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 realistic to know which side of the die in the dice will fall out next. It was this question that two great scientists asked, who laid the foundation for such a science as the theory of probability, in which the probability of an event is studied quite extensively.
Origination
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, so 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 are Pierre Fermat and Blaise Pascal. It was they who were among the first who tried to calculate the outcome of an event using formulas and mathematical calculations. On the whole, the rudiments of this science appeared as early asMiddle Ages. At that time, various thinkers and scientists tried to analyze gambling, such as roulette, craps, and so on, thereby establishing a pattern and percentage of a particular number falling out. The foundation was laid in the seventeenth century by the aforementioned scientists.
At first, their work could not be attributed to the great achievements in this field, because everything they did was simply empirical facts, and the experiments were set visually, without the use of formulas. Over time, it turned out to achieve great results, which appeared as a result of observing the throwing of dice. It was this tool that helped to derive the first intelligible formulas.
Associates
It is impossible not to mention such a person as Christian Huygens, in the process of studying a topic called "probability theory" (the probability of an event is covered precisely in this science). This person is very interesting. He, like the scientists presented above, tried to derive the regularity 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 in any way 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 pioneers' work, or rather, twenty years earlier. Among the designated concepts, the most famous are:
- the concept of probability as a magnitude of chance;
- expectation for discretecases;
- theorems of multiplication and addition of probabilities.
It is also impossible not to remember Jacob Bernoulli, who also made a significant contribution to the study of the problem. Conducting his own tests, independent of anyone, he was able 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. It was from this moment that probability theory began to be used to analyze errors in the course of observations. Russian scientists, or rather Markov, Chebyshev and Dyapunov, could not bypass this science either. 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:
- law of large numbers;
- Markov chain theory;
- central limit theorem.
So, with the history of the birth of science and with the main people who influenced it, everything is more or less clear. Now it's time to concretize all the facts.
Basic concepts
Before touching on laws and theorems, it is worth studying the basic concepts of probability theory. The event takes the leading role in it. 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 an experiment. There are not so many concepts of this phenomenon. So, scientist Lotman,working in this area, said that in this case we are talking about something that “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 phenomena that have occurred. Probability theory indicates that all conditions can be repeated constantly. It was their conduct that was called "experience" or "test".
A certain event is one that will 100% happen in a given test. Accordingly, an impossible event is one that will not happen.
Combination of a pair of actions (conventionally case A and case B) is a phenomenon that occurs simultaneously. They are designated as AB.
The sum of pairs of events A and B is C, in other words, if at least one of them happens (A or B), then C will be obtained. The formula of the described phenomenon is written as follows: C=A + B.
Disjoint events in probability theory imply that two cases are mutually exclusive. They can never happen at the same time. Joint events in probability theory are their antipode. This implies that if A happened, then it does not interfere with B.
Opposite events (probability theory deals with them in great detail) are easy to understand. It is best to deal with them in comparison. They are almost the same asand incompatible events in probability theory. But their difference lies in the fact that one of the many phenomena must happen anyway.
Equivalent events are those actions, the possibility of which is equal. To make it clearer, we can imagine the tossing of a coin: the fall of one of its sides is equally likely to fall of the other.
Auspicious event is easier to see with an example. Let's say there is episode B and episode A. The first is the roll of the 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 only on two or more cases and imply the independence of any action from another. For example, A is the loss of tails when a coin is tossed, and B is the drawing of 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 B can occur only if A has already happened or, on the contrary, 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 happened only once.
Basic Formulas
So, the concepts of "event", "probability theory",the definition of the basic terms of this science was also given. Now it's time to get acquainted directly with the important formulas. These expressions mathematically confirm all the main concepts in such a difficult subject as probability theory. The probability of an event plays a huge role here as well.
Better start with the basic formulas of combinatorics. And before proceeding to them, it is worth considering what it is.
Combinatorics is primarily a branch of mathematics, it deals with the study of 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 branch is important for statistics, computer science and cryptography.
So now we can move on to presenting the formulas themselves and defining them.
The first one will be the expression for the number of permutations, it looks like this:
P_n=n ⋅ (n - 1) ⋅ (n - 2)…3 ⋅ 2 ⋅ 1=n!
Equation applies only if elements differ only in order.
Now the placement formula will be considered, it looks like this:
A_n^m=n ⋅ (n - 1) ⋅ (n-2) ⋅ … ⋅ (n - m + 1)=n!: (n - m)!
This expression applies not only to the order of the element, but also to its composition.
The third equation from combinatorics, and it is also the last one, is called the formula for the number of combinations:
C_n^m=n !: ((n -m))!:m !
Combinations are selections that are not ordered, respectively, and this rule applies to them.
It turned out to be easy to figure out the formulas of combinatorics, now we can move on to the classical definition of probabilities. This expression looks like this:
P(A)=m: n.
In this formula, m is the number of conditions favorable 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 of them, but the most important of them will be touched upon, such as, for example, the probability of the sum of events:
P(A + B)=P(A) + P(B) - this theorem is for adding only incompatible events;
P(A + B)=P(A) + P(B) - P(AB) - and this one is for adding only compatible ones.
Probability of producing events:
P(A ⋅ B)=P(A) ⋅ P(B) – this theorem is for independent events;
(P(A ⋅ B)=P(A) ⋅ P(B∣A); P(A ⋅ B)=P(A) ⋅ P(A∣B)) - and this one is for addicts.
The event formula ends the list. Probability theory tells us about Bayes' theorem, which looks like this:
P(H_m∣A)=(P(H_m)P(A∣H_m)): (∑_(k=1)^n P(H_k)P(A∣H_k)), m=1, …, n
In this formula, H1, H2, …, H is the complete group of hypotheses.
Let's stop here, then examples of applying formulas to solve specific problems from practice will be considered.
Examples
If you carefully study any sectionmathematics, it does not do without exercises and sample solutions. So is the theory of probability: events, examples here are an integral component that confirms scientific calculations.
Formula for the number of permutations
Let's say there are thirty cards in a deck of cards, starting with face value one. Next question. How many ways are there to stack the deck so that cards with a face value of one and two are not next to each other?
The task has been set, now let's move on to solving it. 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 are next. To do this, let's 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 the twenty-ninth, and the second card from the second to the thirtieth, it turns out twenty-nine places for a pair of cards. In turn, the rest can take twenty-eight places, and in any order. That is, for a permutation of 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 over the second, there are 29 ⋅ 28 extra possibilities!=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 ⋅ 28!=29!
It follows that there are 2 ⋅ 29 extra options!, while there are 30 required ways to build a deck! - 2 ⋅ 29!. It remains only to count.
30!=29! ⋅ 30; 30!-2⋅29!=29! ⋅ (30 - 2)=29! ⋅ 28
Now you need to multiply all the numbers from one to twenty-nine together, and then at the end multiply everything by 28. The answer is 2, 4757335 ⋅〖10〗^32
Solution of the example. Formula for Placement Number
In this problem, you need to find out how many ways there are to put fifteen volumes on one shelf, but under the condition that there are thirty volumes in total.
This problem has a slightly easier solution than the previous one. Using the already known formula, it is necessary to calculate the total number of locations from thirty volumes of fifteen.
A_30^15=30 ⋅ 29 ⋅ 28⋅… ⋅ (30 - 15 + 1)=30 ⋅ 29 ⋅ 28 ⋅ … ⋅ 16=202 843 204 931 727 360 000
The answer, respectively, will be 202 843 204 931 727 360 000.
Now let's take the task a little more difficult. You need to find out how many ways there are to 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 problems are solved in several ways, so there are two ways in this one, 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 a shelf with fifteen books for-differently. It turned out A_30^15=30 ⋅ 29 ⋅ 28 ⋅ … ⋅ (30 - 15 + 1)=30 ⋅ 29 ⋅ 28 ⋅ …⋅ 16.
We will calculate the second shelf using the permutation formula, because fifteen books are placed in it, while only fifteen remain. Use the formula P_15=15!.
It turns out that the total will be A_30^15 ⋅ P_15 ways, but, in addition, the product of all numbers from thirty to sixteen will have to be multiplied by the product of numbers from one to fifteen, as a result, the product of all numbers from one to thirty, so the answer is 30!
But this problem can be solved in a different 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 be two shelves, we cut one long one in half, it turns out two fifteen each. From this it turns out that the placement options can be P_30=30!.
Solution of the example. Formula for combination number
Now we will consider a variant of the third problem from combinatorics. You need to find out how many ways there are to arrange fifteen books, provided that you need to choose from thirty absolutely identical.
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 identical 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 it. Using this formula, in the shortest possible time it was possiblesolve such a problem, the answer, respectively, is 155 117 520.
Solution of the example. The classic definition of probability
With the formula above, you can find the answer to a simple problem. But it will help to visually see and follow the course of actions.
It is given in the problem that there are ten absolutely identical balls in the urn. 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 getting the blue ball as event A. This experience can have ten outcomes, which, in turn, are elementary and equally probable. At the same time, out of ten, six are favorable to event A. We solve according to the formula:
P(A)=6: 10=0, 6
Applying this formula, we found out that the probability of getting a blue ball is 0.6.
Solution of the example. Probability of the sum of events
Now a variant will be presented, which is solved using the formula for the probability of the sum of events. So, in the condition given that there are two boxes, the first contains one gray and five white balls, and the second contains 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 balls you get will be gray and white.
To solve this problem, you need to label the events.
- So, A - take a gray ball from the first box: P(A)=1/6.
- A’ – take a white ball also from the first box: P(A')=5/6.
- B – the gray ball has already been taken out of the second box: P(B)=2/3.
- B’ – take a gray ball from the second box: P(B')=1/3.
According to the condition of the problem, one of the phenomena must happen: 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. Next, to find out the answer, you need to apply the equation for their addition:
P=P(AB' + A'B)=P(AB') + P(A'B)=11/18.
This is how, using the formula, you can solve similar problems.
Result
The article provided information on the topic "Probability Theory", in which the probability of an event plays a crucial role. Of course, not everything was taken into account, but, based on the text presented, one can theoretically get acquainted with this section of mathematics. The science in question can be useful not only in professional work, but also in everyday life. With its help, you can calculate any possibility of any event.
The text also touched upon 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 led to the fact that people learned to calculate even random events. Once they were just interested in it, but today everyone already knows about it. And no one will say what awaits us in the future, what other brilliant discoveries related to the theory under consideration will be made. But one thing is for sure - research is not standing still!
