Probability problems with solutions. Probability theory problems with solutions Probability addition and multiplication problems
Probability problems with solutions
1. Combinatorics
Problem 1 . There are 30 students in a group. It is necessary to choose the headman, deputy headman and trade union organizer. How many ways are there to do this?
Solution. Any of the 30 students can be chosen as a headman, any of the remaining 29 students can be chosen as a deputy, and any of the remaining 28 students can be chosen as a trade union leader, i.e. n1 = 30, n2 = 29, n3 = 28. According to the rule of multiplication, the total number N of ways to select the headman, his deputy and the trade union organizer is N = n1´n2´n3 = 30´29´28 = 24360.
Task 2 . Two postmen must deliver 10 letters to 10 addresses. In how many ways can they distribute work?
Solution. The first letter has n1 = 2 alternatives - either the first postman assigns it to the addressee, or the second. For the second letter, there are also n2 = 2 alternatives, etc., that is, n1 = n2 =… = n10 = 2. Consequently, by virtue of the multiplication rule, the total number of ways to distribute letters between two postmen is equal to
Problem 3. In the box there are 100 parts, of which 30 are 1st grade parts, 50 are 2nd grade, the rest are 3rd grade. How many ways are there to retrieve one grade 1 or 2 part from a box?
Solution. A part of the 1st grade can be extracted in n1 = 30 ways, the 2nd grade - n2 = 50 ways. According to the sum rule, there are N = n1 + n2 = 30 + 50 = 80 ways to extract one part of the 1st or 2nd grade.
Problem 5 . The order of performance of 7 participants in the competition is determined by lot. How many different types of drawing are possible?
Solution. Each variant of the draw differs only in the order of the participants in the competition, that is, it is a permutation of 7 elements. Their number is
Problem 6 . 10 films are participating in the competition in 5 nominations. How many options are there for the distribution of prizes, if all nominations are set various awards?
Solution. Each of the options for distributing prizes is a combination of 5 films out of 10, which differs from other combinations both in composition and in their order. Since each film can receive prizes in one or several nominations, the same films can be repeated. Therefore, the number of such combinations is equal to the number of placements with repetitions of 10 elements of 5:
Problem 7 . The chess tournament is attended by 16 people. How many games must be played in a tournament if one game is to be played between any two participants?
Solution. Each game is played by two participants out of 16 and differs from others only in the composition of pairs of participants, that is, it is a combination of 16 elements of 2. Their number is equal 
Problem 8 . Under the conditions of task 6, determine how many options there are for the distribution of prizes, if the same prizes?
Solution. If the same prizes are established for each nomination, then the order of films in a combination of 5 prizes does not matter, and the number of options is the number of combinations with repetitions of 10 elements of 5, determined by the formula
Problem 9. The gardener must plant 6 trees within three days. In how many ways can he distribute the work over the days if he plants at least one tree a day?
Solution. Suppose a gardener plants trees in a row and can make different decisions about which tree to stop after on the first day and after which on the second. Thus, one can imagine that the trees are separated by two partitions, each of which can stand in one of 5 places (between the trees). The partitions should be there one at a time, because otherwise not a single tree will be planted on some day. Thus, you need to select 2 elements out of 5 (no repetitions). Hence, the number of ways.
Problem 10. How many four-digit numbers (possibly starting with zero) are there that add up to 5?
Solution. Let us represent the number 5 as the sum of consecutive units, divided into groups by partitions (each group in total forms the next digit of the number). It is clear that such partitions will need 3. There are 6 places for partitions (before all units, between them and after). Each place can be occupied by one or several partitions (in the latter case, there are no units between them, and the corresponding amount is zero). Consider these places as elements of the set. Thus, you need to select 3 elements out of 6 (with repetitions). Therefore, the required number of numbers 
Assignment 11 . How many ways can you divide a group of 25 students into three subgroups A, B and C of 6, 9 and 10 people, respectively?
Solution. Here n = 25, k = 3, n1 = 6, n2 = 9, n3 = 10..gif "width =" 160 "height =" 41 ">
Problem 1 . There are 5 oranges and 4 apples in a box. Three fruits are chosen at random. What is the probability that all three fruits are oranges?
Solution... Elementary outcomes here are sets of 3 fruits. Since the order of the fruits is irrelevant, we will consider their selection as random (and unrepeatable) .. gif "width =" 21 "height =" 25 src = ">. The number of favorable outcomes is equal to the number of ways to choose 3 oranges from the available 5, that is .. gif "width =" 161 height = 83 "height =" 83 ">.
Task 2 . The teacher asks each of the three students to think of any number from 1 to 10. Considering that the choice of any of the given numbers by each of the students is equally possible, find the probability that one of them will have the same numbers.
Solution. First, let's calculate the total number of outcomes. The first student chooses one of 10 numbers and has n1 = 10 possibilities, the second also has n2 = 10 possibilities, and finally, the third one also has n3 = 10 possibilities. By virtue of the multiplication rule, the total number of ways is: n = n1´n2´n3 = 103 = 1000, that is, the entire space contains 1000 elementary outcomes. To calculate the probability of event A, it is convenient to go to the opposite event, that is, to count the number of those cases when all three students conceive different numbers. The first one still has m1 = 10 ways to choose a number. The second student now has only m2 = 9 possibilities, since he has to take care that his number does not coincide with the intended number of the first student. The third student is even more limited in his choice - he has only m3 = 8 possibilities. Therefore, the total number of combinations of conceived numbers in which there are no matches is m = 10 × 9 × 8 = 720. There are 280 cases in which there are coincidences. Therefore, the desired probability is P = 280/1000 = 0.28.
Problem 3 . Find the probability that in an 8-digit number exactly 4 digits coincide, and the rest are different.
Solution... Event А = (an eight-digit number contains 4 identical digits). From the condition of the problem it follows that there are five different numbers in the number, one of them is repeated. The number of ways to choose it is equal to the number of ways to select one digit from 10 digits..gif "width =" 21 "height =" 25 src = ">. Then the number of favorable outcomes. All in all, ways of composing 8-digit numbers is | W | = 108 The sought probability is
Problem 4 . Six clients randomly contact 5 firms. Find the probability that no one will contact at least one firm.
Solution. Consider the opposite event https://pandia.ru/text/78/307/images/image020_10.gif "width =" 195 "height =" 41 ">. The total number of ways to distribute 6 clients across 5 firms. 
... Hence, .
Problem 5 . Let the urn contain N balls, of which M are white and N – M are black. N balls are taken from the urn. Find the probability that there are exactly m white balls among them.
Solution. Since the order of the elements is unimportant here, the number of all possible sets of volume n of N elements is equal to the number of combinations of m white balls, n – m black ", is equal, and, therefore, the desired probability is P (A) = https: // pandia. ru / text / 78/307 / images / image031_2.gif "width =" 167 "height =" 44 ">.
Problem 7 (meeting problem) ... The two persons A and B agreed to meet at a certain place between 12 and 13 o'clock. The one who came first waits for the other for 20 minutes, after which he leaves. What is the probability of a meeting of persons A and B, if the arrival of each of them can happen at random during the specified hour and the moments of arrival are independent?
Solution. Let's designate the moment of arrival of person A through x and person B - through y. For the meeting to take place, it is necessary and sufficient that ôх-уо £ 20. Let's represent x and y as coordinates on a plane, we will choose a minute as a scale unit. All possible outcomes are represented by the dots of a 60-sided square, and those favorable to the meeting are located in the shaded area. The desired probability is equal to the ratio of the area of the shaded figure (Fig. 2.1) to the area of the whole square: P (A) = (602–402) / 602 = 5/9.
3. Basic formulas of probability theory
Problem 1 ... There are 10 red and 5 blue buttons in the box. Two buttons are taken out at random. What is the probability that the buttons will be of the same color ?
Solution... Event A = (buttons of the same color removed) can be represented as a sum, where events mean the choice of buttons in red and blue, respectively. The probability of pulling out two red buttons is equal, and the probability of pulling out two blue buttons https://pandia.ru/text/78/307/images/image034_2.gif "width =" 19 height = 23 "height =" 23 ">. Gif" width = "249" height = "83">
Task 2 ... Among the company's employees, 28% know English, 30% - German, 42% - French; English and German - 8%, English and French - 10%, German and French - 5%, all three languages - 3%. Find the probability that a randomly selected employee of the firm: a) knows English or German; b) knows English, German or French; c) does not know any of the listed languages.
Solution. Let us denote by A, B and C events where a random employee of the firm speaks English, German, or French, respectively. Obviously, the shares of the firm's employees who speak a particular language determine the likelihood of these events. We get:
a) P (AÈB) = P (A) + P (B) -P (AB) = 0.28 + 0.3-0.08 = 0.5;
b) P (AÈBÈC) = P (A) + P (B) + P (C) - (P (AB) + P (AC) + P (BC)) + P (ABC) = 0.28 + 0, 3 + 0.42-
-(0,08+0,1+0,05)+0,03=0,8;
c) 1-P (AÈBÈC) = 0.2.
Problem 3 . The family has two children. What is the probability that the oldest child is a boy if it is known that there are children of both sexes in the family?
Solution. Let A = (the oldest child is a boy), B = (there are children of both sexes in the family). We will assume that the birth of a boy and the birth of a girl are equally probable events. If the birth of a boy is denoted by the letter M, and the birth of a girl is denoted by D, then the space of all elementary outcomes consists of four pairs:. In this space, only two outcomes (MD and DM) correspond to event B. Event AB means that there are children of both sexes in the family. The oldest child is a boy, therefore the second (youngest) child is a girl. This AB event corresponds to one outcome - MD. Thus, | AB | = 1, | B | = 2, and
Problem 4 . The master, having 10 parts, of which 3 are non-standard, checks the parts one by one until he comes across a standard one. What is the likelihood that he will check exactly two details?
Solution. Event A = (the foreman checked exactly two parts) means that during such a check, the first part turned out to be non-standard, and the second was standard. So, where = (the first part turned out to be non-standard) and = (the second part is standard). It is obvious that the probability of event A1 is also equal to 
, since before taking the second part, the master had 9 parts left, of which only 2 are non-standard and 7 are standard. By the multiplication theorem
Problem 5 . In one box there are 3 white and 5 black balls, in the other box there are 6 white and 4 black balls. Find the probability that a white ball will be removed from at least one box if one ball is removed from each box.
Solution... The event A = (at least a white ball was taken out of at least one box) can be represented as a sum, where events mean the appearance of a white ball from the first and second boxes, respectively .. gif "width =" 91 "height =" 23 "> .. gif "width =" 20 "height =" 23 src = ">. gif" width = "480" height = "23">.
Problem 6 . Three examiners take an exam in a certain subject from a group of 30 people, with the first interviewing 6 students, the second with 3 students, and the third with 21 students (students are selected randomly from the list). The attitude of the three examiners to those who are poorly prepared is different: the chances of such students to pass the exam with the first teacher are 40%, with the second - only 10%, with the third - 70%. Find the probability that a poorly prepared student will pass the exam .
Solution. Let us denote by the hypotheses that the poorly prepared student answered the first, second and third examiners, respectively. By the condition of the problem
, 
, 
.
Let event A = (poorly prepared student passed the exam). Then again, by virtue of the condition of the problem
, 
, 
.
By the formula of total probability we get:
Problem 7 ... The firm has three sources of supply of components - firms A, B, C. Firm A accounts for 50% of the total supply, B - 30% and C - 20%. It is known from practice that 10% of the parts supplied by the company A are defective, by the company B - 5% and by the company C - 6%. What is the probability that a part taken at random will be good?
Solution. Let event G be the appearance of a usable part. The probabilities of the hypotheses that the part was supplied by firms A, B, C are, respectively, P (A) = 0.5, P (B) = 0.3, P (C) = 0.2. The conditional probabilities of the appearance of a good part in this case are P (G | A) = 0.9, P (G | B) = 0.95, P (G | C) = 0.94 (as the probabilities of opposite events to the appearance of a defective one). By the formula of total probability we get:
P (G) = 0.5 x 0.9 + 0.3 x 0.95 + 0.2 x 0.94 = 0.923.
Problem 8 (see problem 6). Let it be known that the student did not pass the exam, that is, received the grade "unsatisfactory". Which of the three teachers most likely answered ?
Solution. The probability of getting "bad" is. It is required to calculate conditional probabilities. Using Bayes' formulas, we get:
https://pandia.ru/text/78/307/images/image059_0.gif "width =" 183 "height =" 44 src = ">, 
.
It follows that, most likely, a poorly prepared student passed the exam to a third examiner.
4. Repeated independent tests. Bernoulli's theorem
Problem 1 . The dice is thrown 6 times. Find the probability that the “six” will be dropped exactly 3 times.
Solution. Throwing the dice six times can be thought of as a series of independent trials with a probability of success (“sixes”) of 1/6 and a probability of failure of 5/6. The required probability is calculated by the formula 
.
Task 2 ... The coin is thrown 6 times. Find the probability that the coat of arms will be drawn no more than 2 times.
Solution. The desired probability is equal to the sum of the probabilities of three events, consisting in the fact that the coat of arms will not fall out even once, either once or twice:
P (A) = P6 (0) + P6 (1) + P6 (2) = https://pandia.ru/text/78/307/images/image063.gif "width =" 445 height = 24 "height = "24">.
Problem 4 . The coin is flipped 3 times. Find the most probable number of successes (coat of arms).
Solution. Possible values for the number of successes in the three trials under consideration are m = 0, 1, 2, or 3. Let Am be the event that the coat of arms appears m times after three tosses of a coin. Using the Bernoulli formula, it is easy to find the probabilities of events Am (see table):
From this table you can see that the most probable values are numbers 1 and 2 (their probabilities are 3/8). The same result can be obtained from Theorem 2. Indeed, n = 3, p = 1/2, q = 1/2. Then
, i.e.
Task 5. As a result of each visit of the insurance agent, the contract is concluded with a probability of 0.1. Find the most likely number of contracts concluded after 25 visits.
Solution. We have n = 10, p = 0.1, q = 0.9. The inequality for the most probable number of successes takes the form: 25 × 0.1–0.9 £ m * £ 25 × 0.1 + 0.1 or 1.6 £ m * £ 2.6. This inequality has only one whole solution, namely, m * = 2.
Problem 6 ... It is known that the scrap rate for a certain part is 0.5%. The inspector checks 1000 parts. What is the probability of finding exactly three defective parts? What is the probability of finding at least three defective parts?
Solution. We have 1000 Bernoulli trials with the probability of "success" p = 0.005. Applying the Poisson approximation with λ = np = 5, we obtain
2) P1000 (m³3) = 1-P1000 (m<3)=1-»1-,
and P1000 (3) "0.14; Р1000 (m³3) "0.875.
Problem 7 . The probability of a purchase when a customer visits a store is p = 0.75. Find the probability that, with 100 visits, a customer will make a purchase exactly 80 times.
Solution... In this case, n = 100, m = 80, p = 0.75, q = 0.25. We find 
, and define j (x) = 0.2036, then the desired probability is equal to Р100 (80) = 
.
Task 8. The insurance company has concluded 40,000 contracts. The probability of an insured event for each of them during the year is 2%. Find the probability that there will be no more than 870 such cases.
Solution. By the condition of the problem, n = 40000, p = 0.02. Find np = 800 ,. To calculate P (m £ 870), we use the Moivre-Laplace integral theorem:
P (0
We find from the table of values of the Laplace function:
P (0 Problem 9
.
The probability of an event occurring in each of the 400 independent trials is 0.8. Find a positive number e such that, with a probability of 0.99, the absolute value of the deviation of the relative frequency of occurrence of an event from its probability does not exceed e. Solution. By the condition of the problem, p = 0.8, n = 400. We use a corollary from the Moivre-Laplace integral theorem: 5.
Discrete random variables Problem 1
.
In a bunch of 3 keys, only one key fits the door. The keys are searched until a suitable key is found. Construct the distribution law for a random variable x - the number of tested keys .
Solution. The number of keys tested can be 1, 2, or 3. If only one key was tested, this means that this first key immediately came to the door, and the probability of such an event is 1/3. So, Further, if there were 2 tested keys, that is, x = 2, this means that the first key did not fit, and the second one did. The probability of this event is 2/3 × 1/2 = 1 / 3..gif "width =" 100 "height =" 21 "> The result is the following distribution series: Task 2
.
Construct the distribution function Fx (x) for a random variable x from Problem 1. Solution. The random variable x has three values 1, 2, 3, which divide the entire number axis into four intervals:. If x<1, то неравенство x£x невозможно (левее x нет значений случайной величины x) и значит, для такого x функция Fx(x)=0. If 1 £ x<2, то неравенство x£x возможно только если x=1, а вероятность такого события равна 1/3, поэтому для таких x функция распределения Fx(x)=1/3. If 2 £ x<3, неравенство x£x означает, что или x=1, или x=2, поэтому в этом случае вероятность P(x And, finally, in the case x³3, the inequality x £ x holds for all values of the random variable x, therefore P (x So we got the following function: Problem 3.
The joint law of distribution of random variables x and h is given using the table Calculate the particular distribution laws of the constituent quantities x and h. Determine if they are dependent..gif "width =" 423 "height =" 23 src = ">; https://pandia.ru/text/78/307/images/image086.gif "width =" 376 "height =" 23 src = ">. The partial distribution for h is obtained in a similar way: https://pandia.ru/text/78/307/images/image088.gif "width =" 229 "height =" 23 src = ">. The obtained probabilities can be written in the same table opposite the corresponding values of the random variables: Now let's answer the question about the independence of the random variables x and h..gif "width =" 108 "height =" 25 src = "> in this cell. For example, in the cell for the values x = -1 and h = 1 there is a probability 1 / 16, and the product of the corresponding partial probabilities 1/4 × 1/4 is 1/16, that is, it coincides with the joint probability. This condition is also checked in the remaining five cells, and it turns out to be true in all. Therefore, the random variables x and h are independent. Note that if our condition were violated in at least one cell, then the quantities should be recognized as dependent. To calculate the probability mark the cells for which the condition is met https://pandia.ru/text/78/307/images/image092.gif "width =" 574 "height =" 23 src = "> Problem 4
.
Let a random variable ξ have the following distribution law: Calculate the expectation Mx, variance Dx, and standard deviation s. Solution... By definition, the expectation of x is Standard deviation https://pandia.ru/text/78/307/images/image097.gif "width =" 51 "height =" 21 ">. Solution. Let's use the formula Problem 6
.
For a pair of random variables from Problem 3, calculate the covariance cov (x, h). Solution. In the previous task, the mathematical expectation was already calculated .
It remains to calculate and .
Using the particular distribution laws obtained in solving Problem 3, we obtain ; ;
and so which is to be expected due to the independence of random variables. Task 7.
The random vector (x, h) takes the values (0,0), (1,0), (–1,0), (0,1) and (0, –1) equally probable. Calculate the covariance of the random variables x and h. Show that they are addicted. Solution... Since P (x = 0) = 3/5, P (x = 1) = 1/5, P (x = –1) = 1/5; P (h = 0) = 3/5, P (h = 1) = 1/5, P (h = –1) = 1/5, then Mx = 3/5´0 + 1/5´1 + 1 / 5´ (–1) = 0 and Мh = 0; M (xh) = 0´0´1 / 5 + 1´0´1 / 5–1´0´1 / 5 + 0´1´1 / 5–0´1´1 / 5 = 0. We get cov (x, h) = М (xh) –МxМh = 0, and the random variables are uncorrelated. However, they are addicted. Let x = 1, then the conditional probability of the event (h = 0) is equal to P (h = 0 | x = 1) = 1 and is not equal to the unconditional P (h = 0) = 3/5, or the probability (ξ = 0, η = 0) is not equal to the product of probabilities: P (x = 0, h = 0) = 1 / 5¹P (x = 0) P (h = 0) = 9/25. Therefore, x and h are dependent. Problem 8
... The random increments in the stock prices of two companies for day x and h have a joint distribution given by the table: Find the correlation coefficient. Solution. First of all, we calculate Mxh = 0.3-0.2-0.1 + 0.4 = 0.4. Next, we find the particular distribution laws of x and h: Determine Mx = 0.5-0.5 = 0; Mh = 0.6-0.4 = 0.2; Dx = 1; Dh = 1-0.22 = 0.96; cov (x, h) = 0.4. We get Problem 9.
The random increments in stock prices of two companies per day have variances Dx = 1 and Dh = 2, and their correlation coefficient is r = 0.7. Find the variance of the price increment for a portfolio of 5 shares of the first company and 3 shares of the second company. Solution... Using the properties of variance, covariance and determination of the correlation coefficient, we obtain: Problem 10
.
The distribution of a two-dimensional random variable is given by the table: Find the conditional distribution and conditional mathematical expectation h at x = 1. Solution. The conditional expectation is From the problem statement, we find the distribution of the components h and x (the last column and the last row of the table). This section contains the first part of the problems in probability theory, which are simple enough to be placed not only in the version of the exam in mathematics of the profile level, but also in the version of the exam in the basic level or in the version of the exam for the 9th grade. Demo versions Unified State Exam 2020
years of assignments to test knowledge of the elements of the theory of probability can be found under the number 10
for basic level and under number 4
for the profile level, as well as number 10 in the OGE version for grade 9. It is better to learn to solve such problems in stages.
The probability of event A is the fraction P (A) = __, m n
In the numerator of which there is a number m
elementary events, favorable
event A, and in the denominator n
- number of all
elementary events. Thus, in order to solve the problem, it is necessary to calculate the number of favorable and the number of all possible elementary events. Pay attention to the highlighted wording. It often happens that the conditions of two problems differ only in one word, and the solutions can be directly opposite. And vice versa, seemingly different questions, but in fact about the same. Be careful!
Do not forget that there cannot be more favorable events than all possible events, which means that the numerator of the fraction will never exceed the denominator. The answer to the question about the probability of an event must contain a number that satisfies the condition 0 ≤ P ≤ 1
... If you get a different answer, it is obviously wrong. Example 1 On board the aircraft there are 12 seats next to emergency exits and 18 seats behind the partitions dividing the cabins. The rest of the seats are inconvenient for a tall passenger. Passenger V. is tall. Find the probability that at check-in at random choice of location passenger V. will get a comfortable seat if there are 300 seats in the plane. Solution If "the rest of the seats are inconvenient," then the mentioned 12 + 18 = 30 seats are convenient. P (A) = ___
30
300
= 0,1. Answer: 0,1
The example presented above implements the simplest concept of an elementary event. Since one person can only take one seat, the events are independent. And since the condition specifically stipulates that during registration the place was chosen randomly, then they are equally possible. Therefore, in fact, we did not count events, but seats on the plane. Example 2 There are 30 people in a group of tourists. They are thrown into a hard-to-reach area by helicopter in several steps, 6 people per flight. The order in which the helicopter transports tourists is random. Find the probability that tourist P. will fly the first helicopter flight. Solution Determine how many flights the helicopter should make P (A) = 1/5 = 0.2. Answer: 0,2
In this example, you should already think about what constitutes an elementary event. Here it is a formed helicopter flight. One person can only get on one flight, i.e. only in one group of 6 people - events are independent. By the condition of the problem, the order of flights is random, i.e. all flights for each group are equally possible. We count flights. Example 3 One number is chosen at random from the set of natural numbers from 10 to 19. What is the probability that it is divisible by 3? Solution Let's write out the given numbers in a row and mark those of them that are divisible by 3. 10, 11, 12
, 13, 14, 15
, 16, 17, 18
, 19 It turns out that out of 10 given numbers, 3 numbers are divisible by 3. P (A) = 3/10 = 0.3. Answer: 0,3
One number is chosen at random from the set of natural numbers from 107 to 198. What is the probability that it is divisible by 3? Then you will have to remember that "every third number in a natural row is divided by 3" (by 4 - every fourth, by 5 every fifth ...) and determine the number of groups of three numbers in the range from 107 to 198. P (A) = 31/92 ≈ 0.337 Now check yourself. Problem 1 There are only 55 tickets in the collection of tickets for biology, 11 of them contain a question about botany. Find the probability that on a ticket randomly selected for the exam the student will get a question about botany. Event A - "Choosing a ticket with a botany question". You can choose only one ticket (events are incompatible in pairs), all tickets are the same (events are equally possible) and all tickets are available to a student (full group). So the event "ticket selection" is elementary. There are as many such events as tickets, i.e. n = 55. There are as many favorable events as tickets with a botany question, ie. m = 11. According to the formula P (A) = 11/55 = 1/5 = 0.2. Answer: 0,2
Comment:
Indeed, the "everyday" situation is so familiar and simple that it is intuitively clear which events are elementary and which are favorable. Further I will not describe in detail this part of the solution, unless it is necessary. Objective 2. In the collection of tickets for mathematics there are only 25 tickets, in 10 of them there is a question on inequalities. Find the probability that on a ticket randomly selected for the exam the student will not get question on inequalities. Method I. Method II. Answer: 0,6
Problem 3 20 athletes participate in the gymnastics championship: 8 from Russia, 7 from the USA, the rest from China. The order in which the gymnasts perform is determined by lot. Find the probability that the first athlete will be from China. Event A - "The first gymnast from China to perform." Answer: 0,25
Problem 4 The shot put competition involves 4 athletes from Finland, 7 athletes from Denmark, 9 athletes from Sweden and 5 from Norway. The order in which the athletes compete is determined by lot. Find the likelihood that the last competitor is from Sweden. Similar to the previous task. Answer: 0,36
Problem 5 25 athletes compete in the diving championship, among them 8 jumpers from Russia and 9 jumpers from Paraguay. The order of performances is determined by drawing lots. Find the likelihood that a jumper from Paraguay will perform in sixth. Similar to the 2-mind of the previous tasks. Answer: 0,36
Comment:
The last three tasks are essentially the same, but at first glance, their questions seem to be different. What for? To confuse a student? No, the compilers have a different task: the exam should have many different options of the same degree of difficulty. So, do not be afraid of the "tricky question", you need to consider the situation described in the problem from all sides. Problem 6 The contest of performers is held in 5 days. A total of 80 performances have been announced - one from each country. On the first day there are 8 performances, the rest are divided equally between the remaining days. The order of performances is determined by drawing lots. What is the likelihood that the speech of the Russian representative will take place on the third day of the competition? Method I. Method II. Answer: 0,225
Comment:
Probability problems are often solved in different ways. Choose for yourself the one that is clearer for you. Problem 7 On average, out of 1000 garden pumps on sale, 5 are leaking. Find the probability that one pump randomly selected to control is not leaking. Event A - "the selected pump is not leaking". Answer: 0,995
Problem 8 The factory makes bags. On average, there are eight bags with hidden defects per 100 quality bags. Find the likelihood that the bag you buy will be of good quality. Round the result to the nearest hundredth. Event A - "quality purchased bag". Answer: 0,93
Note 1:
Compare this with the previous problem. How important it is to be attentive to each word in the condition! Problem 9 Before the start of the first round of the badminton championship, the participants are randomly divided into game pairs by lot. In total, 26 badminton players participate in the championship, including 10 participants from Russia, including Ruslan Orlov. Find the probability that in the first round Ruslan Orlov will play with some badminton player from Russia? Event A - "Ruslan Orlov will play with a badminton player from Russia." Answer: 0,36
Addition rule:
if some object A can be chosen k ways, and object B - l ways ( not like A), then the object "or A or
In "you can choose m + l ways. Multiplication rule:
if object A can be selected k ways, and after each such choice, another object B can be selected ( whatever from object A) l ways, then pairs of objects A and
B can be chosen m l ways. The multiplication rule is also called the "AND-rule", and the addition rule is the "OR-rule". Do not forget to check the independence of ways for "AND" and incompatibility (not so) for "OR". The following tasks can be solved both by enumerating options and using. I give several ways to solve each problem, because in one way it can be solved quickly, while in the other it takes a long time, and because someone understands one approach, and someone else. But this does not mean that it is imperative to disassemble all methods. Better to learn well one beloved. The choice is yours. Example 4 In a random experiment, a symmetrical coin is thrown five times. Find the probability that it hits heads twice. This problem can be solved in several ways. Consider the one that corresponds to the section heading, namely only by applying combinatorial formulas. Solution In each of the five coin tosses, one of the outcomes can be realized - heads or tails - for short, "o" or "p". Thus, the result of a series of tests will be a group of five letters, made up of two original letters, and therefore with repetitions. For example, "ooror" means that two times in a row fell heads, then tails, again heads and again tails. Hence, to calculate the number of all possible outcomes, you need to count the number of placements from n= 2 to k= 5 with repetitions, which is determined by the formula A —
n k = n k; A —
2 5 = 2 5 = 32. Favorable outcomes - heads will fall out exactly two times - are five-letter "words" composed of three letters "p" and two "o", which can be in different positions, for example, "opppo" or "poopp", i.e. these are permutations with repetitions. Their number is determined by the formula P —
n = ______
n!
n o! · n p! = ____
5
! 2! 3! = _______ 1 2 3 4 5 1 2 1 2 3 = 10, Where n= 5 the number of rearranged letters, n o = 2 and n p = 3 - the number of repetitions of the letters "o" and "p", respectively. By the formula of the classical probability, we obtain P = __
10
32
= 0,3125
Answer: 0,3125
However, if you do not know these formulas, do not be afraid of school exams in mathematics. Not only on the OGE and the basic USE, but also on the USE of the profile level, it is usually suggested to consider a short series of tests. In such cases, you will be able to write out and review the outcomes explicitly. Try it. Problem 10 In a random experiment, a symmetrical coin is thrown three times. Find the probability that it will never land heads. Method I. Method II. Answer: 0,125
Assignment 11 In a random experiment, a symmetrical coin is thrown three times. Find the probability that it will be heads exactly once. Method I. Method II. Answer: 0,375
Assignment 12 In a random experiment, a symmetrical coin is thrown three times. Find the probability that heads will come up at least once. Method I. Method II. Method III. Answer: 0,875
Assignment 13 In a random experiment, a symmetrical coin is thrown four times. Find the probability that it will never land heads. Let's use the multiplication rule for independent tests. Answer: 0,0625
Comment:
Of course, this problem could be solved in any of the ways discussed earlier. But the greater the number of possible outcomes, the longer and more senseless it is to decide by enumerating options. For those who do not know how otherwise or want to test a shorter solution with a longer one, I will still write: (oooo, ooop, ooro, oorr, oooo, orop, oro, oppr, pooo, poop, poro, ppp, proo, ppp, pro , ppprr). But to make sure that it is really discharged all possible outcomes, it is still worth counting the number of placements from 2 to 4 with repetitions: A —
n k = n k; A —
2 3 = 2 4 = 16.
The best way for a lot of throws is Bernoulli's formula. Try it yourself in this task. Assignment 14 In a random experiment, two dice are rolled. Find the probability that the total will be 8 points. Round the result to the nearest hundredth. Method I. (More about calculation m: First, we determine which terms the number 7 can consist of, for example, according to the scheme Method II. A detailed solution is best seen in the animation. Answer: 0,07
If your browser supports Flash then you can look animated graphics solutions method II- enumeration of options using tables. ( Click on the icon to view it.) Try to consider these examples not only as a solution to a specific problem, but also as an illustration of the rules for adding and multiplying test outcomes, as well as in order to finally decide on the choice of a solution method for the exam. Which is better - by enumeration of options or by formulas? Output:
problems in the theory of probability of this task can be solved using a single formula in one action, if you can count the number of possible and favorable events "on your fingers", diagrams, tables ... However, the more difficult the experiment is ("... a coin is thrown four times ... "," ... three dice are thrown ... "), the more cumbersome is the" simple "solution and the shorter is the" complex "- using formulas, rules and theorems. But, if you still make mistakes when solving problems on the classical definition of probability, then perhaps they have the same origin as in the famous anecdote about a dinosaur. In this case, follow the link AV: Yulia, you did your diploma in the studio of Sergei Tchoban "Coordination of Movements", where your design object was the D-1 block in Skolkovo. As far as I can tell, your work was probably the most specific: you designed for a place whose context has not yet been faked. How does it feel? Yu.A..: Working without an existing context was really a bit weird. In the Skolkovo area, the master plan for which was developed by Sergei Tchoban's Speech bureau together with David Chipperfield's company, we were given a plot, and we had to figure out what to do with it. In the first semester, we were divided into 3 groups of 4 people each and a competition was announced between us for a planning solution for one quarter. We had to place on the piece of land that we got, twelve, on average - five-story, houses, according to the number of students in the group. It so happened that our team won the competition: Anya Shevchenko, Dima Stolbovoy, Artem Slizunov and me. We got a fairly tough plan, which was limited not only by some cadastral parameters, but also by the terms of reference and design code. What is your master plan? We changed the structure that was in the original version of the general plan: to reduce the scale of the environment, we divided our block into 4 sub-blocks with public space inside each. In addition, each sub-block had its own function: housing, start-ups, a sub-block with a sports function and a major building, and a site with a hostel, a hotel, a museum, and the main square is also located right there. What restrictions did you specify in the design code? The block is very small, and the intentions of each of the participants could strongly influence the others. Therefore, we did not prescribe specific materials, but regulated possible shape changes by setting “footprint” and FAR. For example, if you do "gnawing", your number of storeys grows, which in turn is also limited to a certain level. What was the next step? Further, each of us had to develop one of the buildings on the site, but which one, with what function - determined the lot, we pulled pieces of paper with "lots". This was Sergei Tchoban's plan. And this situation is fundamentally different from the one when you yourself choose the topic of the diploma and design a building with a specific function, which, perhaps, you dreamed of designing for all six years of study. Here we had to come to terms with what was drawn by lot, and, on the one hand, it was rather painful, but on the other hand, this is a situation close to life. What did you get? I was lucky in my opinion. I designed a startup building. With certain dimensions, which could not be changed. The most important principle from which I proceeded was both ideological and functional: today it is a startup, and tomorrow it will probably no longer be. After all, what is Skolkovo in general? No one can intelligibly answer this question. Studying the materials, I concluded that Skolkovo's own development strategy is quite flexible. For me, this became the main condition that my project had to meet. Therefore, with a hull width of 12 meters, it was important for me that there were no extra walls in my building. I left nothing but the cores of stiffness, which are obligatory from the point of view of the design. Inside there is an open, free layout. As for the appearance, I tried to design my building so that it was quite modest, but at the same time expressive. The main facade turned out to be a 12-meter butt facing the boulevard. So I decided to sharpen its shape. The gable roof, which has become the visual accent of the entire building, plays an important role. It is an intermediate link between two "neighbors" of my object, different in height and expressiveness. In the course of work, it changed. At first, the ideological context was a bit dominant. And then we began to perceive Skolkovo not as a phenomenon on the scale of Russia, but to carefully consider the problems of the place itself. After all, today it can be an Innovation Center, and tomorrow it can be something else. Is that why your building has to be demolished? Good architecture can live longer than its original context. She also forms a new one. Was it hard to work in a group? How was the relationship built within the studio when each of you took up his own project? Yes, of course it's hard. After all, it turned out so that the situation as a whole could radically change from the wishes of each person. The site is small enough, and someone's idea to make, say, a console or something else, could affect, for example, the insolation rates. And then we all sat down and started discussing whether it was right or not. The final version pleasantly surprised me. At first it seemed to me that in everyone the desire to make a wow-thesis-project, and not harmonious group work, would outweigh it. But the general plan in the end turned out to be quite balanced. It seems to me that we managed to find a "golden mean" between personal ambitions and the need to follow certain rules of the game. What features did the training with Sergei Tchoban have? It was a pleasure to work with all the heads of our studio. In addition to Sergei, these are Alexei Ilyin and Igor Chlenov from the Speech bureau, and there were also subcontractors who helped to deal with certain nodes. The educational process was structured in a delightfully accurate way, literally in minutes. Although it was probably difficult for Sergey to some extent with us. I think he was counting on the fact that we are almost professionals. And we, I cannot say that we are still children, but the difference between a bureau employee and a student is still incredibly large. He shared his knowledge with us not as a teacher, but as a practicing architect and managed to make us work more independently and with each other than with teachers. It really was "coordination of movements." What did the two years of study at MARSH give you in general? I cannot say that the third eye has opened. But some doubts were resolved, some positions were strengthened. Now I take a more responsible attitude to what I do and what I say. Perhaps thank you very much for this MARSH, perhaps thank you very much for this time. I can say that the most valuable thing in MARSH, the main resource of the school, is people and some kind of special atmosphere. Mostly for the sake of the people, I went there. I went to Sergei Sitar, Kiril Ass, Evgeny Viktorovich, Narine Tyutcheva. In addition, I had you, comrades, who inspired and supported me. I hope we will communicate in the future, I hope we will do something together. Where did you study before? I defended my bachelor's degree at the Moscow Architectural Institute from the most beautiful teacher Irina Mikhailovna Yastrebova. And I can add that I have a very good attitude towards Moscow Architectural Institute and do not think that this is some kind of Soviet relic. He gives the academic basics, and everyone later decides for himself what he wants to do. What do you want to do now? For all the years of my existence in architecture, I have written about it, read about it, talked about it, but I have never created it in the full sense of the word. I was basically doing paper architecture, you know, with a claim to conceptual art. And if before I was completely sure that theory determines practice, now I cannot believe it until I check it out. Therefore, I now need to visit a construction site, I need to understand what it is - when you did something on paper, then fought for it, argued, coordinated, and in the end you stand, look and understand: this is it, it happened! This is my fix idea. Therefore, for the next two years, I plan to practice and try to make my way to the construction site, to implementation as short as possible.
... Hence, 
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... Namely, in each cell of the table we multiply the corresponding values and, the result is multiplied by the probability pij, and all this is summed over all cells of the table. As a result, we get:
.
Tasks only for determining the probability
To solve most of the following problems, it is enough to repeat the classical definition of the probability of an event:
Let's remember - elementary events (test outcomes) pairwise incompatible and equally possible... Sometimes it's obvious, and sometimes it's worth considering. "Pairwise incompatible" means, for example, that one person cannot travel on two buses at the same time. Are not "equally possible", for example, meeting on the street with a dinosaur and a dog.
Passenger V. can get any one seat out of 300 seats on the plane, which means all possible events n= 300. But only those of them will be "favorable" when passenger V. has got to a convenient place, such events as places, m = 30.
30: 6 = 5 (flights).
Tourist P. can fly any, but only one of them will be "favorable" - the first. Hence n = 5, m = 1.
We find the answer according to the general formula
1, 2, ..., 105, 106, 107, 108, ..., 197, 198
, 199, ...
There are only 92 numbers on this site: 198 - 106 = 92.
They make up 30 complete groups and one incomplete (92/3 = 30 integers and 2 in the remainder). Each complete group has one number that is divisible by 3. In an incomplete group, which is the last two numbers, 197 is not divisible by 3, but 198 is divisible. In total, we have 30 + 1 = 31 "favorable" numbers out of "total" 92.
Event A - "ticket selection without question on inequalities". There are 25 tickets in total, if 10 tickets have a question about inequalities, then 25 - 10 = 15 tickets do not. Thus, the total number of possible outcomes n = 25, the number of outcomes favorable to event A, m = 15. According to the formula P (A) = 15/25 = 3/5 = 0.6.
Event A - "choosing a ticket with a question on inequalities". Just as in Problem 1, we get P (A) = 10/25 = 2/5 = 0.4. But the question of this problem is the opposite of the question of problem 1, i.e. we need the probability of the opposite event B - "the choice of the ticket without the question of inequalities." The probability of the opposite event is calculated by the formula P (B) = 1 - P (A) = 1 - 0.4 = 0.6.
To determine the number of selections, let's first think about what is the outcome of the draw? What are we going to take for an elementary event? If we imagine a procedure when one athlete has already pulled out the ball with the number of the performance, and the second has to pull something out of the remaining ones, then there will be a complex solution using conditional probability. The answer can be obtained (see, for example, method II in problem 6). But why involve complex mathematics if you can consider the "everyday" situation from a different point of view?
Let's imagine that the draw is over and each gymnast is already holding a numbered ball in her hand. Each has only one ball, all the balls have different numbers, the ball with the number "1" is only for one of the athletes. Which one? The organizers of the draw are obliged to ensure that all athletes have an equal opportunity to receive this ball, otherwise it will be unfair. This means that the event - "the ball with the number" 1 "for the sportswoman" - is elementary.
There are n = 20 athletes in total, a favorable event is a ball with the number "1" for a Chinese woman, all athletes from China m = 20 - 8 - 7 = 5. According to the formula P (A) = 5/20 = 1/4 = 0.25 ...
Event A - "The last athlete to perform is from Sweden." An elementary event - "the last number went to a specific athlete." Total athletes n = 4 + 7 + 9 + 5 = 25. Favorable event - the athlete who got the last number from Sweden. Total athletes from Sweden m = 9.
According to the formula P (A) = 5/20 = 9/25 = 0.36.
Event A - "The sixth is a jumper from Paraguay." Elementary event - "number six for a particular athlete." Total Athletes n = 25. Auspicious Event - Athlete numbered “6” from Paraguay. Total athletes from Paraguay m = 9.
According to the formula P (A) = 9/25 = 0.36.
Event A - "the speech of the Russian representative will take place on the third day." One performance can be considered an elementary event, since representatives from all countries are equal (one from each country). Total n = 80 performances. On the first day there are 8 performances, in the remaining 5 - 1 = 4 days (80 - 8) / 4 = 18 performances. This means that 18 performances will take place on the third day - these are events favorable for the Russian, m = 18.
According to the formula P (A) = 18/80 = 9/40 = 0.225.
Let event A - "the speech of the representative of Russia will take place on the third day", event B - "the speech of the representative of Russia not will take place on the first day ", event C -" the speech of the Russian representative will take place on the third day on condition that he did not perform on the first day. "
By the definition of the conditional probability, P (A) = P (B) P (C).
80 - 8 = 72 people will not perform on the first day. According to the formula P (B) = 72/80 = 9/10 = 0.9.
If the performance of the representative of Russia does not fall on the first day, then he has the same chances to perform on any of the next 4 days (the rest of the performances are distributed evenly, which means the days are equally possible). According to the formula P (C) = 1/4 = 0.25.
Therefore P (A) = 0.9 0.25 = 0.225.
There are n = 1000 pumps in total. 5 of them are leaking, so m = 1000 - 5 = 995 are not leaking.
According to the formula P (A) = 995/1000 = 0.995.
Total n = 100 + 8 = 108 bags (100 quality and 8 defective). Quality m = 100 bags.
According to the formula P (A) = 100/108 = 0.9259259 ≈ 0.93.
Note 2:
We repeated the rounding rules when solving word problems.
Badminton competitions are usually held with elimination, and only in the first round all 26 badminton players participate. But the number of all possible outcomes is not 26, n = 26 - 1 = 25, because Ruslan Orlov cannot play with himself. For the same reason, m = 10 - 1 = 9, because Ruslan Orlov is one of the 10 participants from Russia.
According to the formula P (A) = 9/25 = 0.36.Tasks using elements of combinatorics
In these problems, the answer is also determined by the formula P (A) = m / n
but counting the number n all possible events and numbers m favorable events are noticeably more difficult than in previous cases. For this, various methods of enumerating options and auxiliary figures, tables, graphs ("tree of possibilities") are used. The rules of addition and multiplication of variants, as well as ready-made recipes for combinatorics: formulas for the number of permutations, combinations, and placements, can facilitate the situation.
You can write down and consider all the possible outcomes of 3 coin tosses: (ooo, oop, oo, opr, roo, pop, ppo, ppr), where o is an abbreviation for "heads", p is an abbreviation for "tails". The listing shows that n = 8, m = 1. (Favorable only ppr).
According to the formula P (A) = 1/8 = 0.125.
It can be noted that the test conditions satisfy the Bernoulli scheme with p = 1/2 and q = 1/2 and use the formula
P (0) = C 0 3 (1/2) 0 (1/2) (3-0) = 1 (1/2) 3 = 1/8 = 0.125.
The test is the same and the outcomes are the same as in the previous case: (ooo, oop, oro, orr, roo, pop, ppo, ppr). It can be seen from the listing that n = 8, m = 3. (Favorable: (op, pop, pp)).
According to the formula P (A) = 3/8 = 0.375.
The test conditions satisfy the Bernoulli scheme with p = 1/2 and q = 1/2, hence the formula
P (1) = C 1 3 (1/2) 1 (1/2) (3-1) = 3 (1/2) 1 (1/2) 2 = 3/8 = 0.375.
The test is the same and the outcomes are the same as in the previous cases: (ooo, oop, oro, orr, roo, poop, ppo, ppr). It can be seen from the listing that n = 8, m = 7. (Favorable all, except ooo).
According to the formula P (A) = 7/8 = 0.875.
According to the Bernoulli formula, taking into account the addition rule (at least 1 of 3 = or 1, or 2, or 3)
P (A) = P (1) + P (2) + P (3) = C 1 3 (1/2) 1 (1/2) (3-1) + C 2 3 (1/2) 2 (1/2) (3-2) + C 3 3 (1/2) 3 (1/2) (3-3) = (3 + 3 + 1) (1/2) 3 = 7 / 8 = 0.875.
The event "heads at least once" is the opposite of the event "heads will not be drawn even once." The probability of the latter is 0.125. We defined it in Problem 10.
Hence P (A) = 1 - 0.125 = 0.875 according to the formula for the probability of the opposite event.
With each throw, 2 outcomes are possible, which means that with 4 throws, 2 · 2 · 2 · 2 = 16 outcomes are possible.
With each throw, the heads will not fall out in one way, which means that with 4 throws it will not fall out 1 · 1 · 1 · 1 = 1 in one way.
According to the formula P (A) = 1/16 = 0.0625.
For one die, there can be 6 different outcomes of the test (loss of points 1,2, ..., 6) and
for the other - 6 outcomes independent from the first. The total number of possible outcomes when throwing two dice is determined by the multiplication rule n= 6 × 6 = 36.
To determine the number of favorable outcomes, let's see from which terms the sum of 8 is obtained:
1 + 7 = 8; 2 + 6 = 8; 3 + 5 = 8; 4 + 4 = 8; 5 + 3 = 8; 6 + 2 = 8; 7 + 1 = 8.
The first and the last options are in our case impossible events, the number 7 is not on ordinary dice. The rest are realized if the first term falls on one bone, and the second on the other. Favorable outcomes ("2; 6", "3; 5", "4; 4", "5; 3", "6; 2"), total m
For one die, there can be 6 different outcomes of the test (loss of points 1,2, ..., 6), and
for the other - 6 outcomes, and
for the third - 6 outcomes, independent apart. The total number of possible outcomes when throwing three dice is determined by the multiplication rule n= 6 × 6 × 6 = 216.
To determine the number of favorable outcomes, let's see which 3 terms can be used to obtain the number 7. Recall that the sum does not change from the rearrangement of the places of the terms.
or 7 = 1 + 1 + 5 (3 permutations) or 1 + 2 + 4 (6 permutations) or 1 + 3 + 3 (3 permutations) or 2 + 2 + 3 (3 permutations).
Thus, according to the addition rule m= 3 + 6 + 3 + 3 = 15 ways to get 7, as the sum of points on 3 dice.
According to the formula P (A) = 15/216 = 0.069444444 ≈ 0.07.
,
and arrange the terms in ascending order (to eliminate errors with unnecessary or missing permutations). And then we accurately count the permutations either by the formulas P k = k!
- permutations without repetitions, P k 1, k 2, ..., k n = k! / (K 1! K 2! ... k n!)
- permutations with repetitions, or reasoning.
According to the formulas: P 3 = 3!
= 1 2 3 = 6 and P 1,2 = 3! / (1! 2!)
= 1 2 3 / (1 1 2) = 3.
Reasoning: if 2 numbers are the same, and the 3rd is different, then it can stand on the 1st, 2nd or 3rd places, you get 3 permutations, if all 3 numbers are different, then each of them can stand on 1- th place, and the remaining two take the 2nd and 3rd or 3rd and 2nd places, respectively, then 3 · 2 = 6 permutations.
For this task, you can also count the options using the plate, but already 3-D! 
Solving problems using tables
Problems on the rules of addition and multiplication of probabilities
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