Pulling out white balls from black ones. Probability addition theorem: the probability of the occurrence of one (no matter which) event from several incompatible events is equal to the sum of their probabilities

Lecture 1 .

Goals, objectives and structure of medical and biological physics. Its place and role in the medical education system, interdisciplinary connections with other biomedical and clinical disciplines.

Probabilistic nature of medical and biological processes. Elements of probability theory. The probability of a random event. The law of addition and multiplication of probabilities.

Principles of probabilistic approaches to problems of diagnosis and prognosis of diseases.

Probability theory

In probability theory, patterns related to random events, quantities, and processes are studied. Doctors rarely think that making a diagnosis is probabilistic in nature and, as wittily noted, only a pathological examination can reliably determine the diagnosis of a deceased person.

§2.1. Random event. Probability

Observing various phenomena, one can notice that there are two types of connections between conditions S and the occurrence or non-occurrence of some event A. In some cases, the implementation of a set of conditions S (test) certainly causes an event A. So, for example, a material point with mass T 0 under the influence of force F (condition S) gains acceleration A= F/ m 0 (event A). In other cases, repeated repetition of the test may or may not lead to the occurrence of event A. Such events are usually called random: These include the appearance of a patient with a given disease in the doctor’s office, the loss of a certain side of a coin when it is thrown, etc.

One should not think of random phenomena as causeless, unconditioned. It is known that many phenomena are interconnected, a separate phenomenon is a consequence of another and itself serves as the cause of the subsequent one. However, it is often difficult or even impossible to quantitatively trace this connection between conditions and events. Thus, when throwing a die (a uniform cube with six numbered sides: 1, 2, 3, 4, 5 and 6), the final position of the cube depends on the movement of the hand at the moment of throwing, air resistance, the position of the cube when it hits the surface, the features of the surface, which the cube fell, and other factors that cannot be taken into account separately.

In everyday life in relation to such random events use the words “possibly”, “probably”, “unlikely”, “incredible”. In some cases, such an assessment characterizes the desire of the speaker more than the true degree of possibility or impossibility of the event. However, even random events, if their number is large enough, are subject to certain patterns. The quantitative assessment of patterns related to random events is given in a branch of mathematics called probability theory.

Probability theory studies the patterns inherent in mass (statistical) random events.

Individual historical facts, “surprises,” “catastrophes” are isolated, seemingly unique, events, and it is impossible to make quantitative probabilistic judgments regarding them. Historically, the theory of probability appeared in connection with attempts to calculate the possibility of various outcomes in gambling. Currently, it is used in science, including biology and medicine, to assess the probability of practically important events. All that remains from the games are illustrative examples that can be conveniently used to illustrate theoretical points.

Statistical definition of probability. Probability P(A)V probability theory acts as a numerical characteristic of the degree of possibility of the occurrence of any specific random event A when tests are repeated many times.

Let's say that in 1000 throws of a die, the number 4 appears 160 times. The ratio 160/1000 = 0.16 shows the relative frequency of the number 4 in a given series of tests. More generally, when a random event A occurs T once in a series P independent tests, relative frequency with being in a given series of tests or simply the frequency of event A is the ratio

With a large number of trials, the event frequency is approximately constant: increasing the number of trials reduces the fluctuation of the event frequency around a constant value.

The probability of a random event is the limit to which the frequency of an event tends with an unlimited increase in the number of trials:

(2.2)

Naturally, no one will ever be able to perform an unlimited number of tests in order to determine the probability. There is no need for this. Practically for the probability [see. (2.2)] we can accept the relative frequency of the event over a large number of trials. For example, from the statistical patterns of birth established over many years of observations, the probability that the newborn will be a boy is estimated at 0.515.

Classic definition of probability. If during testing there are no reasons due to which one random event would appear more often than others (equally possible dogs tiya), the probability can be determined based on theoretical considerations. For example, let’s find out in the case of tossing a coin the frequency of the coat of arms falling out (event A). Various experimenters, over several thousand tests, have shown that the relative frequency of such an event takes values close to 0.5. Considering that the appearance of the coat of arms and the opposite side of the coin (event IN) are equally possible events if the coin is symmetrical, the proposition P(A)= P(B)= 0.5 could be done without determining the frequency of these events. Based on the concept of “equal possibility” of events, another definition of probability is formulated.

Let us assume that as a result of the test only one of the following should occur: P equally possible incompatible events (incompatible events are called if their simultaneous occurrence is impossible). Let the event in question A happens in T cases, which are called favorable A, and does not occur in the rest p - t, unfavorable A. Then probability can be called a favorable attitude existing cases to the total number of equally possible cases local events:

P(A) =m/ n . (2.3)

This classical definition of probability.

Let's look at a few examples.

1. An urn contains 40 balls: 10 black and 30 white. Find the probability that one ball drawn at random will be black.

The number of favorable cases is equal to the number of black balls in the urn: t = 10. The total number of equally possible events (taking out one ball) is equal to the total number of balls in the urn: P= 40. These events are inconsistent, since one and only one ball is drawn. According to formula (2.3) we have:

P(A)= 10/40 = 1/4.

2. Find the probability of getting an even number when throwing a dice.

When throwing a dice, six equally possible incompatible events occur: the appearance of one number 1, 2, 3, 4, 5 or 6, i.e. n = 6. Favorable cases are when one of the numbers 2, 4 or 6 appears: t = 3. The required probability:

P(A) =m/ n – 3/6 = 1/2.

As can be seen from the definitions of event probability (2.2) and (2.3), for all events 0  P(A) 1.

Events that cannot occur during these tests are called impossible: their probability is equal to zero.

So, for example, it is impossible to draw a red ball from an urn with white and black balls, and it is impossible to get the number 7 on a dice.

An event that is mandatory for this test happens, is called reliable, its probability is equal to on 1.

An example of a reliable event is the drawing of a white ball from an urn containing only white balls.

In some cases, it is easier to calculate the probability of an event if you imagine it as a combination of simpler events. Some theorems of probability theory serve this purpose.

Probability addition theorem:probability of occurrence one (no matter what) event from several events local events is equal to the sum of their probabilities. For two incompatible events

P(A or B) = P(A) + P(B).(2.4)

Let's prove this theorem. Let P - total number tests, T 1 - the number of cases favorable to event A, T 2 - number of cases favorable to the event IN. Number of cases favoring the occurrence or event A, or events IN, equals m 1 +m 2 . Then P(A or B) = (t 1 + t 2 )/n = t 1 /p + t 2 /P. Hence, taking (2.3) into account, we have

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

* Find the probability of getting a 1 or a 6 when throwing a die.

Events A(roll 1) and IN ( drop 6) are equally possible: P(A) = P(B) = 1/6, therefore from (2.4) we find P(A or B) =1/6 + 1/6 = 1/3.

The addition of probabilities is valid not only for two, butand for any number of incompatible events.

* An urn contains 50 balls: 10 white, 20 black, 5 red and 15 blue. Find the probability of a white, or black, or red ball appearing during a single operation of removing a ball from the urn.

Probability of drawing a white ball (event A) equal to P(A) = 10/50 = 1/5, black ball (event B) - P(B) = 20/50 = 2/5 and red (event C) - P(C) = 5/50 = 1/10. From here, using the formula for adding probabilities, we get P(A or IN or C) = P(A) + P(B) + P(C)= 1/5 + 2/5 + + 1/10= 7/10.

If two events are the only possible and incompatible, then they are called opposite.

Such events are usually designated, for example, A And .

The sum of the probabilities of two opposite events, as follows from the addition of probabilities theorem, is equal to one face:

(2.5)

*Let us illustrate the validity of (2.5) using the previous example. Let drawing a white or black or red ball be an event A 1 , P(A 1 ) = 7/10. The opposite event is to get the blue ball. Since there are 15 blue balls, and the total number of balls is 50, we get R() = 15/50 = 3/10 and P(A 1 ) + P() = 7/10 + 3/10 = = 1.

*The urn contains white, black and red balls. The probability of getting a black or red ball is 0.4. Find the probability of drawing a white ball from the urn.

Let's denote A event of drawing a black or red ball, P(A) = 0.4; opposite event will be the removal of the white ball, then based on (2.5) the probability of this event R() = 1 - P(A) == 1 - 0,4 = 0,6.

Event system (A 1 , A 2 , ... A k ) is called complete if when tested, one and only one of these events will occur. The sum of the probabilities of events forming a complete systemtopic is equal to one.

* There are 40 balls in the urn: 20 white, 15 black and 5 red. Probability of a white ball appearing (event A) is equal to P(A) = 20/40 = 1/2, for a black ball (event B) - P(B) = 15/40 = 3/8 and for the red ball (event C) - P(S)= 5/40 = 1/8. In this case the event system A 1 , A 2 , A 3 is complete; you can be sure that P(A) + P(B) + P(C) = 1/2 + 3/8 + + 1/8 = 1.

Probability multiplication theorem:probability together probability of occurrence of independent events is equal to the product of their probabilities. For two events

P(A And B) = P(A) P(B).(2.6)

Let's prove this theorem. Since events A And IN independent, then each of T 1 cases favorable A, correspond T 2 cases favorable IN. Thus, the total number of cases favoring the co-occurrence of events A And IN, equals T 1 T 2 . Similarly, the total number of equally possible events is P 1 P 2 , Where P 1 And P 2 - the number of equally possible events, respectively, for A And IN. We have

* One urn contains 5 black and 10 white balls, the other contains 3 black and 17 white balls. Find the probability that when balls are first drawn from each urn, both balls will be:

1) black; 2) white; 3) a black ball will be drawn from the first urn, and a white ball from the second; 4) a white ball will be drawn from the first urn, and a black ball from the second.

The probability of drawing a black ball from the first urn (event A)is equal P(A) =

= 5/15 = 1/3, black ball from the second urn (event IN) -P(B)= 3/20, white ball from the first urn (event A")- P(A") = 10/15 = 2/3 and the white ball from the first urn (event IN")-P(B") = 17/20. We find the probability of the joint occurrence of two independent events using formula (2.6):

1)P(A And B) = P(A) P(B) =(1/3) (3/20) = 3/60 - both balls are black;

2) P(A" and B") = P(A") P(B") =(2/3) (17/20) = 17/30 - both balls are white;

3) P(A" and B") = P(A) P(B") =(1/3) (17/20)= 17/60 - a black ball will be drawn from the first urn, and a white ball from the second;

4) P(A" and B) = P(A") P(B) =(2/3) (3/20) = 1/10 - a white ball will be drawn from the first urn, and a black ball from the second.

All four possible cases A And IN, A" And IN", A And IN", A" And IN form a complete system of events, therefore

P(A And B) + P(A" And B") + P(A And B") + P(A" And IN)= 3/60 + 17/30 + 17/60 + 1/10 = 1.

* Find the probability that in a family with three children, all three are sons. Assume that the probability of having a boy is equal to 0,515 and for each subsequent child does not depend on the gender of previous children.

According to the probability multiplication theorem, P(A And IN And WITH)= 0,515 0,515 0.515  0.14.

The probability multiplication theorem becomes more complicated if The probability of an event consisting of the joint occurrence of two mutually dependent events is determined. Incase when event B occurs under the condition that If A took place, the probability of co-occurrence of these two events is equal

P(A And B) = P(A) P(B/A), (2.8)

Where P(V/A)-conditional probability, i.e. the probability of an event IN provided that the event A took place.

* There are 5 balls in the urn: 3 white and 2 black. Find the probability that black and white balls will be drawn one after the other.

The probability that the black ball will be withdrawn first (event A), equal to P(A) = t/p= 2/5. After removing the black ball, there are 4 balls left in the urn: 3 white and 1 black. In this case, the probability of drawing a white ball (event IN after the event is completed A) equal to P(V/A) = 3/4. Using (2.8), we obtain

P(A And B) =(2/5) (3/4) = 3/10.

Individual tasks in mathematics

Problem 1

There are 6 white balls and 11 black balls in the urn. Two balls are drawn at random at the same time. Find the probability that both balls will be:

Solution

1) The probability that one of the drawn balls will be white is equal to the number of chances of drawing a white ball from the total sum of balls in the urn. These chances are exactly the same as the number of white balls in the urn, and the sum of all chances is equal to the sum of the white and black balls.

The probability that the second ball drawn will also be white is equal to

Since one of the white balls has already been drawn.

Thus, the probability that both balls drawn from the urn will be white is equal to the product of these probabilities, since these possibilities are independent:

or two black balls:

3) The probability that both drawn balls will be different colors is the probability that the first ball will be white and the second black or that the first ball will be black and the second ball will be white. It is equal to the sum of the corresponding probabilities.

Answer: 1)

2) 3) .

Problem 2

The first urn has 6 white balls, 11 black, the second urn has 5 white and 2 black. A ball is drawn at random from each urn. Find the probability that both balls will be:

1) white, 2) the same color, 3) different colors.

Solution

1) The probability that both balls will be white is equal to the product of the probability that the ball drawn from the first urn will be white by the probability that the ball drawn from the second urn will also be white:

2) The probability that both balls drawn will be the same color is the probability that both balls will be either white or black. It is equal to the sum of the probabilities - drawing two white balls or two black balls:

3) The probability that a ball drawn from the first urn will be white, and a ball drawn from the second urn will be black, or on the contrary, the first ball will be black and the second will be white, equal to the sum of the corresponding probabilities:

Answer: 1)

2) 3) .

Problem 3

Among the 24 lottery tickets, there are 11 winning ones. Find the probability that at least one of the 2 tickets purchased will be a winner.

Solution

The probability that at least one of the 24 tickets purchased will be a winner is equal to the difference between one and the probability that none of the tickets purchased will be a winner. And the probability that none of the purchased tickets will be winning is equal to the product of the probability that the first ticket will not be winning by the probability that the second ticket will not be winning:

Hence, the probability that at least one of the 24 tickets purchased will be a winner:

Answer:

Problem 4

The box contains 6 parts of the first grade, 5 of the second and 2 of the third. Two details are taken at random. What is the probability that they will both be of the same variety?

Solution

The required probability is the probability that both parts will be either 1st, 2nd or 3rd grade and is equal to the sum of the corresponding probabilities:

The probability that both parts taken will be first grade:

The probability that both parts taken will be second grade:

The probability that both parts taken will be third grade:

Hence the probability of pulling out 2 parts of the same type is equal to:

Answer:

Problem 5

During the hour 0 ≤ t ≤ 1 (t is time in hours), one and only one bus arrives at the stop.

Solution

The bus can arrive at any moment t, where 0 ≤ t ≤ 1 (where t is time in hours) or, which is the same, 0 ≤ t ≤ 60 (where t is time in minutes).

The passenger arrives at t = 0 and waits no more than 28 minutes.

The chances of a bus arriving at the station during this time or during the remaining 32 minutes are equally probable, so the probability that a passenger arriving at this stop at time t = 0 will have to wait for a bus no more than 28 minutes is

Answer:

Problem 8

The probability of the first shooter hitting the target is 0.2, the second – 0.2 and the third – 0.2. All three shooters fired simultaneously. Find the probability that:

1) only one shooter will hit the target;

2) two shooters will hit the target;

3) at least one will hit the target.

Solution

1) The probability that only one shooter will hit the target is equal to the probability of the first shooter hitting the target and the second and third missing or hitting the target by the second shooter and missing by the first and third or hitting the target by the third shooter and missing by the first and second, and therefore equal to the sum of the corresponding probabilities.

The probability that the first shooter will hit the target, and the second and third will miss, is equal to the product of these probabilities:

Similar probabilities of the second shooter hitting the target and the first and third missing, as well as the third hitting the target and the first and second missing:

, .

Hence, the desired probability:

2) The probability that two shooters will hit the target is equal to the probability of the first and second shooter hitting the target and the third missing or hitting the target by the first and third shooter and missing by the second or hitting the target by the second and third shooter and missing by the first, which means it is equal to the sum of the corresponding probabilities.

The probability that the first and second shooters will hit the target, and the third will miss, is equal to the product of these probabilities:

Similar probabilities of the first and third shooter hitting the target and the second missing, as well as the second and third hitting the target and missing the first.

For two incompatible events A and B, the probabilities of these events are equal to the sum of their probabilities:

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

Example #3:find the probability of getting 1 or 6 when throwing a die.

Events A (rolling 1) and B (rolling 6) are equally possible: P(A) = P(B) = 1/6, therefore P(A or B) = 1/6 + 1/6 = 1/3

The addition of probabilities is valid not only for two, but also for any number of incompatible events.

Example #4:There are 50 balls in the urn: 10 white, 20 black, 5 red and 15 blue. Find the probability of a white, or black, or red ball appearing during a single operation of removing a ball from the urn.

The probability of drawing the white ball (event A) is P(A) = 10/50 = 1/5, the black ball (event B) is P(B) = 20/50 = 2/5 and the red ball (event C) is P (C) = 5/50 = 1/10. From here, using the formula for adding probabilities, we get P(A or B or C) = P(A) + P(B) = P(C) = 1/5 + 2/5 + 1/10 = 7/10

The sum of the probabilities of two opposite events, as follows from the theorem of addition of probabilities, is equal to one:

P(A) + P( ) = 1

In the above example, taking out a white, black and red ball will be the event A 1, P(A 1) = 7/10. The opposite event of 1 is drawing the blue ball. Since there are 15 blue balls, and the total number of balls is 50, we get P(1) = 15/50 = 3/10 and P(A) + P() = 7/10 +3/10 = 1.

If events A 1, A 2, ..., A n form a complete system of pairwise incompatible events, then the sum of their probabilities is equal to 1.

In general, the probability of the sum of two events A and B is calculated as

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

Probability multiplication theorem:

Events A and B are called independent , if the probability of occurrence of event A does not depend on whether event B occurred or not, and vice versa, the probability of occurrence of event B does not depend on whether event A occurred or not.

The probability of joint occurrence of independent events is equal to the product of their probabilities. For two events P(A and B)=P(A)·P(B).

Example: One urn contains 5 black and 10 white balls, the other contains 3 black and 17 white balls. Find the probability that when balls are first drawn from each urn, both balls will be black.

Solution: the probability of drawing a black ball from the first urn (event A) is P(A) = 5/15 = 1/3, a black ball from the second urn (event B) is P(B) = 3/20

P(A and B)=P(A)·P(B) = (1/3)(3/20) = 3/60 = 1/20.

In practice, the probability of event B often depends on whether some other event A occurred or not. In this case they talk about conditional probability , i.e. the probability of event B given that event A occurs. Conditional probability is denoted by P(B/A).

The probability multiplication theorem becomes more complicated if the probability of an event consisting of the joint occurrence of two mutually dependent events is determined. In the case where event B occurs given that event A has occurred, the probability of these two events occurring together is equal to

P(A and B)=P(A)P(B/A).

There are 5 balls in the urn: 3 white and 2 black. Find the probability that black and white balls will be drawn one after the other.

The probability that the black ball will be withdrawn first (event A) is equal to P(A) = m/n = 2/5. After removing the black ball, there are 4 balls left in the urn: 3 white and 1 black. In this case, the probability of drawing the white ball (event B after event A) is equal to P(B/A) = ¾. We get P(A and B) = P(A)P(B/A) = (2/5)(3/4) = 3/10.

If event A can occur only with one of the events Н 1 , Н 2 ,…Н n , which form a complete system of pairwise incompatible events, then the probability of event A is determined by total probability formula

P(A) = P(A/N 1)P(N 1)+P(A/N 2)P(H 2)+...+P(A/N n)P(N n).

To calculate the probability P(H i /A) in this case, use Bayes formula:

Control questions

1. Define the probability of events.

2.What events are called equally possible?

3.What events are called reliable?

4.What events are called impossible?

5.What events are called opposite?

6.Formulate the classical definition of probability.

7.What is the probability of a reliable event? An impossible event?

8.Name the formulas for adding and multiplying probabilities.

Homework

Fill in workbook lesson 11-12.

Lecture No. 6

Subject: : Basic concepts of probability theory and mathematical statistics

Individual tasks in mathematics

There are 6 white balls and 11 black balls in the urn. Two balls are drawn at random at the same time. Find the probability that both balls will be:

The probability that the second ball drawn will also be white is equal to

Since one of the white balls has already been drawn.

Thus, the probability that both balls drawn from the urn will be white is equal to the product of these probabilities, since these possibilities are independent:

3) The probability that both drawn balls will be different colors is the probability that the first ball will be white and the second black or that the first ball will be black and the second white. It is equal to the sum of the corresponding probabilities.

Answer: 1) 2) 3) .

The first urn has 6 white balls, 11 black, the second urn has 5 white and 2 black. A ball is drawn at random from each urn. Find the probability that both balls will be:

1) white, 2) the same color, 3) different colors.

2) The probability that both balls drawn will be the same color is the probability that both balls will be either white or black. It is equal to the sum of the probabilities of drawing two white balls or two black balls:

3) The probability that a ball drawn from the first urn will be white and a ball drawn from the second urn will be black, or vice versa - the first ball will be black and the second white, is equal to the sum of the corresponding probabilities:

Answer: 1) 2) 3) .

Among the 24 lottery tickets, there are 11 winning ones. Find the probability that at least one of the 2 tickets purchased will be a winner.

Hence, the probability that at least one of the 24 tickets purchased will be a winner:

Answer:

The box contains 6 parts of the first grade, 5 of the second and 2 of the third. Two details are taken at random. What is the probability that they will both be of the same variety?

The required probability is the probability that both parts will be either 1st, 2nd or 3rd grade and is equal to the sum of the corresponding probabilities:

The probability that both parts taken will be first grade:

The probability that both parts taken will be second grade:

The probability that both parts taken will be third grade:

Hence the probability of pulling out 2 parts of the same type is equal to:

During the hour 0 ≤ t ≤ 1 (t is time in hours), one and only one bus arrives at the stop.

The bus can arrive at any moment t, where 0 ≤ t ≤ 1 (where t is time in hours) or, which is the same, 0 ≤ t ≤ 60 (where t is time in minutes).

The passenger arrives at t = 0 and waits no more than 28 minutes.

Answer:

The probability of the first shooter hitting the target is 0.2, the second – 0.2 and the third – 0.2. All three shooters fired simultaneously. Find the probability that:

1) only one shooter will hit the target;

2) two shooters will hit the target;

3) at least one will hit the target.

1) The probability that only one shooter will hit the target is equal to the probability of the first shooter hitting the target and the second and third missing, or the second shooter hitting the target and missing the first and third, or the third shooter hitting the target and missing the first and second, and therefore equal the sum of the corresponding probabilities.

The probability that the first shooter will hit the target, and the second and third will miss, is equal to the product of these probabilities:

Similar probabilities of the second shooter hitting the target and the first and third missing, as well as the third hitting the target and the first and second missing:

Hence, the desired probability:

2) The probability that two shooters will hit the target is equal to the probability of the first and second shooter hitting the target and the third missing, or the first and third shooter hitting the target and the second missing, or the second and third shooter hitting the target and the first missing, and therefore equals the sum corresponding probabilities.

The probability that the first and second shooters will hit the target, and the third will miss, is equal to the product of these probabilities:

Similar probabilities of the first and third shooter hitting the target and the second missing, as well as the second and third hitting the target and missing the first:

Hence, the desired probability:

3) The probability that at least one shooter will hit the target is equal to the difference between one and the probability that no shooter will hit the target. The probability that no shooter will hit the target is equal to the product of these probabilities:

Answer: 1) , 2) , 3) .

The student knows 11 questions out of 24 questions in the program. Each exam paper contains three questions. Find the probability that: 1) the student knows all three questions; 2) only two questions; 3) only one question on the exam card.

1) The probability that a student knows all three questions on the ticket is equal to the product of the probabilities of knowing each of them. Since all three questions are different and not repeated, then:

2) The probability that a student knows only two questions on a ticket is equal to the probability that he knows the first and second questions, but does not know the third, or that he knows the first and third questions, but does not know the second, or that he knows the second and third questions, but does not know the first. That is, this probability is equal to the sum of all these probabilities.

The first term of this sum:

The second term of this sum:

And the third term of this sum:

Hence the desired probability:

3) The probability that a student knows only one question out of three is equal to the difference of one and the probability that he does not know a single question:

Answer: 1) , 2) , 3) .

The first urn contains 6 white balls and 11 black ones, the second urn contains 5 white and 2 black balls. One ball was transferred from the first urn to the second, then one ball was removed from the second urn. Find the probability that the ball taken from the second urn turns out to be: 1) white, 2) black.

1) The probability that a ball taken at random from the first urn and transferred to the second will turn out to be white:

If the ball transferred from the first urn to the second turns out to be white, then there will be six white balls in the second urn. Then, the probability that the ball taken from the second urn will be white:

The probability that a ball taken at random from the first urn and transferred to the second will turn out to be black:

If the ball moved from the first urn to the second turns out to be black, then there will be three black balls in the second urn.

Then, the probability that the ball taken from the second urn will be black is:

And the probability of both of these events is equal to the product of these probabilities:

Answer: 1) , 2) .

The first urn contains 6 white and 11 black balls, the second urn contains 5 white and 2 black balls, and the third urn contains 7 white balls. An urn is chosen at random and a ball is drawn from it at random. Find the probability that the drawn ball is:

1) white, 2) black.

1) The probability of choosing one of the three urns is 1/3.

Probability of drawing a white ball from the first urn:

This means that the probability of choosing the first urn and drawing a white ball from it is:

Similarly, the probability of choosing the second urn and drawing a white ball from it is:

Probability of choosing the third urn and drawing a white ball from it:

The probability of drawing a white ball from a randomly chosen urn is equal to the sum of these probabilities:

Probability of choosing the first urn and drawing a black ball from it:

Similarly, the probability of choosing the second urn and drawing a black ball from it is:

Probability of choosing the third urn and drawing a black ball from it:

since in the third urn all the balls are white.

The probability of drawing a black ball from a randomly chosen urn is equal to the sum of these probabilities:

Answer: 1) , 2) .

One of the three urns contains 6 white and 11 black balls, the second contains 5 white and 2 black balls, and the third contains 7 white balls. A ball is chosen at random from three urns, and one ball is again chosen at random from it. He turned out to be white. What is the probability that: 1) a ball is drawn from the first urn, 2) a ball is drawn from the second urn, 3) a ball is drawn from the third urn?

To solve this problem, we apply the Bayes formula, the essence of which is as follows: if before the experiment the probabilities of the hypotheses H 1, H 2, ... N n were equal to P(H 1), P(H 2), ..., P(H n), and as a result, event A occurred, then the new (conditional) probabilities of the hypotheses are calculated using the formula:

Where Р(Н i) is the probability of the hypothesis Н i, Р(А|Н i) is the conditional probability of event A under this hypothesis.

Let's denote the hypotheses:

N 1 – selection of the first urn, N 2 – selection of the second urn, N 3 – selection of the third urn.

Before action begins, all these hypotheses are equally probable:

After choosing, it turned out that a white ball was drawn. Let's find the conditional probabilities:

;

1) According to Bayes’ formula, the posterior (after experiment) probability that the ball was drawn from the first urn is equal to:

2) Similarly, the probability that the ball was drawn from the second urn is:

3) Similarly, the probability that the ball was drawn from the third urn is:

1) ,

2) ,

3) .

Of the 24 students who took the mathematics exam, 6 were excellently prepared, 11 were well prepared, 5 were mediocre, and 2 were poorly prepared. IN exam papers 20 questions. A well-prepared student can answer all 20 questions, a well-prepared student can answer 16 questions, a mediocre student can answer 10 questions, and a poorly prepared student can answer 5 questions. A student called at random answered all three randomly assigned questions. Find the probability that this student is prepared: 1) excellent, 2) poorly.

To solve this problem, we apply Bayes’ formula:

Where Р(Н i) is the probability of hypothesis Н i,

Р(А|Н i) – conditional probability of event A under this hypothesis.

Let's denote the hypotheses:

N 1 – the student is well prepared, N 2 – the student is well prepared,

N 3 – the student is mediocrely prepared, N 4 – the student is poorly prepared.

Before the exam, the prior probabilities of these hypotheses are:

, , ,

After an examination test, one of the students found that he answered all three questions. Let's find the conditional probabilities, that is, the probabilities of answering all three questions by a student from each performance group:

, ,

, .

1) According to Bayes’ formula, the posterior (after the exam) probability that the called student was perfectly prepared is equal to:

2) Likewise, the probability that the called student was poorly prepared is:

1) The probability that the called student was perfectly prepared:

2) The likelihood that the called student was poorly prepared:

The coin is tossed 11 times. Find the probability that the coat of arms will appear: 1) 2 times, 2) no more than 2 times, 3) no less than once and no more than 2 times.

If an experiment is carried out n times, and an event appears each time with probability p (and, accordingly, does not appear with probability 1 – p = q), then the probability of this event occurring m times is estimated using the binomial distribution formula:

The number of combinations of n elements of m.

1) B in this case p = 0.5 (probability of the coat of arms falling out),

q = 1 – p =0.5 (probability of getting heads),

Hence, the probability of getting a coat of arms is 2 times:

2) in this case, the event (coat of arms) can appear 0 times, 1 time or 2 times, which means the required probability:

3) in this case, the event (coat of arms) can appear 1 time or 2 times, which means the required probability:

Probability of getting the coat of arms:

1) exactly 2 times equal

2) no more than 2 times:

3) at least once and no more than 2 times:

11 messages are transmitted over the communication channel, each of which, independently of the others, is distorted by interference with probability p = 0.2. Find the probability that: 1) out of 11 messages exactly 2 will be distorted by interference,

2) all messages will be received without distortion, 3) at least two messages will be distorted.

1) here p = 0.2 (probability of distortion),

q = 1 – p =0.8 (probability of non-distortion),

2) The probability of accepting all 11 messages without distortion is equal to the product of all the probabilities of accepting each of them without distortion:

3) Distortion of at least two messages means that two or one or no messages may be distorted:

The probability that:

1) out of 11 messages exactly 2 will be distorted, equal to ,

Nothing else, except again events and... Indeed, we have: *=, *=, =, =. Another example of an event algebra L is a collection of four events: . Indeed: *=,*=,=,. 2.Probability. Probability theory studies random events. This means that until a certain point in time, generally speaking, it is impossible to say in advance about a random event A whether this event will occur or not. Only...