Probability of an event a. Bernoulli scheme

then we say that the random variable ξ It has probability distribution density f(x).

Definition. Let be . For a continuous distribution function F theoretical α-quantile is called the solution of the equation.

This solution may not be the only one.

Level quantile ½ called theoretical median , level quantiles ¼ and ¾ -lower and upper quartiles respectively.

In actuarial applications, an important role is played by Chebyshev's inequality:

for any

Mathematical expectation symbol.

It reads like this: the probability that modulus is greater than less than or equal to the expectation of modulus divided by .

Lifetime as a random variable

The uncertainty of the moment of death is a major risk factor in life insurance.

Nothing definite can be said about the moment of death of an individual. However, if we are dealing with a large homogeneous group of people and are not interested in the fate of individual people from this group, then we are within the framework of probability theory as a science of mass random phenomena with the frequency stability property.

Respectively, we can talk about life expectancy as a random variable T.

survival function

In probability theory, they describe the stochastic nature of any random variable T distribution function F(x), which is defined as the probability that the random variable T less than number x:

.

In actuarial mathematics, it is pleasant to work not with a distribution function, but with an additional distribution function . In terms of longevity, it is the probability that a person will live to the age x years.

called survival function(survival function):

The survival function has the following properties:

In life tables, it is usually assumed that there is some age limit (limiting age) (as a rule, years) and, accordingly, at x>.

When describing mortality by analytical laws, it is usually assumed that the life time is unlimited, however, the type and parameters of the laws are selected so that the probability of life over a certain age is negligible.

The survival function has a simple statistical meaning.

Let's say that we are observing a group of newborns (usually ) whom we observe and can record the moments of their death.

Let us denote the number of living representatives of this group in age through . Then:

.

Symbol E here and below is used to denote the mathematical expectation.

So, the survival function is equal to the average proportion of those who survived to age from a certain fixed group of newborns.

In actuarial mathematics, one often works not with a survival function, but with a value just introduced (having fixed the initial group size).

The survival function can be reconstructed from the density:

Life span characteristics

From a practical point of view, the following characteristics are important:

1 . The average lifetime

,
2 . Dispersion lifetime

,
where
,

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 will fall 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.

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, 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, and 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 beginnings of this science appeared in the Middle 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.

Like-minded people

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 brought out

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 magnitude of chance;
• mathematical expectation for discrete cases;
• theorems of multiplication and addition of probabilities.

It is also impossible not to remember who also made a significant contribution to the study of the problem. Conducting his own tests, 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. 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;
• theory of Markov chains;
• 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, the scientist Lotman, who works 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 phenomena that have occurred. Probability theory indicates that all conditions can be repeated constantly. It was their conduct that was called "experiment" or "test".

A certain event is one that will 100% occur in a given test. 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 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 \u003d A + B.

Disjoint events in probability theory imply that the 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 prevent B in any way.

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 as incompatible events in probability theory. But their difference lies in the fact that one of the many phenomena in any case must occur.

Equally probable events are those actions, the possibility of repetition of which is equal. To make it clearer, we can imagine the tossing of a coin: the loss of one of its sides is equally likely to fall out of the other.

A favorable 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 die 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 the theory of probability are projected only on two or more cases and imply the independence of any action from another. For example, A - dropping tails when throwing a coin, and B - getting a jack from the deck. They are independent events in probability theory. At this point, 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" were considered above, the definition of the main 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 too.

It is better to start with the main ones. 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 you can move on to the presentation of the formulas themselves and their definition.

The first of these will be an expression for the number of permutations, it looks like this:

P_n = n ⋅ (n - 1) ⋅ (n - 2)…3 ⋅ 2 ⋅ 1 = n!

The equation applies only if the elements differ only in their 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 is applicable 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!

A combination is called a selection that is 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:

In this formula, m is the number of conditions favorable to the 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 dependents.

The event formula will end 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, H 1 , H 2 , …, H n is the full group of hypotheses.

Examples

If you carefully study any branch of mathematics, it is not complete without exercises and sample solutions. So is the theory of probability: events, examples here are an integral component that confirms scientific calculations.

Formula for 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 is 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 only 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 above 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!

From this it follows that there are 2 ⋅ 29! extra options, while there are 30 necessary ways to build the 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 among themselves, and then at the end multiply everything by 28. The answer is 2.4757335 ⋅〖10〗^32

Example solution. 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 on the condition that there are thirty volumes in total.

In this problem, the solution is slightly simpler than in the previous one. Using the already known formula, it is necessary to calculate the total number of arrangements 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 equal to 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 in different ways. It turned out A_30^15 = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ...⋅ 16.

We calculate the second shelf according to the permutation formula, because fifteen books are placed in it, while only fifteen remain. We use the formula P_15 = 15!.

It turns out that in total there will be A_30^15 ⋅ 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, as a result, the product of all numbers from one to thirty will be obtained, that is, the answer equals 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!.

Example solution. 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 ones.

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 all. Using this formula, in the shortest possible time it was possible to solve such a problem, the answer, respectively, is 155 117 520.

Example solution. The classical definition of probability

Using the formula above, you can find the answer in a simple problem. But it will help to visually see and trace the course of actions.

The problem is given 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, six out of ten are favorable for event A. We solve using the formula:

P(A) = 6: 10 = 0.6

By applying this formula, we found out that the probability of getting a blue ball is 0.6.

Example solution. 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. It is necessary to find out what is the chance that the balls taken out will be gray and white.

To solve this problem, it is necessary to designate events.

• So, A - take a gray ball from the first box: P(A) = 1/6.
• A '- they took a white ball also from the first box: P (A ") \u003d 5/6.
• B - a gray ball was taken out already from the second box: P(B) = 2/3.
• B' - they took a gray ball from the second box: P(B") = 1/3.

According to the condition of the problem, it is necessary that one of the phenomena occur: AB 'or A'B. Using the formula, we get: P(AB") = 1/18, P(A"B) = 10/18.

Now the formula for multiplying the probability 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.

So, using the formula, you can solve similar problems.

Outcome

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 the theory of probability 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 does not stand still!

Whether we like it or not, our life is full of all kinds of accidents, both pleasant and not very. Therefore, each of us would do well to know how to find the probability of an event. This will help you make the right decisions under any circumstances that are associated with uncertainty. For example, such knowledge will be very useful when choosing investment options, evaluating the possibility of winning a stock or lottery, determining the reality of achieving personal goals, etc., etc.

Probability Formula

In principle, the study of this topic does not take too much time. In order to get an answer to the question: "How to find the probability of a phenomenon?", you need to understand the key concepts and remember the basic principles on which the calculation is based. So, according to statistics, the events under study are denoted by A1, A2,..., An. Each of them has both favorable outcomes (m) and the total number of elementary outcomes. For example, we are interested in how to find the probability that an even number of points will be on the top face of the cube. Then A is roll m - rolling 2, 4, or 6 (three favorable choices), and n is all six possible choices.

The calculation formula itself is as follows:

With one outcome, everything is extremely easy. But how to find the probability if the events go one after the other? Consider this example: one card is shown from a deck of cards (36 pieces), then it is hidden again in the deck, and after mixing, the next one is pulled out. How to find the probability that at least in one case the Queen of Spades was drawn? There is the following rule: if a complex event is considered, which can be divided into several incompatible simple events, then you can first calculate the result for each of them, and then add them together. In our case, it will look like this: 1/36 + 1/36 = 1/18. But what about when several occur at the same time? Then we multiply the results! For example, the probability that when two coins are tossed at the same time, two tails will fall out will be equal to: ½ * ½ = 0.25.

Now let's take an even more complex example. Suppose we enter a book lottery in which ten out of thirty tickets are winning. It is required to determine:

1. The probability that both will win.
2. At least one of them will bring a prize.
3. Both will be losers.

So let's consider the first case. It can be broken down into two events: the first ticket will be lucky, and the second one will also be lucky. Let's take into account that the events are dependent, since after each pulling out the total number of options decreases. We get:

10 / 30 * 9 / 29 = 0,1034.

In the second case, you need to determine the probability of a losing ticket and take into account that it can be both the first in a row and the second one: 10 / 30 * 20 / 29 + 20 / 29 * 10 / 30 = 0.4598.

Finally, the third case, when even one book cannot be obtained from the lottery: 20 / 30 * 19 / 29 = 0.4368.

Do you want to know what are the mathematical chances of your bet being successful? Then we have two good news for you. First: to calculate the patency, you do not need to carry out complex calculations and spend a lot of time. It is enough to use simple formulas, which will take a couple of minutes to work with. Second, after reading this article, you will easily be able to calculate the probability of passing any of your trades.

To correctly determine the patency, you need to take three steps:

• Calculate the percentage of the probability of the outcome of an event according to the bookmaker's office;
• Calculate the probability from statistical data yourself;
• Find out the value of a bet given both probabilities.

Let us consider in detail each of the steps, using not only formulas, but also examples.

Fast passage

Calculation of the probability embedded in the betting odds

The first step is to find out with what probability the bookmaker evaluates the chances of a particular outcome. After all, it is clear that bookmakers do not bet odds just like that. For this we use the following formula:

PB=(1/K)*100%,

where P B is the probability of the outcome according to the bookmaker's office;

K - bookmaker odds for the outcome.

Let's say the odds are 4 for the victory of the London Arsenal in a duel against Bayern. This means that the probability of its victory by the BC is regarded as (1/4) * 100% = 25%. Or Djokovic is playing against South. The multiplier for Novak's victory is 1.2, his chances are equal to (1/1.2)*100%=83%.

This is how the bookmaker itself evaluates the chances of success for each player and team. Having completed the first step, we move on to the second.

Calculation of the probability of an event by the player

The second point of our plan is our own assessment of the probability of the event. Since we cannot mathematically take into account such parameters as motivation, game tone, we will use a simplified model and use only the statistics of previous meetings. To calculate the statistical probability of an outcome, we use the formula:

PAnd\u003d (UM / M) * 100%,

wherePAnd- the probability of the event according to the player;

UM - the number of successful matches in which such an event took place;

M is the total number of matches.

To make it clearer, let's give examples. Andy Murray and Rafael Nadal have played 14 matches. In 6 of them, total under 21 games were recorded, in 8 - total over. It is necessary to find out the probability that the next match will be played for a total over: (8/14)*100=57%. Valencia played 74 matches at the Mestalla against Atlético, in which they scored 29 victories. Probability of Valencia winning: (29/74)*100%=39%.

And we all know this only thanks to the statistics of previous games! Naturally, such a probability cannot be calculated for some new team or player, so this betting strategy is only suitable for matches in which opponents meet not for the first time. Now we know how to determine the betting and own probabilities of outcomes, and we have all the knowledge to go to the last step.

Determining the value of a bet

The value (valuability) of the bet and the passability are directly related: the higher the valuation, the higher the chance of a pass. The value is calculated as follows:

V=PAnd*K-100%,

where V is the value;

P I - the probability of an outcome according to the better;

K - bookmaker odds for the outcome.

Let's say we want to bet on Milan to win the match against Roma and we calculated that the probability of the Red-Blacks winning is 45%. The bookmaker offers us a coefficient of 2.5 for this outcome. Would such a bet be valuable? We carry out calculations: V \u003d 45% * 2.5-100% \u003d 12.5%. Great, we have a valuable bet with good chances of passing.

Let's take another case. Maria Sharapova plays against Petra Kvitova. We want to make a deal for Maria to win, which, according to our calculations, has a 60% probability. Bookmakers offer a multiplier of 1.5 for this outcome. Determine the value: V=60%*1.5-100=-10%. As you can see, this bet is of no value and should be refrained from.