The theorem of addition of probabilities and the theorem of multiplication of probabilities. Addition and multiplication of probabilities: examples of solutions and theory The probability of a product is calculated by the formula

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)

Definition. Product or intersection events A and B call an event consisting in the simultaneous occurrence of events both A and B. The designation of the work: AB or A B.

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 A and B are called incompatible if their product is an impossible event, i.e. AB = V.

Events A - the loss of the coat of arms and B - the loss of a number during a single toss of a coin cannot occur simultaneously, their product is an impossible event, events A and B are incompatible.

The concepts of sum and product of events have a clear geometric interpretation.

Rice. 6.4. Geometric interpretation of the product (a) and sum (b) of two joint events

Let event A be a set of points in area A; event B is a set of points in area B. The shaded area corresponds to the event AB in Fig. 6.4, a; event in Fig. 6.4, b.

For incompatible events A and B we have: AB = V (Fig. 6.5, a). Event A + B corresponds to the shaded area in Fig. 6.5, b.

Rice. 6.5. Geometric interpretation of the product (a) and sum (b) of two incompatible events

Events and call opposite, if they are incompatible and in total constitute a reliable event, i.e.

For example, let's make one shot at the target: event - the shooter hit the target, did not hit; a coin is tossed: the event is heads, − numbers are rolled; schoolchildren write a test: event - not a single mistake in the test, - there are errors in the test; the student came to take the test: event A - passed the test, - did not pass the test.

There are boys and girls in the class, excellent students, good students and three students studying English and German. Let event M be a boy, O be an excellent student, and A be 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. A dice is thrown - 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 event A be the appearance of an odd number, and event B be the appearance of a number that is a multiple of three. Find the outcomes that make up each of the events: U, A, A + B, 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 makes up the space of elementary events. It is clear that the event , event

An event is the occurrence of either an odd number or a multiple of three. When listing the outcomes, it is taken into account that each outcome in the set can be contained only once.

An event is the occurrence of both an odd number and a multiple of three.

Example. The homework of three students was checked. Let the event be the completion of the task by the th student, What is the meaning of the events: and ?

Solution. An event is the completion of a 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 - the task was not completed by any student: neither the first, nor the second, nor the third. An event is the completion of a task by three students: the first, the second, and the third.

When considering the joint occurrence of several events, there may be 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.

Definition. Event A is called independent from event B, if the probability of event A does not change depending on whether event B occurred or not. Otherwise, event A is called dependent on event B. Two events A and B 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.

Theorem. (Probability multiplications) The probability of the product of two independent events is equal to the product of the probabilities of these events:

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

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

Solution: Denote A - the event student did not pass at least one exam. Then P(A) = 1- P(ùA), where ùA is the opposite event the student has passed all exams. Since the passing of each exam does not depend on other exams, then Р(А)=1-Р(ùА)= 1-0.9*0.65*0.35=0.7953.

Definition. The probability of event A, calculated assuming that event B occurs, is called conditional probability event A, subject to the occurrence of B and is denoted by P 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:

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

Example. 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 A is that the first time you got a bad ticket, event B - the second time you get a good ticket. Then A·B - the student will pass the exam (under the specified circumstances). Events A and B are dependent, because the probability of choosing a successful ticket on the second attempt depends on the outcome of the first choice. Therefore, we use formula (6):

P (A B) \u003d P (A) RA (B) \u003d (4/34) * (30/33) \u003d 20/187

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?!

Theorem. (Extended addition theorem) 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 (product):

P(A+B)=P(A)+P(B)-P(A B).

Example. Two students solve a problem. The probability that the first student will solve the problem (event A) is 0.9; the probability that the second student will solve the problem (event B) is 0.8. What is the probability that the problem will be solved?

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

Addition theorem

Consider incompatible random events.

It is known that incompatible random events $A$ and $B$ in the same trial have probabilities $P\left(A\right)$ and $P\left(B\right)$ respectively. Let us find the probability of the sum $A+B$ of these events, i.e. the probability of occurrence of at least one of them.

Suppose that in this test the number of all equally possible elementary events is $n$. Of these, events $A$ and $B$ are favored by $m_(A)$ and $m_(B)$ elementary events, respectively. Since the events $A$ and $B$ are incompatible, the event $A+B$ is favored by $m_(A) +m_(B)$ elementary events. We have $P\left(A+B\right)=\frac(m_(A) +m_(B) )(n) =\frac(m_(A) )(n) +\frac(m_(B) ) (n) =P\left(A\right)+P\left(B\right)$.

Theorem 1

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

Note 1

Consequence 1. The probability of the sum of any number of incompatible events is equal to the sum of the probabilities of these events.

Consequence 2. The sum of the probabilities of a complete group of incompatible events (the sum of the probabilities of all elementary events) is equal to one.

Consequence 3. The sum of the probabilities of opposite events is equal to one, since they form a complete group of incompatible events.

Example 1

The probability that it will never rain in the city for some time is $p=0.7$. Find the probability $q$ that during the same time it will rain in the city at least once.

The events "for some time it never rained in the city" and "for some time it rained in the city at least once" are opposite. Therefore $p+q=1$, whence $q=1-p=1-0.7=0.3$.

Consider joint random events.

It is known that joint random events $A$ and $B$ in the same trial have probabilities $P\left(A\right)$ and $P\left(B\right)$ respectively. Let us find the probability of the sum $A+B$ of these events, i.e. the probability of occurrence of at least one of them.

Suppose that in this test the number of all equally possible elementary events is $n$. Of these, events $A$ and $B$ are favored by $m_(A)$ and $m_(B)$ elementary events, respectively. Since the events $A$ and $B$ are joint, then out of the total number of $m_(A) +m_(B)$ elementary events, a certain number $m_(AB)$ favors both the event $A$ and the event $B$, that is, their joint occurrence (the product of events $A\cdot B$). This quantity $m_(AB)$ entered both $m_(A)$ and $m_(B)$. So event $A+B$ is favored by $m_(A) +m_(B) -m_(AB) $ elementary events. We have: $P\left(A+B\right)=\frac(m_(A) +m_(B) -m_(AB) )(n) =\frac(m_(A) )(n) +\frac (m_(B) )(n) -\frac(m_(AB) )(n) =P\left(A\right)+P\left(B\right)-P\left(A\cdot B\right )$.

Theorem 2

The probability of the sum of two joint events is equal to the sum of the probabilities of these events minus the probability of their product.

Comment. If the events $A$ and $B$ are incompatible, then their product $A\cdot B$ is an impossible event whose probability is $P\left(A\cdot B\right)=0$. Therefore, the formula for adding the probabilities of incompatible events is a special case of the formula for adding the probabilities of joint events.

Example 2

Find the probability that when two dice are thrown at the same time, the number 5 will come up at least once.

When throwing two dice at the same time, the number of all equally possible elementary events is equal to $n=36$, since six digits of the second die can fall on each digit of the first dice. Of these, the event $A$ - the number 5 rolled on the first die - occurs 6 times, the event $B$ - the number 5 rolled on the second die - also occurs 6 times. Of all twelve times, the number 5 appears once on both dice. So $P\left(A+B\right)=\frac(6)(36) +\frac(6)(36) -\frac(1)(36) =\frac(11)(36) $.

Probability multiplication theorem

Consider independent events.

Events $A$ and $B$ that occur in two successive trials are called independent if the probability of occurrence of event $B$ does not depend on whether event $A$ took place or did not take place.

For example, suppose there are 2 white and 2 black balls in an urn. The test is to extract the ball. The event $A$ is "a white ball is drawn in the first trial". Probability $P\left(A\right)=\frac(1)(2) $. After the first test, the ball was put back and a second test was carried out. Event $B$ -- ``white ball drawn in second trial''. Probability $P\left(B\right)=\frac(1)(2) $. The probability $P\left(B\right)$ does not depend on whether the event $A$ took place or not, hence the events $A$ and $B$ are independent.

It is known that independent random events $A$ and $B$ of two consecutive trials have probabilities $P\left(A\right)$ and $P\left(B\right)$ respectively. Let us find the probability of the product $A\cdot B$ of these events, that is, the probability of their joint occurrence.

Suppose that in the first trial the number of all equally possible elementary events is $n_(1) $. Of these, $A$ is favored by $m_(1)$ elementary events. Let us also assume that in the second test the number of all equally possible elementary events is $n_(2) $. Of these, event $B$ is favored by $m_(2)$ elementary events. Now consider a new elementary event, which consists in the successive occurrence of events from the first and second trials. The total number of such equally probable elementary events is equal to $n_(1) \cdot n_(2) $. Since the events $A$ and $B$ are independent, then from this number the joint occurrence of the event $A$ and the event $B$ (the product of the events $A\cdot B$) is favored by $m_(1) \cdot m_(2) $ events . We have: $P\left(A\cdot B\right)=\frac(m_(1) \cdot m_(2) )(n_(1) \cdot n_(2) ) =\frac(m_(1) ) (n_(1) ) \cdot \frac(m_(2) )(n_(2) ) =P\left(A\right)\cdot P\left(B\right)$.

Theorem 3

The probability of the product of two independent events is equal to the product of the probabilities of these events.

Consider dependent events.

In two consecutive trials, events $A$ and $B$ occur. An event $B$ is said to be dependent on the event $A$ if the probability of occurrence of the event $B$ depends on whether the event $A$ took place or not. Then the probability of the event $B$, which was calculated under the condition that the event $A$ took place, is called the conditional probability of the event $B$ under the condition $A$ and is denoted by $P\left(B/A\right)$.

For example, suppose there are 2 white and 2 black balls in an urn. The test is the extraction of the ball. The event $A$ is "a white ball is drawn in the first trial". Probability $P\left(A\right)=\frac(1)(2) $. After the first test, the ball is not put back and the second test is performed. Event $B$ -- ``white ball drawn in second trial''. If a white ball was drawn in the first trial, then the probability is $P\left(B/A\right)=\frac(1)(3) $. If a black ball was drawn in the first trial, then the probability is $P\left(B/\overline(A)\right)=\frac(2)(3) $. Thus, the probability of the event $B$ depends on whether the event $A$ took place or not, therefore, the event $B$ depends on the event $A$.

Assume that events $A$ and $B$ occur in two consecutive trials. It is known that the event $A$ has the probability of occurrence $P\left(A\right)$. It is also known that the event $B$ is dependent on the event $A$ and its conditional probability under condition $A$ is equal to $P\left(B/A\right)$.

Theorem 4

The probability of the product of the event $A$ and the event $B$ dependent on it, that is, the probability of their joint occurrence, can be found by the formula $P\left(A\cdot B\right)=P\left(A\right)\cdot P\left(B/A\right)$.

The symmetric formula $P\left(A\cdot B\right)=P\left(B\right)\cdot P\left(A/B\right)$ is also valid, where the event $A$ is assumed to be dependent on the event $ B$.

For the conditions of the last example, we find the probability that the white ball will be drawn in both trials. Such an event is a product of the events $A$ and $B$. Its probability is $P\left(A\cdot B\right)=P\left(A\right)\cdot P\left(B/A\right)=\frac(1)(2) \cdot \frac( 1)(3) =\frac(1)(6) $.

The sum of all event probabilities in the sample space is 1. For example, if the experiment is a coin toss with Event A = "heads" and Event B = "tails", then A and B represent the entire sample space. Means, P(A) + P(B) = 0.5 + 0.5 = 1.

Example. In the previously proposed example of calculating the probability of extracting a red pen from the pocket of a bathrobe (this is event A), in which there are two blue and one red pen, P(A) = 1/3 ≈ 0.33, the probability of the opposite event - extracting a blue pen - will be

Before moving on to the main theorems, we introduce two more more complex concepts - the sum and the product of events. These concepts are different from the usual concepts of sum and product in arithmetic. Addition and multiplication in probability theory are symbolic operations subject to certain rules and facilitating the logical construction of scientific conclusions.

sum of several events is an event consisting in the occurrence of at least one of them. That is, the sum of two events A and B is called event C, which consists in the appearance of either event A, or event B, or events A and B together.

For example, if a passenger is waiting at a tram stop for one of the two routes, then the event he needs is the appearance of a tram of the first route (event A), or a tram of the second route (event B), or a joint appearance of trams of the first and second routes (event FROM). In the language of probability theory, this means that the event D necessary for the passenger consists in the appearance of either event A, or event B, or event C, which is symbolically written as:

D=A+B+C

The product of two eventsBUT And IN is an event consisting in the joint occurrence of events BUT And IN. The product of several events the joint occurrence of all these events is called.

In the passenger example above, the event FROM(joint appearance of trams of two routes) is a product of two events BUT And IN, which is symbolically written as follows:

Assume that two physicians are separately examining a patient in order to identify a specific disease. During inspections, the following events may occur:

Detection of diseases by the first physician ( BUT);

Failure to detect the disease by the first doctor ();

Detection of the disease by the second doctor ( IN);

Non-detection of the disease by the second doctor ().

Consider the event that the disease is detected exactly once during the examinations. This event can be implemented in two ways:

The disease is detected by the first doctor ( BUT) and will not find the second ();

Diseases will not be detected by the first doctor () and will be detected by the second ( B).

Let us denote the event under consideration by and write it symbolically:

Consider the event that the disease is discovered in the process of examinations twice (both by the first and the second doctor). Let's denote this event by and write: .

The event, which consists in the fact that neither the first nor the second doctor detects the disease, will be denoted by and we will write: .

Basic theorems of probability theory

The probability of the sum of two incompatible events is equal to the sum of the probabilities of these events.

Let's write the addition theorem symbolically:

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

where R- the probability of the corresponding event (the event is indicated in brackets).

Example . The patient has stomach bleeding. This symptom is recorded in ulcerative vessel erosion (event A), rupture of varicose veins of the esophagus (event B), stomach cancer (event C), gastric polyp (event D), hemorrhagic diathesis (event F), obstructive jaundice (event E) and end gastritis (eventG).

The doctor, based on the analysis of statistical data, assigns a probability value to each event:

In total, the doctor had 80 patients with gastric bleeding (n= 80), of which 12 had ulcerative vessel erosion (), at6 - rupture of varicose veins of the esophagus (), 36 had stomach cancer () etc.

To prescribe an examination, the doctor wants to determine the likelihood that stomach bleeding is associated with stomach disease (event I):

The likelihood that gastric bleeding is associated with stomach disease is quite high, and the doctor can determine the tactics of examination based on the assumption of stomach disease, justified at a quantitative level using probability theory.

If joint events are considered, 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.

Symbolically, this is written as follows:

If we imagine that the event BUT consists in hitting a target shaded with horizontal stripes while shooting, and the event IN- in hitting a target shaded with vertical stripes, then in the case of incompatible events, according to the addition theorem, the probability of the sum is equal to the sum of the probabilities of individual events. If these events are joint, then there is some probability corresponding to the joint occurrence of events BUT And IN. If you do not introduce a correction for the deductible P(AB), i.e. on the probability of the joint occurrence of events, then this probability will be taken into account twice, since the area shaded by both horizontal and vertical lines is an integral part of both targets and will be taken into account both in the first and in the second summand.

On fig. 1 a geometric interpretation is given that clearly illustrates this circumstance. In the upper part of the figure there are non-intersecting targets, which are an analogue of incompatible events, in the lower part - intersecting targets, which are an analogue of joint events (one shot can hit both target A and target B at once).

Before moving on to the multiplication theorem, it is necessary to consider the concepts of independent and dependent events and conditional and unconditional probabilities.

Independent an event B is an event A whose probability of occurrence does not depend on the occurrence or non-occurrence of event B.

addicted An event B is an event A whose probability of occurrence depends on the occurrence or non-occurrence of event B.

Example . An urn contains 3 balls, 2 white and 1 black. When choosing a ball at random, the probability of choosing a white ball (event A) is: P(A) = 2/3, and black (event B) P(B) = 1/3. We are dealing with a scheme of cases, and the probabilities of events are calculated strictly according to the formula. When the experiment is repeated, the probabilities of occurrence of events A and B remain unchanged if after each choice the ball is returned to the urn. In this case, events A and B are independent. If the ball chosen in the first experiment is not returned to the urn, then the probability of the event (A) in the second experiment depends on the occurrence or non-occurrence of the event (B) in the first experiment. So, if event B appeared in the first experiment (a black ball was chosen), then the second experiment is carried out if there are 2 white balls in the urn and the probability of the occurrence of event A in the second experiment is: P(A) = 2/2= 1.

If the event B did not appear in the first experiment (a white ball is chosen), then the second experiment is carried out if there are one white and one black balls in the urn and the probability of the occurrence of event A in the second experiment is equal to: P(A) = 1/2. Obviously, in this case, events A and B are closely related and the probabilities of their occurrence are dependent.

Conditional Probability event A is the probability of its occurrence, provided that event B has appeared. The conditional probability is symbolically denoted P(A/B).

If the probability of an event occurring BUT does not depend on the occurrence of the event IN, then the conditional probability of the event BUT is equal to the unconditional probability:

If the probability of occurrence of event A depends on the occurrence of event B, then the conditional probability can never be equal to the unconditional probability:

Revealing the dependence of various events among themselves is of great importance in solving practical problems. So, for example, an erroneous assumption about the independence of the appearance of certain symptoms in the diagnosis of heart defects using a probabilistic method developed at the Institute of Cardiovascular Surgery. A. N. Bakuleva, caused about 50% of erroneous diagnoses.

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 may be 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.