In fact, formulas (1) and (2) are a short record of conditional probability based on a contingency table of characteristics. Let's return to the example discussed (Fig. 1). Suppose we learn that a family is planning to buy a wide-screen television. What is the probability that this family will actually buy such a TV?

Rice. 1. Widescreen TV Buying Behavior

In this case, we need to calculate the conditional probability P (purchase completed | purchase planned). Since we know that the family is planning to buy, the sample space does not consist of all 1000 families, but only those planning to buy a wide-screen TV. Of the 250 such families, 200 actually bought this TV. Therefore, the probability that a family will actually buy a wide-screen TV if they have planned to do so can be calculated using the following formula:

P (purchase completed | purchase planned) = number of families who planned and bought a wide-screen TV / number of families planning to buy a wide-screen TV = 200 / 250 = 0.8

Formula (2) gives the same result:

where is the event A is that the family is planning to purchase a widescreen TV, and the event IN- that she will actually buy it. Substituting real data into the formula, we get:

Decision tree

In Fig. 1 families are divided into four categories: those who planned to buy a wide-screen TV and those who did not, as well as those who bought such a TV and those who did not. A similar classification can be performed using a decision tree (Fig. 2). The tree shown in Fig. 2 has two branches corresponding to families who planned to purchase a widescreen TV and families who did not. Each of these branches splits into two additional branches corresponding to households that did and did not purchase a widescreen TV. The probabilities written at the ends of the two main branches are the unconditional probabilities of events A And A'. The probabilities written at the ends of the four additional branches are the conditional probabilities of each combination of events A And IN. Conditional probabilities are calculated by dividing the joint probability of events by the corresponding unconditional probability of each of them.

Rice. 2. Decision tree

For example, to calculate the probability that a family will buy a wide-screen television if it has planned to do so, one must determine the probability of the event purchase planned and completed, and then divide it by the probability of the event purchase planned. Moving along the decision tree shown in Fig. 2, we get the following (similar to the previous) answer:

Statistical independence

In the example of buying a wide-screen TV, the probability that a randomly selected family purchased a wide-screen TV given that they planned to do so is 200/250 = 0.8. Recall that the unconditional probability that a randomly selected family purchased a wide-screen TV is 300/1000 = 0.3. This leads to a very important conclusion. Prior information that the family was planning a purchase influences the likelihood of the purchase itself. In other words, these two events depend on each other. In contrast to this example, there are statistically independent events whose probabilities do not depend on each other. Statistical independence is expressed by the identity: P(A|B) = P(A), Where P(A|B)- probability of event A provided that the event occurred IN, P(A)- unconditional probability of event A.

Please note that events A And IN P(A|B) = P(A). If in a contingency table of characteristics having a size of 2×2, this condition is satisfied for at least one combination of events A And IN, it will be valid for any other combination. In our example events purchase planned And purchase completed are not statistically independent because information about one event affects the probability of another.

Let's look at an example that shows how to test the statistical independence of two events. Let's ask 300 families who bought a widescreen TV if they were satisfied with their purchase (Fig. 3). Determine whether the degree of satisfaction with the purchase and the type of TV are related.

Rice. 3. Data characterizing the degree of satisfaction of buyers of widescreen TVs

Judging by these data,

At the same time,

P (customer satisfied) = 240 / 300 = 0.80

Therefore, the probability that the customer is satisfied with the purchase and that the family purchased an HDTV TV are equal, and these events are statistically independent because they are not related in any way.

Probability multiplication rule

The formula for calculating conditional probability allows you to determine the probability of a joint event A and B. Having resolved formula (1)

relative to joint probability P(A and B), we obtain a general rule for multiplying probabilities. Probability of event A and B equal to the probability of the event A provided that the event occurs IN IN:

(3) P(A and B) = P(A|B) * P(B)

Let's take as an example 80 families who bought a widescreen HDTV television (Fig. 3). The table shows that 64 families are satisfied with the purchase and 16 are not. Let us assume that two families are randomly selected from among them. Determine the probability that both customers will be satisfied. Using formula (3), we obtain:

P(A and B) = P(A|B) * P(B)

where is the event A is that the second family is satisfied with their purchase, and the event IN- that the first family is satisfied with their purchase. The probability that the first family is satisfied with their purchase is 64/80. However, the likelihood that the second family is also satisfied with their purchase depends on the first family's response. If the first family does not return to the sample after the survey (selection without return), the number of respondents is reduced to 79. If the first family is satisfied with their purchase, the probability that the second family will also be satisfied is 63/79, since there are only 63 left in the sample families satisfied with their purchase. Thus, substituting specific data into formula (3), we obtain the following answer:

P(A and B) = (63/79)(64/80) = 0.638.

Therefore, the probability that both families are satisfied with their purchases is 63.8%.

Suppose that after the survey the first family returns to the sample. Determine the probability that both families will be satisfied with their purchase. In this case, the probabilities that both families are satisfied with their purchase are the same and equal to 64/80. Therefore, P(A and B) = (64/80)(64/80) = 0.64. Thus, the probability that both families are satisfied with their purchases is 64.0%. This example shows that the choice of the second family does not depend on the choice of the first. Thus, replacing the conditional probability in formula (3) P(A|B) probability P(A), we obtain a formula for multiplying the probabilities of independent events.

The rule for multiplying the probabilities of independent events. If events A And IN are statistically independent, the probability of an event A and B equal to the probability of the event A, multiplied by the probability of the event IN.

(4) P(A and B) = P(A)P(B)

If this rule is true for events A And IN, which means they are statistically independent. Thus, there are two ways to determine the statistical independence of two events:

1. Events A And IN are statistically independent of each other if and only if P(A|B) = P(A).
2. Events A And B are statistically independent of each other if and only if P(A and B) = P(A)P(B).

If in a 2x2 contingency table, one of these conditions is met for at least one combination of events A And B, it will be valid for any other combination.

Unconditional probability of an elementary event

(5) P(A) = P(A|B 1)P(B 1) + P(A|B 2)P(B 2) + … + P(A|B k)P(B k)

where the events B 1, B 2, ... B k are mutually exclusive and exhaustive.

Let us illustrate the application of this formula using the example of Fig. 1. Using formula (5), we obtain:

P(A) = P(A|B 1)P(B 1) + P(A|B 2)P(B 2)

Where P(A)- the likelihood that the purchase was planned, P(B 1)- the probability that the purchase is made, P(B 2)- the probability that the purchase is not completed.

BAYES' THEOREM

The conditional probability of an event takes into account information that some other event has occurred. This approach can be used both to refine the probability taking into account newly received information, and to calculate the probability that the observed effect is a consequence of some specific reason. The procedure for refining these probabilities is called Bayes' theorem. It was first developed by Thomas Bayes in the 18th century.

Let's assume that the company mentioned above is researching the market for a new TV model. In the past, 40% of the TVs created by the company were successful, while 60% of the models were not recognized. Before announcing the release of a new model, marketing specialists carefully research the market and record demand. In the past, 80% of successful models were predicted to be successful, while 30% of successful predictions turned out to be wrong. The marketing department gave a favorable forecast for the new model. What is the likelihood that a new TV model will be in demand?

Bayes' theorem can be derived from the definitions of conditional probability (1) and (2). To calculate the probability P(B|A), take formula (2):

and substitute instead of P(A and B) the value from formula (3):

P(A and B) = P(A|B) * P(B)

Substituting formula (5) instead of P(A), we obtain Bayes’ theorem:

where events B 1, B 2, ... B k are mutually exclusive and exhaustive.

Let us introduce the following notation: event S - TV is in demand, event S’ - TV is not in demand, event F - favorable prognosis, event F’ - poor prognosis. Let’s assume that P(S) = 0.4, P(S’) = 0.6, P(F|S) = 0.8, P(F|S’) = 0.3. Applying Bayes' theorem we get:

The probability of demand for a new TV model, given a favorable forecast, is 0.64. Thus, the probability of lack of demand given a favorable forecast is 1–0.64=0.36. The calculation process is shown in Fig. 4.

Rice. 4. (a) Calculations using the Bayes formula to estimate the probability of demand for televisions; (b) Decision tree when studying demand for a new TV model

Let's look at an example of using Bayes' theorem for medical diagnostics. The probability that a person suffers from a particular disease is 0.03. A medical test can check if this is true. If a person is truly sick, the probability of an accurate diagnosis (saying that the person is sick when he really is sick) is 0.9. If a person is healthy, the probability of a false positive diagnosis (saying that a person is sick when he is healthy) is 0.02. Let's assume that medical test gave positive result. What is the probability that a person is actually sick? What is the likelihood of an accurate diagnosis?

Let us introduce the following notation: event D - the person is sick, event D’ - the person is healthy, event T - diagnosis is positive, event T’ - diagnosis negative. From the conditions of the problem it follows that P(D) = 0.03, P(D’) = 0.97, P(T|D) = 0.90, P(T|D’) = 0.02. Applying formula (6), we obtain:

The probability that with a positive diagnosis a person is really sick is 0.582 (see also Fig. 5). Please note that the denominator of the Bayes formula is equal to the probability of a positive diagnosis, i.e. 0.0464.

Probability of the opposite event

Consider some random event A, and let its probability p(A) known. Then the probability of the opposite event is determined by the formula

. (1.8)

Proof. Let us remember that according to axiom 3 for non-joint events

p(A+B) = p(A) + p(B).

Due to incompatibility A And

Consequence., that is, the probability of an impossible event is zero.

Using formula (1.8), for example, the probability of missing is determined if the probability of a hit is known (or, conversely, the probability of a hit if the probability of a miss is known; for example, if the probability of a hit for a gun is 0.9, the probability of a miss for it is (1 – 0, 9 = 0.1).

1. Probability of the sum of two events

It would be appropriate to recall here that for non-joint events this formula looks like:

Example. The plant produces 85% of first-grade products and 10% of second-grade products. The remaining products are considered defective. What is the probability that if we take a product at random, we will get a defect?

Solution. P = 1 – (0.85 + 0.1) = 0.05.

Probability of the sum of any two random events equal to

Proof. Let's imagine an event A + B as a sum of incompatible events

Given the incompatibility A and , we obtain according to axiom 3

Similarly we find

Substituting the latter into the previous formula, we obtain the desired (1.10) (Figure 2).

Example. Of the 20 students, 5 passed the exam in history with a bad mark, 4 - in English language, and 3 students received bad marks in both subjects. What is the percentage of students in the group who do not have failures in these subjects?

Solution. P = 1 – (5/20 + 4/20 – 3/20) = 0.7 (70%).

1. Conditional probability

In some cases it is necessary to determine the probability of a random event B provided that a random event occurred A, which has a non-zero probability. What is the event A happened, narrows the space of elementary events to a set A corresponding to this event. We will carry out further discussions using the example classical scheme. Let W consist of n equally possible elementary events (outcomes) and the event A favors m(A), and the event AB - m(AB) outcomes. Let us denote the conditional probability of the event B provided that A happened - p(B|A). By definition,

= .

If A happened, then one of the m(A) outcomes and event B can only happen if one of the outcomes favoring AB; such outcomes m(AB). Therefore, it is natural to put the conditional probability of the event B provided that A happened, equal to the ratio

To summarize, let's give general definition: conditional probability of event B, provided that event A occurs with non-zero probability , called

. (1.11)

It is easy to check that the definition introduced in this way satisfies all the axioms and, therefore, all the previously proven theorems are valid.

Often conditional probability p(B|A) can be easily found from the problem statement, in more difficult cases we have to use definition (1.11).

Example. An urn contains N balls, n of which are white and N-n black. A ball is taken out of it and, without putting it back ( sample without return ), they take out another one. What is the probability that both balls are white?

Solution. When solving this problem, we apply both the classical definition of probability and the product rule: let us denote by A the event that the white ball was drawn first (then the black ball was drawn first), and by B the event that the second was drawn white ball; Then

.

It is easy to see that the probability that three balls drawn in a row (without replacement) are white:

etc.

Example. Out of 30 exam tickets, the student prepared only 25. If he refuses to answer the first ticket taken (which he does not know), then he is allowed to take the second one. Determine the probability that the second ticket will be lucky.

Solution. Let the event A is that the first ticket pulled out turned out to be “bad” for the student, and B- the second - ²good². Because after the event A one of the “bad” ones has already been removed, then only 29 tickets remain, of which the student knows 25. Hence, the desired probability, assuming that the appearance of any ticket is equally possible and they do not return, is equal to .

1. Product probability

Relation (1.11), assuming that p(A) or p(B) are not equal to zero, can be written in the form

This ratio is called the theorem on the probability of the product of two events , which can be generalized to any number of factors, for example, for three it has the form

Example. Using the conditions of the previous example, find the probability of successfully passing the exam if for this the student must answer the first ticket or, without answering the first, must answer the second.

Solution. Let events A And B are that, respectively, the first and second tickets are ²good². Then – the appearance of a “bad” ticket for the first time. The exam will be taken if the event occurs A or at the same time B. That is, the desired event C - successful passing of the exam - is expressed as follows: C = A+ .From here

Here we took advantage of the incompatibility A and, and therefore, incompatibility A and , theorems on the probability of a sum and a product and the classical definition of probability when calculating p(A) And .

This problem can be solved more simply if we use the theorem on the probability of the opposite event:

1. Independence of events

Random events A and Blet's callindependent, If

For independent events, it follows from (1.11) that ; The converse is also true.

Independence of eventsmeans that the occurrence of event A does not change the probability of the occurrence of event B, that is, the conditional probability is equal to the unconditional probability .

Example. Let's consider the previous example with an urn containing N balls, of which n are white, but let's change the experiment: having taken out a ball, we put it back and only then take out the next one ( sample with return ).

A is the event that the white ball is drawn first, the event that the black ball is drawn first, and B is the event that the white ball is drawn second; Then

that is, in this case, events A and B are independent.

Thus, in sampling with return, the events of the second draw of the ball are independent of the events of the first drawing, but in sampling without return this is not the case. However, for large N and n these probabilities are very close to each other. This is used because sometimes sampling without return is performed (for example, during quality control, when testing an object leads to its destruction), and calculations are carried out using formulas for sampling with return, which are simpler.

In practice, when calculating probabilities, they often use the rule according to which from the physical independence of events follows their independence in the theoretical-probabilistic sense .

Example. The probability that a person aged 60 will not die in the next year is 0.91. Insurance company insures the lives of two 60-year-old people for a year.

Probability that neither of them will die: 0.91 × 0.91 = 0.8281.

Probability that they both die:

(1 – 0.91) × (1 – 0.91) = 0.09 × 0.09 = 0.0081.

Probability of dying at least one:

1 – 0.91 × 0.91 = 1 – 0,8281 = 0,1719.

Probability of dying one:

0.91 × 0.09 + 0.09 × 0.91 = 0.1638.

Event system A 1 , A 2 ,..., A n We call it independent in the aggregate if the probability of the product is equal to the product of the probabilities for any combination of factors from this system. In this case, in particular,

Example. The safe code consists of seven decimal digits. What is the probability that a thief will type it correctly the first time?

In each of the 7 positions you can dial any of the 10 digits 0,1,2,...,9, a total of 10 7 numbers, starting from 0000000 and ending with 9999999.

Example. The safe code consists of a Russian letter (there are 33 of them) and three numbers. What is the probability that a thief will type it correctly the first time?

P = (1/33) × (1/10) 3 .

Example. In more general view insurance problem: the probability that a person aged ... will not die in the next year is p. An insurance company insures the lives of n people of this age for a year.

The probability that none of them will not die: pn (no one will have to pay an insurance premium).

Probability of dying at least one: 1 – p n (payments are coming).

The likelihood that they All will die: (1 – p) n (largest payouts).

Probability of dying one: n × (1 – p) × p n-1 (if people are numbered, then the one who dies may have number 1, 2,…,n is n various events, each of which has probability (1 – p) × p n-1).

1. Formula full probability

Let events H 1 , H 2 , ... , H n meet the conditions

If , and .

Such a collection is called full group of events.

Let us assume that the probabilities are known p(H i), p(A/H i). In this case it is applicable total probability formula

. (1.14)

Proof. Let's use the fact that H i(they are usually called hypotheses ) are pairwise incompatible (hence incompatible and H i× A), and their sum is a reliable event

This scheme always occurs when we can talk about dividing the entire space of events into several, generally speaking, heterogeneous regions. In economics, this is the division of a country or region into regions different sizes And different conditions, when the share of each region is known p(Hi) and the probability (share) of some parameter in each region (for example, the percentage of unemployed - each region has its own) - p(A/H i). The warehouse may contain products from three different factories supplying different quantities products with different percentages of defects, etc.

Example. Casting in blanks comes from two workshops to the third: 70% from the first and 30% from the second. At the same time, the products of the first workshop have 10% defects, and the second – 20%. Find the probability that one blank taken at random has a defect.

Solution: p(H 1) = 0.7; p(H 2) = 0.3; p(A/H 1) = 0.1; p(A/H 2) = 0.2;

P = 0.7 × 0.1 + 0.3 × 0.2 = 0.13 (on average, 13% of the ingots in the third workshop are defective).

A mathematical model could be, for example, like this: there are several urns of different composition; the first urn contains n 1 balls, of which m 1 are white, etc. Using the total probability formula, we look for the probability of choosing an urn at random and drawing a white ball from it.

The same scheme is used to solve problems in the general case.

Example. Let's return to the example of an urn containing N balls, of which n are white. We take out two balls from it (without returning). What is the probability that the second ball is white?

Solution. H 1 – the first ball is white; p(H 1)=n/N;

H 2 – the first ball is black; p(H 2)=(N-n)/N;

B - the second ball is white; p(B|H 1)=(n-1)/(N-1); p(B|H 2)=n/(N-1);

The same model can be applied to solve the following problem: out of N tickets, a student has learned only n. What is more profitable for him - to draw the ticket first or second? It turns out that in any case he is likely n/N will draw a good ticket and with probability ( N-n)/N – bad.

Example. Determine the probability that a traveler starting from point A will end up at point B if at a fork in the road he randomly chooses any road (except the return one). The road map is shown in Fig. 1.3.

Solution. Let the traveler's arrival at points H 1, H 2, H 3 and H 4 be the corresponding hypotheses. Obviously, they form a complete group of events and according to the conditions of the problem

p(H 1) = p(H 2) = p(H 3) = p(H 4) = 0,25.

(All directions from A are equally possible for the traveler). According to the road map, the conditional probabilities of getting into B, provided that the traveler has passed through Hi, are equal to:

Applying the total probability formula, we get

1. Bayes formula

Let us assume that the conditions of the previous paragraph are met and it is additionally known that the event A happened. Let's find the probability that the hypothesis was realized H k. By definition of conditional probability

. (1.15)

The resulting relationship is called Bayes formula. It allows according to known
(before the experiment) a priori probabilities of the hypotheses p(Hi) and conditional probabilities p(A|H i) determine conditional probability p(H k |A) which is called a posteriori (that is, obtained under the condition that as a result of the experience the event A has already happened).

Example. 30% of patients admitted to the hospital belong to the first social group, 20% to the second and 50% to the third. The probability of contracting tuberculosis for a representative of each social group, respectively, is equal to 0.02, 0.03 and 0.01. Tests performed on a randomly selected patient showed the presence of tuberculosis. Find the probability that this is a representative of the third group.

In economics, as well as in other areas human activity or in nature, we constantly have to deal with events that cannot be accurately predicted. Thus, the sales volume of a product depends on demand, which can vary significantly, and on a number of other factors that are almost impossible to take into account. Therefore, when organizing production and carrying out sales, you have to predict the outcome of such activities on the basis of either your own previous experience, or similar experience of other people, or intuition, which to a large extent also relies on experimental data.

In order to somehow evaluate the event in question, it is necessary to take into account or specially organize the conditions in which this event is recorded.

The implementation of certain conditions or actions to identify the event in question is called experience or experiment.

The event is called random, if as a result of experience it may or may not occur.

The event is called reliable, if it necessarily appears as a result of a given experience, and impossible, if it cannot appear in this experience.

For example, snowfall in Moscow on November 30 is a random event. The daily sunrise can be considered a reliable event. Snowfall at the equator can be considered an impossible event.

One of the main tasks in probability theory is the task of determining a quantitative measure of the possibility of an event occurring.

Algebra of events

Events are called incompatible if they cannot be observed together in the same experience. Thus, the presence of two and three cars in one store for sale at the same time are two incompatible events.

Amount events is an event consisting of the occurrence of at least one of these events

An example of the sum of events is the presence of at least one of two products in the store.

The work events is an event consisting of the simultaneous occurrence of all these events

An event consisting of the appearance of two goods in a store at the same time is a product of events: - the appearance of one product, - the appearance of another product.

Events form a complete group of events if at least one of them is sure to occur in experience.

Example. The port has two berths for receiving ships. Three events can be considered: - the absence of ships at the berths, - the presence of one ship at one of the berths, - the presence of two ships at two berths. These three events form a complete group of events.

Opposite two unique possible events that form a complete group are called.

If one of the events that is opposite is denoted by , then the opposite event is usually denoted by .

Classical and statistical definitions of event probability

Each of the equally possible results of tests (experiments) is called an elementary outcome. They are usually designated by letters. For example, he rushes dice. There can be a total of six elementary outcomes based on the number of points on the sides.

From elementary outcomes you can create a more complex event. Thus, the event of an even number of points is determined by three outcomes: 2, 4, 6.

A quantitative measure of the possibility of the occurrence of the event in question is probability.

The most widely used definitions of the probability of an event are: classic And statistical.

The classical definition of probability is associated with the concept of a favorable outcome.

The outcome is called favorable to a given event if its occurrence entails the occurrence of this event.

In the above example, the event in question—an even number of points on the rolled side—has three favorable outcomes. In this case, the general
quantity possible outcomes. This means that the classical definition of the probability of an event can be used here.

Classic definition equals the ratio of the number of favorable outcomes to the total number of possible outcomes

where is the probability of the event, is the number of outcomes favorable to the event, total number possible outcomes.

In the considered example

The statistical definition of probability is associated with the concept of the relative frequency of occurrence of an event in experiments.

The relative frequency of occurrence of an event is calculated using the formula

where is the number of occurrences of an event in a series of experiments (tests).

Statistical definition. The probability of an event is the number around which the relative frequency stabilizes (sets) with an unlimited increase in the number of experiments.

In practical problems, the probability of an event is taken to be the relative frequency for a sufficiently large number of trials.

From these definitions of the probability of an event it is clear that the inequality is always satisfied

To determine the probability of an event based on formula (1.1), combinatorics formulas are often used, which are used to find the number of favorable outcomes and the total number of possible outcomes.

Everything in the world happens deterministically or by chance...
Aristotle

Probability: Basic Rules

Probability theory calculates the probabilities of various events. Fundamental to probability theory is the concept of a random event.

For example, you throw a coin, it randomly lands on a head or a tail. You don't know in advance which side the coin will land on. You enter into an insurance contract; you do not know in advance whether payments will be made or not.

In actuarial calculations, you need to be able to estimate the probability of various events, so probability theory plays a key role. No other branch of mathematics can deal with the probabilities of events.

Let's take a closer look at tossing a coin. There are 2 mutually exclusive outcomes: the coat of arms falls out or the tails fall out. The outcome of the throw is random, since the observer cannot analyze and take into account all the factors that influence the result. What is the probability of the coat of arms falling out? Most will answer ½, but why?

Let it be formal A indicates the loss of the coat of arms. Let the coin toss n once. Then the probability of the event A can be defined as the proportion of those throws that result in a coat of arms:

Where n total quantity throws, n(A) number of coat of arms drops.

Relation (1) is called frequency events A in a long series of tests.

It turns out that in various series of tests the corresponding frequency at large n clusters around some constant value P(A). This quantity is called probability of an event A and is designated by the letter R- abbreviation for English word probability - probability.

Formally we have:

(2)

This law is called law of large numbers.

If the coin is fair (symmetrical), then the probability of getting a coat of arms is equal to the probability of getting heads and equals ½.

Let A And IN some events, for example, whether an insured event occurred or not. The union of two events is an event consisting of the execution of an event A, events IN, or both events together. The intersection of two events A And IN called an event consisting in the implementation as an event A, and events IN.

Basic rules The calculus of event probabilities is as follows:

1. The probability of any event lies between zero and one:

2. Let A and B be two events, then:

It reads like this: the probability of two events combining is equal to the sum of the probabilities of these events minus the probability of the events intersecting. If the events are incompatible or non-overlapping, then the probability of the union (sum) of two events is equal to the sum of the probabilities. This law is called the law addition probabilities.

We say that an event is reliable if its probability is equal to 1. When analyzing certain phenomena, the question arises of how the occurrence of an event affects IN upon the occurrence of an event A. To do this, enter conditional probability :

(4)

It reads like this: probability of occurrence A given that IN equals the probability of intersection A And IN, divided by the probability of the event IN.
Formula (4) assumes that the probability of an event IN more than zero.

Formula (4) can also be written as:

(5)

This is the formula multiplying probabilities.

Conditional probability is also called a posteriori probability of an event A- probability of occurrence A after the attack IN.

In this case, the probability itself is called a priori probability. There are several other important formulas that are intensively used in actuarial calculations.

Total Probability Formula

Let us assume that an experiment is being carried out, the conditions of which can be determined in advance mutually mutually exclusive assumptions (hypotheses):

We assume that there is either a hypothesis, or... or. The probabilities of these hypotheses are known and equal:

Then the formula holds full probabilities :

(6)

Probability of an event occurring A equal to the sum of the products of the probability of occurrence A for each hypothesis on the probability of this hypothesis.

Bayes formula

Bayes formula allows you to recalculate the probability of hypotheses in the light new information which gave the result A.

Bayes' formula in a certain sense is the inverse of the total probability formula.

Consider the following practical problem.

Problem 1

Suppose there is a plane crash and experts are busy investigating its causes. 4 reasons why the disaster occurred are known in advance: either the cause, or, or, or. According to available statistics, these reasons have the following probabilities:

When examining the crash site, traces of fuel ignition were found; according to statistics, the probability of this event for one reason or another is as follows:

Question: what is the most likely cause of the disaster?

Let's calculate the probabilities of causes under the conditions of the occurrence of an event A.

From this it can be seen that the most likely reason is the first one, since its probability is maximum.

Problem 2

Consider an airplane landing at an airfield.

When landing, weather conditions may be as follows: no low clouds (), low clouds present (). In the first case, the probability of a safe landing is P1. In the second case - P2. It's clear that P1>P2.

Devices that provide blind landing have a probability of trouble-free operation R. If there is low cloud cover and the blind landing instruments have failed, the probability of a successful landing is P3, and P3<Р2 . It is known that for a given airfield the proportion of days in a year with low clouds is equal to .

Find the probability of the plane landing safely.

We need to find the probability.

There are two mutually exclusive options: the blind landing devices are working, the blind landing devices have failed, so we have:

Hence, according to the total probability formula:

Problem 3

An insurance company provides life insurance. 10% of those insured by this company are smokers. If the insured person does not smoke, the probability of his death during the year is 0.01. If he is a smoker, then this probability is 0.05.

What is the proportion of smokers among those insured who died during the year?

Possible answers: (A) 5%, (B) 20%, (C) 36%, (D) 56%, (E) 90%.

Solution

Let's enter the events:

The condition of the problem means that

In addition, since the events form a complete group of pairwise incompatible events, then .
The probability we are interested in is .

Using Bayes' formula, we have:

therefore the correct option is ( IN).

Problem 4

The insurance company sells life insurance contracts in three categories: standard, preferred and ultra-privileged.

50% of all insured are standard, 40% are preferred and 10% are ultra-privileged.

The probability of death within a year for a standard insured is 0.010, for a privileged one - 0.005, and for an ultra-privileged one - 0.001.

What is the probability that the deceased insured is ultra-privileged?

Solution

Let us introduce the following events into consideration:

In terms of these events, the probability we are interested in is . According to the condition:

Since the events , , form a complete group of pairwise incompatible events, using Bayes' formula we have:

Random variables and their characteristics

Let it be some random variable, for example, damage from a fire or the amount of insurance payments.
A random variable is completely characterized by its distribution function.

Definition. Function called distribution function random variable ξ .

Definition. If there is a function such that for arbitrary a completed

then they say that the random variable ξ has probability density function f(x).

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

This solution may not be the only one.

Quantile level ½ called theoretical median , quantile levels ¼ And ¾ -lower and upper quartiles respectively.

Plays an important role in actuarial applications Chebyshev's inequality:

at any

Symbol of mathematical expectation.

It reads like this: the probability that the modulus is greater than or equal to the mathematical expectation of the 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 regarding 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 the science of mass random phenomena that have the property of frequency stability.

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

Survival function

Probability theory describes 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 nice to work not with the distribution function, but with the additional distribution function . In terms of longevity, this is the probability that a person will live to 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) (usually years) and, accordingly, at x>.

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

The survival function has 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 at age by . Then:

.

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

So, the survival function is equal to the average proportion of those who survive to age from some fixed group of newborns.

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

The survival function can be reconstructed from density:

Lifespan Characteristics

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

1 . Average life time

,
2 . Dispersion lifetime

,
Where
,

Want to know the mathematical odds of your bet being successful? Then there is two good news for you. First: to calculate cross-country ability, you don’t 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 can easily calculate the probability of any of your trades passing.

To correctly determine cross-country ability, you need to take three steps:

• Calculate the percentage of probability of the outcome of an event according to the bookmaker's office;
• Calculate the probability using statistical data yourself;
• Find out the value of the bet, taking into account both probabilities.

Let's look at each of the steps in detail, using not only formulas, but also examples.

Quick Jump

Calculating the probability included in bookmaker odds

The first step is to find out with what probability the bookmaker himself estimates the chances of a particular outcome. It’s clear that bookmakers don’t set odds just like that. To do 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 that the odds for London Arsenal’s victory in the match against Bayern Munich are 4. This means that the probability of their victory is assessed by the bookmaker as (1/4)*100%=25%. Or Djokovic plays against Youzhny. The multiplier for Novak's victory is 1.2, his chances are (1/1.2)*100%=83%.

This is how the bookmaker itself evaluates the chances of success of 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 and game tone, we will use a simplified model and use only statistics from previous meetings. To calculate the statistical probability of an outcome, we use the formula:

PAND=(UM/M)*100%,

WherePAND– probability of an event according to the player;

UM – the number of successful matches in which such an event occurred;

M – total number of matches.

To make it clearer, let's give examples. Andy Murray and Rafael Nadal played 14 matches between themselves. In 6 of them the total was less than 21 in games, in 8 the total was more. You need to find out the probability that the next match will be played with a higher total: (8/14)*100=57%. Valencia played 74 matches against Atlético at Mestalla, in which they won 29 victories. Probability of Valencia winning: (29/74)*100%=39%.

And we learn all this only thanks to the statistics of previous games! Naturally, it will not be possible to calculate such a probability for any new team or player, so this betting strategy is only suitable for matches in which the opponents meet more than once. Now we know how to determine the bookmaker's and our own probabilities of outcomes, and we have all the knowledge to move on to the last step.

Determining the value of a bet

The value (value) of a bet and the passability have a direct connection: the higher the value, the higher the chance of passing. The value is calculated as follows:

V=PAND*K-100%,

where V is value;

P I – probability of outcome according to the bettor;

K – bookmaker odds for the outcome.

Let’s say we want to bet on Milan’s victory in the match against Roma and we calculate that the probability of the “red-blacks” winning is 45%. The bookmaker offers us odds of 2.5 for this outcome. Would such a bet be valuable? We carry out calculations: V=45%*2.5-100%=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, the probability of which, according to our calculations, is 60%. Bookmakers offer a 1.5 multiplier for this outcome. We determine the value: V=60%*1.5-100=-10%. As you can see, this bet is of no value and should be avoided.