = 155 117 520
That's all. Using this formula, in the shortest time it was possible to solve such a problem, the answer, respectively, is 155 117 520.
Solution example. Classical Definition of Probability
Using the formula above, you can find the answer in a simple task. But this will help to visually see and follow the course of action.
In the task it is given that in the urn there are ten absolutely identical balls. Of these, four are yellow and six are blue. One ball is taken from the urn. You need to find out the probability of getting blue.
To solve the problem, it is necessary to designate the delivery of the blue ball as event A. This experience can have ten outcomes, which, in turn, are elementary and equally possible. At the same time, out of ten, six are favorable to event A. We decide by the formula:
P (A) = 6: 10 = 0.6
Applying this formula, we learned that the possibility of getting a blue ball is 0.6.
Solution example. The probability of the sum of events
Now an option will be presented, which is solved using the probability formula for the sum of events. So, in the condition it is given that there are two boxes, in the first one there is one gray and five white balls, and in the second there are eight gray and four white balls. As a result, one of them was taken from the first and second boxes. You need to find out what is the chance that the delivered balls will be gray and white.
To solve this problem, it is necessary to identify events.
- So, A - took the gray ball from the first box: P (A) = 1/6.
- A '- they also took a white ball from the first box: P (A') = 5/6.
- B - they removed the gray ball from the second box: P (B) = 2/3.
- In '- they took a gray ball from the second box: P (B') = 1/3.
According to the conditions of the problem, it is necessary that one of the phenomena occurs: AB 'or A'B. Using the formula, we get: P (AB ') = 1/18, P (A'B) = 10/18.
Now the probability multiplication formula has been used. Further, to find out the answer, it is necessary to apply the equation of their addition:
P = P (AB '+ A'B) = P (AB') + P (A'B) = 11/18.
So, using the formula, you can solve such problems.
Total
The article presented information on the topic "Probability Theory", the probability of an event in which plays a crucial role. Of course, not everything was taken into account, but, based on the text presented, you can theoretically familiarize yourself with this section of mathematics. The science in question can be useful not only in the professional field, but also in everyday life. With its help, you can calculate any possibility of any event.
The text also touched on significant dates in the history of the formation of probability theory as a science, and the names of people whose works were invested in it. This is how human curiosity has led people to learn to count even random events. - , . , , , , . - !