Theorem. The probability of the sum of incompatible events and is equal to the sum of the probabilities of these events:

Consequence 1. Using the method of mathematical induction, formula (3.10) can be generalized to any number of pairwise incompatible events:

Consequence 2. Since the opposite events are incompatible, and their sum is a reliable event, then, using (3.10), we have:

Often, when solving problems, formula (3.12) is used in the form:

(3.13)

Example 3.29. In the experiment with throwing a dice, find the probabilities of getting more than 3 and less than 6 on the upper bound of the number of points.

Let us denote the events associated with the loss of one point on the upper face of the dice through U 1 , two points across U 2 ,…, six points through U 6 .

Let the event U- drop on the upper face of the die number of points more than 3 and less than 6. This event will occur if at least one of the events occurs U 4 or U 5 , therefore, it can be represented as the sum of these events: . Because events U 4 And U 5 are inconsistent, then to find the probability of their sum we use formula (3.11). Considering that the probabilities of events U 1 , U 2 ,…,U 6 are equal, we get:

Comment. Previously, problems of this type were solved by counting the number of favorable outcomes. Indeed, the event U is favored by two outcomes, and only six elementary outcomes, therefore, using the classical approach to the concept of probability, we obtain:

However, the classical approach to the concept of probability, in contrast to the theorem on the probability of the sum of incompatible events, is applicable only for equally possible outcomes.

Example 3.30. The shooter's chance of hitting the target is 0.7. What is the probability that the shooter misses the target?

Let the event be the shooter hitting the target, then the event that the shooter does not hit the target is the opposite event to the event, because as a result of each test, one and only one of these events always occurs. Using formula (3.13), we obtain:

3.2.10. Probability of producing events

Definition. The event is called dependent from the event if the probability of an event depends on whether the event occurred or not.

Definition. The probability of an event, given that the event has occurred, is called conditional probability events and is marked

Theorem. The probability of a product of events is equal to the product of the probability of one of them by the conditional probability of the other, calculated under the condition that the first took place:

Condition of independence of an event from an event can be written in the form It follows from this statement that for independent events the following relation holds:

i.e., the probability of the product of independent events and is equal to the product of their probabilities.

Comment. The probability of the product of several events is equal to the product of the probabilities of these events, and the probability of each next event in order is calculated under the condition that all previous ones took place:

If the events are independent, then we have:

Example 3.31. There are 5 white and 3 black balls in a box. Two balls are drawn at random from it without replacement. Find the probability that both balls are white.

Let the event be the appearance of a white ball on the first draw, the appearance of a white ball on the second draw. Considering that, (the probability of a second white ball appearing, provided that the first ball drawn was white and was not returned to the box). Since the events and are dependent, we can find the probability of their product using the formula (3.15):

Example 3.32. The probability of hitting the target by the first shooter is 0.8; the second - 0.7. Each shooter fired at a target. What is the probability that at least one shooter hits the target? What is the probability that one shooter will hit the target?

Let the event be a hit on the target by the first shooter, - by the second. All possible options can be represented as tables 3.5, where "+" means that the event happened, and "-" - did not happen.

Table 3.5

Let an event be a hit by at least one shooter on the target. Then the event is the sum of independent events and, therefore, it is impossible to apply the theorem on the probability of the sum of incompatible events in this situation.

Consider an event opposite to the event that will occur when no shooter hits the target, i.e., is the product of independent events. Using formulas (3.13) and (3.15), we obtain:

Let the event be a hit by one shooter on the target. This event can be represented as follows:

The events and are independent, the events and are also independent. Events that are products of events are incompatible. Using formulas (3.10) and (3.15) we get:

Properties of operations of addition and multiplication of events:

3.2.11. Total Probability Formula. Bayes formula

Let an event occur only together with one of pairwise incompatible events (hypotheses),,...,, forming a complete group, i.e.

The probability of an event is found by the formula full probability:

If the event has already happened, then the probabilities of the hypotheses can be overestimated by the formula Bayes:

(3.17)

Example 3.33. There are two identical urns with balls. The first urn contains 5 white and 10 black balls, and the second urn contains 3 white and 7 black balls. One urn is chosen at random and one ball is drawn from it.

Find the probability that this ball is white.

A white ball is drawn from an urn at random. Find the probability that the ball was drawn from the first urn.

\(\blacktriangleright\) If the execution of the event \(C\) requires the execution of both simultaneous (which can occur simultaneously) events \(A\) and \(B\) (\(C=\(A\) and \( B\)\) ), then the probability of the event \(C\) is equal to the product of the probabilities of the events \(A\) and \(B\) .

Note that if the events are incompatible, then the probability of their simultaneous occurrence is \(0\) .

\(\blacktriangleright\) Each event can be represented as a circle. Then if the events are joint, then the circles must intersect. The probability of the event \(C\) is the probability of getting into both circles at the same time.

\(\blacktriangleright\) For example, when tossing a dice, find the probability \(C=\) (rolling the number \(6\) ).
The event \(C\) can be formulated as \(A=\) (an even number) and \(B=\) (a number divisible by three).
Then \(P\,(C)=P\,(A)\cdot P\,(B)=\dfrac12\cdot \dfrac13=\dfrac16\).

Task 1 #3092

Task level: Equal to the Unified State Examination

The store sells sneakers from two brands: Dike and Ananas. The probability that a randomly selected pair of sneakers is Dike is \(0.6\) . Each company can make a mistake in writing its name on sneakers. The probability that Dike spells the name wrong is \(0.05\) ; The probability that Ananas spells the name wrong is \(0.025\) . Find the probability that a randomly purchased pair of sneakers will have the correct spelling of the company name.

Event A: “the pair of sneakers will be with the correct name” is equal to the sum of the events B: “the pair of sneakers will be from Dike and with the correct name” and C: “the pair of sneakers will be from Ananas and with the correct name.”
The probability of event B is equal to the product of the probabilities of the events “sneakers will be made by Dike” and “the name of the company Dike spelled correctly”: \ Similarly for event C: \ Consequently, \

Answer: 0.96

Task 2 #166

Task level: Equal to the Unified State Examination

If Timur plays with white checkers, then he beats Vanya with a probability of 0.72. If Timur plays with black checkers, then he beats Vanya with a probability of 0.63. Timur and Vanya play two games, and in the second game they change the color of the checkers. Find the probability that Vanya wins both times.

Vanya wins with white with probability \(0.37\) , and with black with probability \(0.28\) . The events “from two games Vanya won with white”\(\ \) and “from two games Vanya won with black”\(\ \) are independent, then the probability of their simultaneous occurrence is equal to \

Answer: 0.1036

Task 3 #172

Task level: Equal to the Unified State Examination

The entrance to the museum is guarded by two guards. The probability that the oldest of them will forget the walkie-talkie is \(0,2\) , and the probability that the youngest of them will forget the walkie-talkie is \(0,1\) . What is the probability that they will not have any radios?

Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. Then the desired probability is equal to \

Answer: 0.02

Task 4 #167

Task level: Equal to the Unified State Examination

Jumping from a height of 1 meter, Kostya breaks his leg with a probability of \(0.05\) . Jumping from a height of 1 meter, Vanya breaks his leg with a probability \(0.01\) . Jumping from a height of 1 meter, Anton breaks his leg with a probability of \(0.01\) . Kostya, Vanya and Anton simultaneously jump from a height of 1 meter. What is the probability that only Kostya will break his leg? Round your answer to thousandths.

Events “when jumping from a height of 1 meter, Kostya broke his leg”\(,\ \) “when jumping from a height of 1 meter, Vanya did not break his leg”\(\ \) and “when jumping from a height of 1 meter, Anton did not break his leg”\( \ \) are independent, therefore, the probability of their simultaneous occurrence is equal to the product of their probabilities: \ After rounding, we finally get \(0,049\) .

Answer: 0.049

Task 5 #170

Task level: Equal to the Unified State Examination

Maxim and Vanya decided to go bowling. Maxim rightly estimated that on average he hits a strike once every eight throws. Vanya rightly estimated that on average he knocks out a strike once every five throws. Maxim and Vanya make exactly one throw each (regardless of the result). What is the probability that there will be no strikes among them?

Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. In this case, the probability that Maxim will not hit a strike is equal to \ The probability that Vanya won't hit a strike is \(1 - 0.2 = 0.8\) . Then the desired probability is equal to \[\dfrac(7)(8)\cdot 0.8 = 0.7.\]

Answer: 0.7

Task 6 #1646

Task level: Equal to the Unified State Examination

Anton and Kostya are playing table tennis. The probability that Kostya will hit the table with his signature blow is \(0.9\) . The probability that Anton will win the rally in which Kostya tried to deliver a signature blow is \(0,3\) . Kostya tried to hit the table with his signature blow. What is the probability that Kostya really hits with his signature blow and eventually wins this draw?

Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. At the same time, the probability that Anton will not win the rally in which Kostya tried to deliver his signature blow is \(1 - 0.3 = 0.7\) . Then the desired probability is equal to \

Let be BUT And IN are the two events considered in this test. In this case, the occurrence of one of the events may affect the possibility of the occurrence of another. For example, the occurrence of an event BUT can influence the event IN or vice versa. To take into account such dependence of some events on others, the concept of conditional probability is introduced.

Definition. If the probability of an event IN is located under the condition that the event BUT happened, then the resulting probability of the event IN called conditional probability developments IN. The following symbols are used to denote such a conditional probability: R BUT ( IN) or R(IN / BUT).

Remark 2. In contrast to the conditional probability, the “unconditional” probability is also considered, when any conditions for the occurrence of some event IN missing.

Example. An urn contains 5 balls, 3 of which are red and 2 are blue. In turn, one ball is drawn from it with a return and without a return. Find the conditional probability of drawing a red ball for the second time, provided that the first time taken is: a) a red ball; b) a blue ball.

Let the event BUT is drawing the red ball for the first time, and the event IN– extracting the red ball for the second time. It's obvious that R(BUT) = 3 / 5; then in the case when the ball taken out for the first time is returned to the urn, R(IN)=3/5. In the case when the drawn ball is not returned, the probability of drawing a red ball R(IN) depends on which ball was drawn for the first time - red (event BUT) or blue (event). Then in the first case R BUT ( IN) = 2 / 4, and in the second ( IN) = 3 / 4.

The theorem of multiplication of the probabilities of events, one of which takes place under the condition of the other

The probability of the product of two events is equal to the product of the probability of one of them by the conditional probability of the other, found under the assumption that the first event occurred:

R(A ∙ B) = R(BUT) ∙ R BUT ( IN) . (1.7)

Proof. Indeed, let n- the total number of equally probable and incompatible (elementary) outcomes of the test. Let it go n 1 - the number of outcomes that favor the event BUT, which occurs at the beginning, and m- the number of outcomes in which the event occurs IN assuming that the event BUT has come. In this way, m is the number of outcomes that favor the event IN. Then we get:

Those. the probability of the product of several events is equal to the product of the probability of one of these events by the conditional probabilities of the others, and the conditional probability of each subsequent event is calculated on the assumption that all previous events have occurred.

Example. There are 4 masters of sports in a team of 10 athletes. By drawing lots, 3 athletes are selected from the team. What is the probability that all the selected athletes are masters of sports?

Solution. Let us reduce the problem to the “urn” model, i.e. Let's assume that there are 4 red balls and 6 white ones in an urn containing 10 balls. 3 balls are drawn at random from this urn (selection S= 3). Let the event BUT consists in extracting 3 balls. The problem can be solved in two ways: by the classical scheme and by formula (1.9).

The first method based on the combinatorics formula:

The second method (by formula (1.9)). 3 balls are drawn consecutively from the urn without replacement. Let be BUT 1 - the first drawn ball is red, BUT 2 - the second drawn ball is red, BUT 3 - the third drawn ball is red. Let also the event BUT means that all 3 drawn balls are red. Then: BUT = BUT 1 ∙ (BUT 2 / BUT 1) ∙ BUT 3 / (BUT 1 ∙ BUT 2), i.e.

Example. Let from the set of cards a, a, r, b, o, t cards are drawn one at a time. What is the probability of getting the word " Job” when sequentially folding them into one line from left to right?

Let be IN- the event at which the declared word is obtained. Then by formula (1.9) we get:

R(IN) = 1/6 ∙ 2/5 ∙ 1/4 ∙ 1/3 ∙ 1/2 ∙ 1/1 = 1/360.

The probability multiplication theorem takes on its simplest form when the product is formed by events independent of each other.

Definition. Event IN called independent from the event BUT if its probability does not change regardless of whether the event occurred BUT or not. Two events are called independent (dependent) if the occurrence of one of them does not change (changes) the probability of occurrence of the other. Thus, for independent events p(B/A) = R(IN) or = R(IN), and for dependent events R(IN/A)

Let's start with the task.

Let's say that you have a 0.5 chance of getting an A on a test and a 0.3 chance of getting a B. What is the probability that you will get a 4 or 5 on a test?

Some will immediately blurt out: “0.8”, but why exactly? Why, for example, not 0.15 (multiplyed, not added)? Let's figure it out.

Suppose there is some experience that has outcomes. Of these, the onset of the event is favorable, and the event is favorable. It is not difficult to find the probabilities of the occurrence of each of the events using the formula - these are, respectively, and . But what is the probability that either the first event or the second will occur? In other words, we are looking for the probability of combining these events. To do this, we need to find out how many favorable outcomes we have. ? Not really. After all, it may happen that these events are executed simultaneously.

Then suppose that the events are non-overlapping, that is, they cannot be executed simultaneously. Then we get that favorable outcomes for the association -. So the probability of joining will be:

The probability of combining incompatible events is equal to the sum of their probabilities.

Let's pay attention: here we are talking about ONE experiment, as a result of which either the first event or the second can occur, but not both at once.

In particular, in the example with the control, we understand that the student cannot get both 5 and 4 on the control at the same time (we are talking about the same grade for the same control), which means that the probability that he will get 4 or 5 is equal to the sum of the probabilities, that is, after all, 0.8.

Answer: 0,8.

But what if the events intersect, that is, there are outcomes that are favorable for both of them? This situation will be discussed at the end of the lesson.

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Homework

1. Two shooters shoot at a target. The first shooter hits the target with a probability of 0.9. The second shooter hits the target with a probability of 0.8. Find the probability that the target will be hit.

2. A random experiment consists of tossing two dice. One of the dice is colored blue, the other is red. Find the probability that a 3 is rolled on a blue dice and a 4 is rolled on a red dice.

The product, or intersection, of events A and B is an event consisting in the simultaneous occurrence of events and A, and IN. Designation of the work AB or L and V.

For example, hitting the target twice is the product of two events, the answer to both questions of the ticket in the exam is the product of two events.

Events L and IN are called inconsistent if their product is an impossible event, i.e. LV = V.

For example, events L - the loss of the coat of arms and IN- loss of numbers during a single toss of a coin cannot occur simultaneously, their product is an impossible event, events L and B are incompatible.

The concepts of the sum and product of events have a clear geometric interpretation (Fig. 6.4).

Rice. 6.4. Geometric interpretation of the work (but) and amounts (b) two joint events

Let the event A be the set of points in the area L, and the event B be the set of points in the area B. The shaded area corresponds to the event LP in fig. 6 La and event A + B in Fig. 6.46.

For incompatible events A and B we have LP = V(Fig. 6.5a). The L + B event corresponds to the shaded area in Fig. 6.56.

Rice. 6.5. The geometric interpretation of the product ( but) and sums (b) two incompatible events

Developments BUT And BUT are called opposite if they are incompatible and in total constitute a reliable event, i.e.

A A = V; A+A=U.

For example, let's fire one shot at a target: event BUT- the shooter hit the target BUT- missed; coin tossed:

event BUT- eagle fall, BUT- loss of numbers; schoolchildren write a test: an event BUT- none

errors in control work, BUT- there are errors in the control work; student came to take the test: event BUT- passed

offset, BUT- did not submit a report.

There are boys and girls in the class, excellent students, good students and three students studying English and German. Let the event M be a boy, O an excellent student, A an English learner. Can a student who accidentally left the class be both a boy, and an excellent student, and an English learner? This will be the product or intersection of MOA events.

Example 6.15. Throw a dice - a cube made of a homogeneous material, the faces of which are numbered. Observe the number (number of points) falling on the top face. Let the event BUT - appearance of an odd number, event IN - appearance of a multiple of three. Find the outcomes that make up each of the events (?/, A, A + V U AB) and indicate their meaning.

Solution. Outcome - the appearance on the upper face of any of the numbers 1, 2, 3, 4, 5, 6. The set of all outcomes is the space of elementary events U= (1, 2, 3, 4, 5, 6). It is clear that the event A =(1, 3, 5), event B = {3, 6}.

Event BUT + B =(1, 3, 5, 6) - the appearance of either an odd number or a number that is a multiple of three. When listing the outcomes, it is taken into account that each outcome in the set can be contained only once.

Event AB =(3) - the appearance of both an odd number and a multiple of three.

Example 6.16. The homework of three students was checked. Let the event BUT ( - performance of the task by the i-th student, G = 1, 2, 3.

What is the meaning of the events: A = A t + A 2+ L 3, BUT And B \u003d A t A 2 A 3?

Solution. Event BUT = A x + A 2 + A 3 - completion of the task by at least one student, i.e. or any one student (or first, or second, or third), or any two, or all three.

Event A \u003d A x -A 2 -A 3- the task was not completed by any student - neither the first, nor the second, nor the third. Event B \u003d A ( A 2 A 3 - performance of the task by three students - and the first, and the second, and the third.

When considering the joint occurrence of several events, there are cases when the occurrence of one of them affects the possibility of the occurrence of another. For example, if the day is sunny in autumn, then the weather is less likely to deteriorate (it starts to rain). If the sun is not visible, then it is more likely that it will rain.

Event L called event independent IN, if the probability of an event BUT does not change depending on whether the event occurred or not IN. Otherwise the event BUT is called event dependent IN. Two events A andIN are called independent if the probability of one of them does not depend on the occurrence or non-occurrence of the other, dependent - otherwise. Events are called pairwise independent if every two of them are independent of each other.

The probability multiplication theorem is formulated as follows. The probability of the product of two independent events is equal to the product of the probabilities of these events:

This theorem is valid for any finite number of events, as long as they are collectively independent, i.e. the probability of any of them does not depend on whether the other of these events occurred or not.

Example 6.17. The student takes three exams. The probability of passing the first exam is 0.9, the second - 0.65, the third - 0.35. Find the probability that he fails at least one exam.

Solution. Denote BUT event - the student did not pass at least one exam. Then P(A) = 1 - /-’(1/1), where BUT- the opposite event - the student passed all the exams. Since the passing of each exam does not depend on other exams, then P(A)= 1 - P(1/1) = = 1 - 0.9 0.65 0.35 = 0.7953.

Event Probability BUT, computed assuming an event occurs IN, called conditional probability developments BUT subject to the appearance IN and denoted R B (A) or P(A/B).

Theorem.The probability of occurrence of the product of two events is equal to the product of the probability of one of them by the conditional probability of the second, calculated under the condition that the first event occurred:

Example 6.18. The student draws one ticket out of 34 twice. What is the probability that he will pass the exam if he has prepared 30 tickets and the first time he takes out an unsuccessful ticket?

Solution. Let the event BUT consists in the fact that for the first time you got an unsuccessful ticket, an event IN- the second time a successful ticket is drawn. Then BUT?IN- the student will pass the exam (under the specified circumstances). Developments BUT And IN are dependent, since the probability of choosing a successful ticket on the second attempt depends on the outcome of the first choice. Therefore, we use formula (6.6):

Note that the probability obtained in the solution is “0.107. Why is the probability of passing the exam so small if 30 tickets out of 34 are learned and two attempts are given?!

Extended addition theorem is formulated as follows. The probability of the sum of two events is equal to the sum of the probabilities of these events without the probability of their joint occurrence (works):

Example 6.19. Two students solve a problem. The probability that the first student solves the problem (event BUT), equal to 0.9; the probability that the second student solves the problem (event IN), equals 0.8. What is the probability that the problem will be solved?

Solution. We are interested in the event C, which consists in the fact that the problem will be solved, i.e. the first, or the second student, or two students at the same time. Thus, the event of interest to pass C \u003d A +IN. Developments BUT And IN are joint, then the probability addition theorem is applicable for the case of joint events: P(A + IN) = P(A) + P(B) - P(AB). For our case P(A + B) = = 0.9 + 0.8 + 0.9 0.8 = 0.98 (events BUT And IN joint but independent).

Example 6.20. The student knows 20 questions out of 25. What is the probability of answering 3 questions out of 25?

Solution. Let's introduce the event A, - the student knows the answer to i-th proposed question, i= 1,2,3. Events L, L 2 , L 3 - dependent. That's why

When finding the probabilities of events, the classical definition of probability was used.