Initially, being just a collection of information and empirical observations of the game of dice, the theory of probability has become a solid science. Fermat and Pascal were the first to give it a mathematical framework.

From reflections on the eternal to the theory of probability

The two personalities to whom the theory of probability owes many fundamental formulas, Blaise Pascal and Thomas Bayes, are known as deeply religious people, the latter was a Presbyterian minister. Apparently, the desire of these two scientists to prove the fallacy of the opinion about a certain Fortune, bestowing good luck on her favorites, gave impetus to research in this area. After all, in fact, any game of chance, with its wins and losses, is just a symphony of mathematical principles.

Thanks to the excitement of the Chevalier de Mere, who was equally a gambler and a person who was not indifferent to science, Pascal was forced to find a way to calculate the probability. De Mere was interested in this question: "How many times do you need to throw two dice in pairs so that the probability of getting 12 points exceeds 50%?". The second question that interested the gentleman extremely: "How to divide the bet between the participants in the unfinished game?" Of course, Pascal successfully answered both questions of de Mere, who became the unwitting initiator of the development of the theory of probability. It is interesting that the person of de Mere remained known in this area, and not in literature.

Previously, no mathematician has yet made an attempt to calculate the probabilities of events, since it was believed that this was only a guesswork solution. Blaise Pascal gave the first definition of the probability of an event and showed that this is a specific figure that can be justified mathematically. Probability theory has become the basis for statistics and is widely used in modern science.

What is randomness

If we consider a test that can be repeated an infinite number of times, then we can define a random event. This is one of the possible outcomes of the experience.

Experience is the implementation of specific actions in constant conditions.

In order to be able to work with the results of experience, events are usually denoted by the letters A, B, C, D, E ...

Probability of a random event

To be able to proceed to the mathematical part of probability, it is necessary to define all its components.

The probability of an event is a numerical measure of the possibility of the occurrence of some event (A or B) as a result of an experience. The probability is denoted as P(A) or P(B).

Probability theory is:

• reliable the event is guaranteed to occur as a result of the experiment Р(Ω) = 1;
• impossible the event can never happen Р(Ø) = 0;
• random the event lies between certain and impossible, that is, the probability of its occurrence is possible, but not guaranteed (the probability of a random event is always within 0≤P(A)≤1).

Relationships between events

Both one and the sum of events A + B are considered when the event is counted in the implementation of at least one of the components, A or B, or both - A and B.

In relation to each other, events can be:

• Equally possible.
• compatible.
• Incompatible.
• Opposite (mutually exclusive).
• Dependent.

If two events can happen with equal probability, then they equally possible.

If the occurrence of event A does not nullify the probability of occurrence of event B, then they compatible.

If events A and B never occur at the same time in the same experiment, then they are called incompatible. Tossing a coin is a good example: coming up tails is automatically not coming up heads.

The probability for the sum of such incompatible events consists of the sum of the probabilities of each of the events:

P(A+B)=P(A)+P(B)

If the occurrence of one event makes the occurrence of another impossible, then they are called opposite. Then one of them is designated as A, and the other - Ā (read as "not A"). The occurrence of event A means that Ā did not occur. These two events form a complete group with a sum of probabilities equal to 1.

Dependent events have mutual influence, decreasing or increasing each other's probability.

Relationships between events. Examples

It is much easier to understand the principles of probability theory and the combination of events using examples.

The experiment that will be carried out is to pull the balls out of the box, and the result of each experiment is an elementary outcome.

An event is one of the possible outcomes of an experience - a red ball, a blue ball, a ball with the number six, etc.

Test number 1. There are 6 balls, three of which are blue with odd numbers, and the other three are red with even numbers.

Test number 2. There are 6 blue balls with numbers from one to six.

Based on this example, we can name combinations:

• Reliable event. In Spanish No. 2, the event "get the blue ball" is reliable, since the probability of its occurrence is 1, since all the balls are blue and there can be no miss. Whereas the event "get the ball with the number 1" is random.
• Impossible event. In Spanish No. 1 with blue and red balls, the event "get the purple ball" is impossible, since the probability of its occurrence is 0.
• Equivalent events. In Spanish No. 1, the events “get the ball with the number 2” and “get the ball with the number 3” are equally likely, and the events “get the ball with an even number” and “get the ball with the number 2” have different probabilities.
• Compatible events. Getting a six in the process of throwing a die twice in a row are compatible events.
• Incompatible events. In the same Spanish No. 1 events "get the red ball" and "get the ball with an odd number" cannot be combined in the same experience.
• opposite events. The most striking example of this is coin tossing, where drawing heads is the same as not drawing tails, and the sum of their probabilities is always 1 (full group).
• Dependent events. So, in Spanish No. 1, you can set yourself the goal of extracting a red ball twice in a row. Extracting it or not extracting it the first time affects the probability of extracting it the second time.

It can be seen that the first event significantly affects the probability of the second (40% and 60%).

Event Probability Formula

The transition from fortune-telling to exact data occurs by transferring the topic to the mathematical plane. That is, judgments about a random event like "high probability" or "minimum probability" can be translated to specific numerical data. It is already permissible to evaluate, compare and introduce such material into more complex calculations.

From the point of view of calculation, the definition of the probability of an event is the ratio of the number of elementary positive outcomes to the number of all possible outcomes of experience with respect to a certain event. Probability is denoted by P (A), where P means the word "probability", which is translated from French as "probability".

So, the formula for the probability of an event is:

Where m is the number of favorable outcomes for event A, n is the sum of all possible outcomes for this experience. The probability of an event is always between 0 and 1:

0 ≤ P(A) ≤ 1.

Calculation of the probability of an event. Example

Let's take Spanish. No. 1 with balls, which is described earlier: 3 blue balls with numbers 1/3/5 and 3 red balls with numbers 2/4/6.

Based on this test, several different tasks can be considered:

• A - red ball drop. There are 3 red balls, and there are 6 variants in total. This is the simplest example, in which the probability of an event is P(A)=3/6=0.5.
• B - dropping an even number. There are 3 (2,4,6) even numbers in total, and the total number of possible numerical options is 6. The probability of this event is P(B)=3/6=0.5.
• C - loss of a number greater than 2. There are 4 such options (3,4,5,6) out of the total number of possible outcomes 6. The probability of the event C is P(C)=4/6=0.67.

As can be seen from the calculations, event C has a higher probability, since the number of possible positive outcomes is higher than in A and B.

Incompatible events

Such events cannot appear simultaneously in the same experience. As in Spanish No. 1, it is impossible to get a blue and a red ball at the same time. That is, you can get either a blue or a red ball. In the same way, an even and an odd number cannot appear in a die at the same time.

The probability of two events is considered as the probability of their sum or product. The sum of such events A + B is considered to be an event that consists in the appearance of an event A or B, and the product of their AB - in the appearance of both. For example, the appearance of two sixes at once on the faces of two dice in one throw.

The sum of several events is an event that implies the occurrence of at least one of them. The product of several events is the joint occurrence of them all.

In probability theory, as a rule, the use of the union "and" denotes the sum, the union "or" - multiplication. Formulas with examples will help you understand the logic of addition and multiplication in probability theory.

Probability of the sum of incompatible events

If the probability of incompatible events is considered, then the probability of the sum of events is equal to the sum of their probabilities:

P(A+B)=P(A)+P(B)

For example: we calculate the probability that in Spanish. No. 1 with blue and red balls will drop a number between 1 and 4. We will calculate not in one action, but by the sum of the probabilities of the elementary components. So, in such an experiment there are only 6 balls or 6 of all possible outcomes. The numbers that satisfy the condition are 2 and 3. The probability of getting the number 2 is 1/6, the probability of the number 3 is also 1/6. The probability of getting a number between 1 and 4 is:

The probability of the sum of incompatible events of a complete group is 1.

So, if in the experiment with a cube we add up the probabilities of getting all the numbers, then as a result we get one.

This is also true for opposite events, for example, in the experiment with a coin, where one of its sides is the event A, and the other is the opposite event Ā, as is known,

Р(А) + Р(Ā) = 1

Probability of producing incompatible events

Multiplication of probabilities is used when considering the occurrence of two or more incompatible events in one observation. The probability that events A and B will appear in it at the same time is equal to the product of their probabilities, or:

P(A*B)=P(A)*P(B)

For example, the probability that in No. 1 as a result of two attempts, a blue ball will appear twice, equal to

That is, the probability of an event occurring when, as a result of two attempts with the extraction of balls, only blue balls will be extracted, is 25%. It is very easy to do practical experiments on this problem and see if this is actually the case.

Joint Events

Events are considered joint when the appearance of one of them can coincide with the appearance of the other. Despite the fact that they are joint, the probability of independent events is considered. For example, throwing two dice can give a result when the number 6 falls on both of them. Although the events coincided and appeared simultaneously, they are independent of each other - only one six could fall out, the second die has no influence on it.

The probability of joint events is considered as the probability of their sum.

The probability of the sum of joint events. Example

The probability of the sum of events A and B, which are joint in relation to each other, is equal to the sum of the probabilities of the event minus the probability of their product (that is, their joint implementation):

R joint. (A + B) \u003d P (A) + P (B) - P (AB)

Assume that the probability of hitting the target with one shot is 0.4. Then event A - hitting the target in the first attempt, B - in the second. These events are joint, since it is possible that it is possible to hit the target both from the first and from the second shot. But the events are not dependent. What is the probability of the event of hitting the target with two shots (at least one)? According to the formula:

0,4+0,4-0,4*0,4=0,64

The answer to the question is: "The probability of hitting the target with two shots is 64%."

This formula for the probability of an event can also be applied to incompatible events, where the probability of the joint occurrence of an event P(AB) = 0. This means that the probability of the sum of incompatible events can be considered a special case of the proposed formula.

Probability geometry for clarity

Interestingly, the probability of the sum of joint events can be represented as two areas A and B that intersect with each other. As you can see from the picture, the area of ​​their union is equal to the total area minus the area of ​​their intersection. This geometric explanation makes the seemingly illogical formula more understandable. Note that geometric solutions are not uncommon in probability theory.

The definition of the probability of the sum of a set (more than two) of joint events is rather cumbersome. To calculate it, you need to use the formulas that are provided for these cases.

Dependent events

Dependent events are called if the occurrence of one (A) of them affects the probability of the occurrence of the other (B). Moreover, the influence of both the occurrence of event A and its non-occurrence is taken into account. Although events are called dependent by definition, only one of them is dependent (B). The usual probability was denoted as P(B) or the probability of independent events. In the case of dependents, a new concept is introduced - the conditional probability P A (B), which is the probability of the dependent event B under the condition that the event A (hypothesis) has occurred, on which it depends.

But event A is also random, so it also has a probability that must and can be taken into account in the calculations. The following example will show how to work with dependent events and a hypothesis.

Example of calculating the probability of dependent events

A good example for calculating dependent events is a standard deck of cards.

On the example of a deck of 36 cards, consider dependent events. It is necessary to determine the probability that the second card drawn from the deck will be a diamond suit, if the first card drawn is:

1. Tambourine.
2. Another suit.

Obviously, the probability of the second event B depends on the first A. So, if the first option is true, which is 1 card (35) and 1 diamond (8) less in the deck, the probability of event B:

P A (B) \u003d 8 / 35 \u003d 0.23

If the second option is true, then there are 35 cards in the deck, and the total number of tambourines (9) is still preserved, then the probability of the following event is B:

P A (B) \u003d 9/35 \u003d 0.26.

It can be seen that if event A is conditional on the fact that the first card is a diamond, then the probability of event B decreases, and vice versa.

Multiplication of dependent events

Based on the previous chapter, we accept the first event (A) as a fact, but in essence, it has a random character. The probability of this event, namely the extraction of a tambourine from a deck of cards, is equal to:

P(A) = 9/36=1/4

Since the theory does not exist by itself, but is called upon to serve practical purposes, it is fair to note that most often the probability of producing dependent events is needed.

According to the theorem on the product of the probabilities of dependent events, the probability of occurrence of jointly dependent events A and B is equal to the probability of one event A multiplied by the conditional probability of event B (depending on A):

P (AB) \u003d P (A) * P A (B)

Then in the example with a deck, the probability of drawing two cards with a suit of diamonds is:

9/36*8/35=0.0571 or 5.7%

And the probability of extracting not diamonds at first, and then diamonds, is equal to:

27/36*9/35=0.19 or 19%

It can be seen that the probability of occurrence of event B is greater, provided that a card of a suit other than a diamond is drawn first. This result is quite logical and understandable.

Total probability of an event

When a problem with conditional probabilities becomes multifaceted, it cannot be calculated by conventional methods. When there are more than two hypotheses, namely A1, A2, ..., A n , .. forms a complete group of events under the condition:

• P(A i)>0, i=1,2,…
• A i ∩ A j =Ø,i≠j.
• Σ k A k =Ω.

So, the formula for the total probability for event B with a complete group of random events A1, A2, ..., A n is:

A look into the future

The probability of a random event is essential in many areas of science: econometrics, statistics, physics, etc. Since some processes cannot be described deterministically, since they themselves are probabilistic, special methods of work are needed. The probability of an event theory can be used in any technological field as a way to determine the possibility of an error or malfunction.

It can be said that, by recognizing the probability, we somehow take a theoretical step into the future, looking at it through the prism of formulas.

In the economy, as well as in other areas of human activity or in nature, we constantly have to deal with events that cannot be accurately predicted. Thus, the volume of sales of goods 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 sales, one has to predict the outcome of such activities on the basis of either one's own previous experience, or similar experience of other people, or intuition, which is also largely based on experimental data.

In order to somehow evaluate the event under consideration, 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 the experiment, it may or may not occur.

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

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

One of the main problems in probability theory is the problem 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.

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

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

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

An event consisting in the appearance of two goods at the same time in the store 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 necessarily occurs in the experience.

Example. The port has two berths for ships. Three events can be considered: - the absence of vessels at the berths, - the presence of one vessel at one of the berths, - the presence of two vessels 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 are opposite is denoted by , then the opposite event is usually denoted by .

Classical and statistical definitions of the probability of an event

Each of the equally possible test results (experiments) is called an elementary outcome. They are usually denoted by letters . For example, a dice is thrown. There can be six elementary outcomes according to the number of points on the sides.

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

A quantitative measure of the possibility of occurrence of the event under consideration is the probability.

Two definitions of the probability of an event are most widely used: classic and statistical.

The classical definition of probability is related to the notion of a favorable outcome.

Exodus is called favorable this event, if its occurrence entails the occurrence of this event.

In the given example, the event under consideration is an even number of points on the dropped edge, has three favorable outcomes. In this case, the general
the number of possible outcomes. So, here you can use the classical definition of the probability of an event.

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 favorable outcomes for the event, is the total number of 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 by the formula

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

Statistical definition. The probability of an event is the number relative to which the relative frequency is stabilized (established) with an unlimited increase in the number of experiments.

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

From these definitions of the probability of an event, it can be seen that the inequality always holds

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

In fact, formulas (1) and (2) are a short record of the conditional probability based on the contingency table of features. Let's return to the example considered (Fig. 1). Let's say we know that a certain family is going to buy a widescreen TV. What is the probability that this family will actually buy such a TV?

Rice. 1. Widescreen TV Buyer Behavior

In this case, we need to calculate the conditional probability P (the purchase was made | the purchase was planned). Since we know that a family is planning to buy, the sample space does not consist of all 1,000 families, but only those that are planning to buy a widescreen TV. Of the 250 such families, 200 actually bought this TV. Therefore, the probability that a family will actually buy a widescreen TV, if they planned to do so, can be calculated using the following formula:

P (purchase made | purchase planned) = number of families planning and purchasing a widescreen TV / number of families planning to buy a widescreen TV = 200 / 250 = 0.8

The same result is given by formula (2):

where is the event BUT is that the family plans to buy a widescreen TV, and the event AT- that she will actually buy it. Substituting real data into the formula, we get:

decision tree

On fig. 1 families were divided into four categories: those who planned to buy a widescreen TV and those who did not, and those who bought such a TV and those who did not. A similar classification can be done 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 is divided into two additional branches, corresponding to families who bought and did not buy a widescreen TV. The probabilities written at the ends of the two main branches are the unconditional probabilities of events BUT and BUT'. The probabilities written at the ends of the four additional branches are the conditional probabilities of each combination of events BUT and AT. 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 widescreen TV, if they planned to do so, one should 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 one) answer:

Statistical independence

In the example of buying a widescreen TV, the probability that a randomly selected family purchased a widescreen 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 widescreen TV is 300/1000 = 0.3. A very important conclusion follows from this. A priori information that the family was planning a purchase affects the probability 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)- event probability BUT assuming an event has occurred AT, P(A) is the unconditional probability of event A.

Please note that the events BUT and AT P(A|B) = P(A). If in the feature contingency table, which has a size of 2 × 2, this condition is satisfied for at least one combination of events BUT and AT, it will be valid for any other combination. In our example, the 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 whether they are satisfied with their purchase (Fig. 3). Determine if the degree of satisfaction with the purchase and the type of TV are related.

Rice. 3. Customer Satisfaction Data for Widescreen TVs

According to these data,

In 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 has bought an HDTV is equal, and these events are statistically independent, since they are not related to each other.

Probability multiplication rule

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

with respect to the joint probability P(A and B), we obtain the general rule for multiplication of probabilities. Event Probability A and B is equal to the probability of the event BUT provided that the event AT AT:

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

Consider, for example, 80 households who purchased a widescreen HDTV (Figure 3). The table shows that 64 families are satisfied with the purchase and 16 are not. Suppose that two families are randomly selected among them. Determine the probability that both buyers will be satisfied. Using formula (3), we obtain:

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

where is the event BUT is that the second family is satisfied with their purchase, and the event AT- 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 probability that the second family is also satisfied with their purchase depends on the response of the first family. If the first family is not returned to the sample after the survey (selection without return), the number of respondents drops to 79. If the first family was satisfied with their purchase, the probability that the second family will also be satisfied is 63/79, since only 63 remained in the sample families satisfied with their purchase. Thus, substituting specific data into formula (3), we get 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 is returned 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 in formula (3) the conditional probability P(A|B) probability P(A), we obtain a formula for multiplying the probabilities of independent events.

Rule for multiplying the probabilities of independent events. If events BUT and AT are statistically independent, the probability of an event A and B is equal to the probability of the event BUT multiplied by the probability of the event AT.

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

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

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

If in the feature contingency table, which has a size of 2 × 2, one of these conditions is satisfied for at least one combination of events BUT and B, it will be valid for any other combination.

Unconditional probability of an elementary event

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

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

We illustrate the application of this formula on 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 probability 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 made.

BAYES' THEOREM

The conditional probability of an event takes into account the 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 the result of some specific cause. The procedure for refining these probabilities is called Bayes' theorem. It was first developed by Thomas Bayes in the 18th century.

Suppose 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, and 60% of the models were not recognized. Before announcing the release of a new model, marketers carefully research the market and capture demand. In the past, the success of 80% of models that received recognition was predicted in advance, while 30% of favorable forecasts turned out to be wrong. For the new model, the marketing department gave a favorable forecast. 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 Р(В|А), we take the 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 the Bayes theorem:

where the 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 not in demand, event F - favorable prognosis, event F' - poor prognosis. Let's say 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, subject to a favorable forecast, is 0.64. Thus, the probability of lack of demand under the condition of a favorable forecast is 1–0.64=0.36. The calculation process is shown in fig. 4.

Rice. 4. (a) Bayesian calculations to estimate the probability of TV demand; (b) Decision tree for researching demand for a new TV model

Let's consider an example of application of Bayes' theorem for medical diagnostics. The probability that a person suffers from a certain disease is 0.03. A medical test allows you to check if this is so. If a person is really sick, the probability of an accurate diagnosis (stating that a person is sick when he is really sick) is 0.9. If a person is healthy, the probability of a false positive diagnosis (stating that a person is sick when they are healthy) is 0.02. Let's say a medical test came back positive. What is the probability that the person is actually sick? What is the likelihood of an accurate diagnosis?

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

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

Probability event is the ratio of the number of elementary outcomes that favor a given event to the number of all equally possible outcomes of experience in which this event may occur. The probability of an event A is denoted by P(A) (here P is the first letter of the French word probabilite - probability). According to the definition
(1.2.1)
where is the number of elementary outcomes favoring event A; - the number of all equally possible elementary outcomes of experience, forming a complete group of events.
This definition of probability is called classical. It arose at the initial stage of the development of probability theory.

The probability of an event has the following properties:
1. The probability of a certain event is equal to one. Let's designate a certain event by the letter . For a certain event, therefore
(1.2.2)
2. The probability of an impossible event is zero. We denote the impossible event by the letter . For an impossible event, therefore
(1.2.3)
3. The probability of a random event is expressed as a positive number less than one. Since the inequalities , or are satisfied for a random event, then
(1.2.4)
4. The probability of any event satisfies the inequalities
(1.2.5)
This follows from relations (1.2.2) -(1.2.4).

Example 1 An urn contains 10 balls of the same size and weight, of which 4 are red and 6 are blue. One ball is drawn from the urn. What is the probability that the drawn ball is blue?

Decision. The event "the drawn ball turned out to be blue" will be denoted by the letter A. This test has 10 equally possible elementary outcomes, of which 6 favor the event A. In accordance with formula (1.2.1), we obtain

Example 2 All natural numbers from 1 to 30 are written on identical cards and placed in an urn. After thoroughly mixing the cards, one card is removed from the urn. What is the probability that the number on the card drawn is a multiple of 5?

Decision. Denote by A the event "the number on the taken card is a multiple of 5". In this test, there are 30 equally possible elementary outcomes, of which 6 outcomes favor event A (numbers 5, 10, 15, 20, 25, 30). Hence,

Example 3 Two dice are thrown, the sum of points on the upper faces is calculated. Find the probability of the event B, consisting in the fact that the top faces of the cubes will have a total of 9 points.

Decision. There are 6 2 = 36 equally possible elementary outcomes in this trial. Event B is favored by 4 outcomes: (3;6), (4;5), (5;4), (6;3), so

Example 4. A natural number not exceeding 10 is chosen at random. What is the probability that this number is prime?

Decision. Denote by the letter C the event "the chosen number is prime". In this case, n = 10, m = 4 (primes 2, 3, 5, 7). Therefore, the desired probability

Example 5 Two symmetrical coins are tossed. What is the probability that both coins have digits on the top sides?

Decision. Let's denote by the letter D the event "there was a number on the top side of each coin". There are 4 equally possible elementary outcomes in this test: (G, G), (G, C), (C, G), (C, C). (The notation (G, C) means that on the first coin there is a coat of arms, on the second - a number). Event D is favored by one elementary outcome (C, C). Since m = 1, n = 4, then

Example 6 What is the probability that the digits in a randomly chosen two-digit number are the same?

Decision. Two-digit numbers are numbers from 10 to 99; there are 90 such numbers in total. 9 numbers have the same digits (these are the numbers 11, 22, 33, 44, 55, 66, 77, 88, 99). Since in this case m = 9, n = 90, then
,
where A is the "number with the same digits" event.

Example 7 From the letters of the word differential one letter is chosen at random. What is the probability that this letter will be: a) a vowel b) a consonant c) a letter h?

Decision. There are 12 letters in the word differential, of which 5 are vowels and 7 are consonants. Letters h this word does not. Let's denote the events: A - "vowel", B - "consonant", C - "letter h". The number of favorable elementary outcomes: - for event A, - for event B, - for event C. Since n \u003d 12, then
, and .

Example 8 Two dice are tossed, the number of points on the top face of each dice is noted. Find the probability that both dice have the same number of points.

Decision. Let us denote this event by the letter A. Event A is favored by 6 elementary outcomes: (1;]), (2;2), (3;3), (4;4), (5;5), (6;6). In total there are equally possible elementary outcomes that form a complete group of events, in this case n=6 2 =36. So the desired probability

Example 9 The book has 300 pages. What is the probability that a randomly opened page will have a sequence number that is a multiple of 5?

Decision. It follows from the conditions of the problem that there will be n = 300 of all equally possible elementary outcomes that form a complete group of events. Of these, m = 60 favor the occurrence of the specified event. Indeed, a number that is a multiple of 5 has the form 5k, where k is a natural number, and , whence . Hence,
, where A - the "page" event has a sequence number that is a multiple of 5".

Example 10. Two dice are thrown, the sum of points on the upper faces is calculated. What is more likely to get a total of 7 or 8?

Decision. Let's designate the events: A - "7 points fell out", B - "8 points fell out". Event A is favored by 6 elementary outcomes: (1; 6), (2; 5), (3; 4), (4; 3), (5; 2), (6; 1), and event B - by 5 outcomes: (2; 6), (3; 5), (4; 4), (5; 3), (6; 2). There are n = 6 2 = 36 of all equally possible elementary outcomes. Hence, and .

So, P(A)>P(B), that is, getting a total of 7 points is a more likely event than getting a total of 8 points.

Tasks

1. A natural number not exceeding 30 is chosen at random. What is the probability that this number is a multiple of 3?
2. In the urn a red and b blue balls of the same size and weight. What is the probability that a randomly drawn ball from this urn is blue?
3. A number not exceeding 30 is chosen at random. What is the probability that this number is a divisor of zo?
4. In the urn a blue and b red balls of the same size and weight. One ball is drawn from this urn and set aside. This ball is red. Then another ball is drawn from the urn. Find the probability that the second ball is also red.
5. A natural number not exceeding 50 is chosen at random. What is the probability that this number is prime?
6. Three dice are thrown, the sum of points on the upper faces is calculated. What is more likely - to get a total of 9 or 10 points?
7. Three dice are tossed, the sum of the dropped points is calculated. What is more likely to get a total of 11 (event A) or 12 points (event B)?

Answers

1. 1/3. 2 . b/(a+b). 3 . 0,2. 4 . (b-1)/(a+b-1). 5 .0,3.6 . p 1 \u003d 25/216 - the probability of getting 9 points in total; p 2 \u003d 27/216 - the probability of getting 10 points in total; p2 > p1 7 . P(A) = 27/216, P(B) = 25/216, P(A) > P(B).

Questions

1. What is called the probability of an event?
2. What is the probability of a certain event?
3. What is the probability of an impossible event?
4. What are the limits of the probability of a random event?
5. What are the limits of the probability of any event?
6. What definition of probability is called classical?