V-6-2014 (all 56 prototypes from the Unified State Exam bank)
Be able to build and explore the simplest mathematical models(probability theory)
1.In a random experiment, two are thrown dice. Find the probability that the total will be 8 points. Round the result to hundredths. Solution: The number of outcomes in which 8 points will appear as a result of throwing the dice is 5: 2+6, 3+5, 4+4, 5+3, 6+2. Each dice has six possible rolls, so the total number of outcomes is 6 6 = 36. Therefore, the probability of rolling a total of 8 is 5: 36 = 0.138... = 0.14
2. In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will appear exactly once. Solution: There are 4 equally possible outcomes of the experiment: heads-heads, heads-tails, tails-heads, tails-tails. Heads appear exactly once in two cases: heads-tails and tails-heads. Therefore, the probability that heads will appear exactly 1 time is 2: 4 = 0.5.
3. 20 athletes are participating 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 athlete competing first is from China. Solution: Participates in the championshipathletes from China. Then the probability that the athlete competing first will be from China is 5: 20 = 0.25
4. On average, out of 1000 garden pumps sold, 5 leak. Find the probability that one pump randomly selected for control does not leak. Solution: On average, out of 1000 garden pumps sold, 1000 − 5 = 995 do not leak. This means that the probability that one pump randomly selected for control does not leak is equal to 995: 1000 = 0.995
5. The factory produces bags. On average, for every 100 quality bags, there are eight bags with hidden defects. Find the probability that the purchased bag will be of high quality. Round the result to hundredths. Solution: According to the condition, for every 100 + 8 = 108 bags there are 100 quality bags. This means that the probability that the purchased bag will be of high quality is 100: 108 =0.925925...= 0.93
6. 4 athletes from Finland, 7 athletes from Denmark, 9 athletes from Sweden and 5 from Norway are participating in the shot put competition. The order in which the athletes compete is determined by lot. Find the probability that the athlete competing last is from Sweden. Solution: In total, 4 + 7 + 9 + 5 = 25 athletes take part in the competition. This means that the probability that the athlete who competes last will be from Sweden is 9: 25 = 0.36
7.Scientific Conference carried out in 5 days. A total of 75 reports are planned - the first three days contain 17 reports, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by drawing lots. What is the probability that Professor M.'s report will be scheduled for the last day of the conference? Solution: In the first three days, 51 reports will be read, and 24 reports are planned for the last two days. Therefore, 12 reports are planned for the last day. This means that the probability that Professor M.’s report will be scheduled for the last day of the conference is 12: 75 = 0.16
8. The competition of performers is held over 5 days. A total of 80 performances have been announced - one from each country. There are 8 performances on the first day, the rest are distributed equally between the remaining days. The order of performances is determined by drawing lots. What is the probability that a Russian representative will perform on the third day of the competition? Solution: Scheduled for the third dayspeeches. This means that the probability that the performance of a representative from Russia will be scheduled on the third day of the competition is 18: 80 = 0.225
9. 3 scientists from Norway, 3 from Russia and 4 from Spain came to the seminar. The order of reports is determined by drawing lots. Find the probability that the eighth report will be a report by a scientist from Russia. Solution: In total, 3 + 3 + 4 = 10 scientists take part in the seminar, which means that the probability that the scientist who speaks eighth will be from Russia is 3:10 = 0.3.
10.Before the start of the first round of the badminton championship, the participants are divided into playing pairs randomly by lot. In total, 26 badminton players are participating in the championship, including 10 participants from Russia, including Ruslan Orlov. Find the probability that in the first round Ruslan Orlov will play with any badminton player from Russia? Solution: In the first round, Ruslan Orlov can play with 26 − 1 = 25 badminton players, of which 10 − 1 = 9 are from Russia. This means that the probability that in the first round Ruslan Orlov will play with any badminton player from Russia is 9: 25 = 0.36
11. In the collection of tickets for biology there are only 55 tickets, 11 of them contain a question on botany. Find the probability that a student will get a question on botany on a randomly selected exam ticket. Solution: 11: 55 = 0.2
12. 25 athletes are performing at 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 probability that a Paraguayan jumper will be sixth.
13.Two factories produce the same glass for car headlights. The first factory produces 30% of these glasses, the second - 70%. The first factory produces 3% of defective glass, and the second - 4%. Find the probability that glass accidentally purchased in a store turns out to be defective.
Solution. Convert %% to fractions.
Event A - "Glass from the first factory was purchased." P(A)=0.3
Event B - "Glass from the second factory was purchased." P(B)=0.7
Event X - "Defective glass".
P(A and X) = 0.3*0.03=0.009
P(B and X) = 0.7*0.04=0.028 According to the formula full probability:P = 0.009+0.028 = 0.037
14.If grandmaster A. plays white, then he wins against grandmaster B. with probability 0.52. If A. plays black, then A. wins against B. with probability 0.3. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times. Solution: 0,52 * 0,3 = 0,156.
15. Vasya, Petya, Kolya and Lyosha cast lots as to who should start the game. Find the probability that Petya will have to start the game.
Solution: Random experiment - casting lots.
In this experiment, the elementary event is the participant who wins the lot.
Let us list the possible elementary events:
(Vasya), (Petya), (Kolya), (Lyosha).
There will be 4 of them, i.e. N=4. The lot implies that all elementary events are equally possible.
The event A= (Petya won the lot) is favored by only one elementary event (Petya). Therefore N(A)=1.
Then P(A)=0.25 Answer: 0.25.
16. 16 teams participate in the World Championship. Using lots, they need to be divided into four groups of four teams each. There are cards with group numbers mixed in the box: 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4. Team captains draw one card each. What is the probability that the Russian team will be in the second group? Solution: Total outcomes - 16. Of these, favorable, i.e. with number 2, it will be 4. So 4: 16=0.25
17. On a geometry exam, a student gets one question from the list exam questions. The probability that this is an inscribed circle question is 0.2. The probability that this is a question on the topic “Parallelogram” is 0.15. There are no questions that simultaneously relate to these two topics. Find the probability that a student will get a question on one of these two topics in the exam.
= (question on the topic “Inscribed circle”),
= (question on the topic “Parallelogram”).
Events And are incompatible, since by condition the list does not contain questions related to these two topics at the same time.
Event= (question on one of these two topics) is a combination of them:.
Let us apply the formula for adding the probabilities of incompatible events:
.
18.B mall two identical machines sell coffee. The probability that the machine will run out of coffee by the end of the day is 0.3. The probability that both machines will run out of coffee is 0.12. Find the probability that at the end of the day there will be coffee left in both machines.
Let's define events
= (coffee will run out in the first machine),
= (coffee will run out in the second machine).
According to the conditions of the problem And .
Using the formula for adding probabilities, we find the probability of an event
And = (coffee will run out in at least one of the machines):
.
Therefore, the probability of the opposite event (coffee will remain in both machines) is equal to
.
19. A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hits the targets the first three times and misses the last two. Round the result to hundredths.
In this problem it is assumed that the result of each next shot does not depend on the previous ones. Therefore, the events “hit on the first shot,” “hit on the second shot,” etc. independent.
The probability of each hit is equal. This means that the probability of each miss is equal to. Let's use the formula for multiplying the probabilities of independent events. We find that the sequence
= (hit, hit, hit, missed, missed) has a probability
=
= . Answer: .
20. There are two payment machines in the store. Each of them can be faulty with probability 0.05, regardless of the other machine. Find the probability that at least one machine is working.
This problem also assumes that the automata operate independently.
Let's find the probability of the opposite event
= (both machines are faulty).
To do this, we use the formula for multiplying the probabilities of independent events:
.
This means the probability of the event= (at least one machine is working) is equal to. Answer: .
21. The room is illuminated by a lantern with two lamps. The probability of one lamp burning out within a year is 0.3. Find the probability that at least one lamp will not burn out during the year. Solution: Both will burn out (the events are independent and we use the formula for the product of probabilities) with probability p1=0.3⋅0.3=0.09
Opposite event(NOT both will burn out = at least ONE will not burn out)
will happen with probability p=1-p1=1-0.09=0.91
ANSWER: 0.91
22. The probability that a new electric kettle will last more than a year is 0.97. The probability that it will last more than two years is 0.89. Find the probability that it will last less than two years but more than a year
Solution.
Let A = “the kettle will last more than a year, but less than two years”, B = “the kettle will last more than two years”, then A + B = “the kettle will last more than a year”.
Events A and B are joint, the probability of their sum is equal to the sum of the probabilities of these events, reduced by the probability of their occurrence. The probability of these events occurring, consisting in the fact that the kettle will fail in exactly two years - exactly on the same day, hour and second - is zero. Then:
P(A + B) = P(A) + P(B) − P(A B) = P(A) + P(B),
from where, using the data from the condition, we obtain 0.97 = P(A) + 0.89.
Thus, for the desired probability we have: P(A) = 0.97 − 0.89 = 0.08.
23. An agricultural company purchases chicken eggs from two households. 40% of eggs from the first farm are eggs of the highest category, and from the second farm - 20% of eggs of the highest category. Total highest category receives 35% of eggs. Find the probability that an egg purchased from this agricultural company will come from the first farm. Solution: Let the agricultural firm purchase from the first farm eggs, including eggs of the highest category, and in the second farm - eggs, including eggs of the highest category. Thus, the total amount the agroform purchases eggs, including eggs of the highest category. According to the condition, 35% of eggs have the highest category, then:
Therefore, the probability that the purchased egg will be from the first farm is equal to =0,75
24. There are 10 digits on the telephone keypad, from 0 to 9. What is the probability that a randomly pressed digit will be even?
25.What is the probability that a randomly selected natural number from 10 to 19 is divisible by three?
26.Cowboy John hits a fly on the wall with a probability of 0.9 if he shoots from a zeroed revolver. If John fires an unfired revolver, he hits the fly with probability 0.2. There are 10 revolvers on the table, only 4 of which have been shot. Cowboy John sees a fly on the wall, randomly grabs the first revolver he comes across and shoots the fly. Find the probability that John misses. Solution: John hits a fly if he grabs a zeroed revolver and hits with it, or if he grabs an unshot revolver and hits with it. According to the formula conditional probability, the probabilities of these events are respectively 0.4·0.9 = 0.36 and 0.6·0.2 = 0.12. These events are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: 0.36 + 0.12 = 0.48. The event that John misses is the opposite. Its probability is 1 − 0.48 = 0.52.
27. There are 5 people in the group of tourists. Using lots, they choose two people who must go to the village to buy food. Tourist A. would like to go to the store, but he obeys the lot. What is the probability that A. will go to the store? Solution: There are five tourists in total, two of them are chosen at random. The probability of being selected is 2: 5 = 0.4. Answer: 0.4.
28.Before the start of a football match, the referee throws a coin to determine which team will start the game with the ball. The Fizik team plays three matches with different teams. Find the probability that in these games “Physicist” will win the lot exactly twice. Solution: Let’s denote “1” the side of the coin that is responsible for the “Physicist” winning the lot, and let’s denote the other side of the coin “0”. Then there are three favorable combinations: 110, 101, 011, and there are 2 combinations in total 3 = 8: 000, 001, 010, 011, 100, 101, 110, 111. Thus, the required probability is equal to:
29. The dice are thrown twice. How many elementary outcomes of the experiment favor the event “A = the sum of points is 5”? Solution: The sum of points can be equal to 5 in four cases: “3 + 2”, “2 + 3”, “1 + 4”, “4 + 1”. Answer: 4.
30. In a random experiment, a symmetrical coin is tossed twice. Find the probability that the OP outcome will occur (heads the first time, tails the second time). Solution: There are four possible outcomes: heads-heads, heads-tails, tails-heads, tails-tails. One is favorable: heads and tails. Therefore, the desired probability is 1: 4 = 0.25. Answer: 0.25.
31. Bands perform at the rock festival - one from each of the declared countries. The order of performance is determined by lot. What is the probability that a group from Denmark will perform after a group from Sweden and after a group from Norway? Round the result to hundredths. Solution: The total number of groups performing at the festival is not important to answer the question. No matter how many there are, there are 6 ways for these countries relative position among the speakers (D - Denmark, W - Sweden, N - Norway):
D...SH...N..., ...D...N...SH..., ...SH...N...D..., ...W. ..D...N..., ...N...D...W..., ...N...W...D...
Denmark is ranked behind Sweden and Norway in two cases. Therefore, the probability that the groups will be randomly distributed in this way is equal to Answer: 0.33.
32. During artillery fire, the automatic system fires a shot at the target. If the target is not destroyed, the system fires a second shot. Shots are repeated until the target is destroyed. The probability of destroying a certain target with the first shot is 0.4, and with each subsequent shot it is 0.6. How many shots will be required to ensure that the probability of destroying the target is at least 0.98? Solution: You can solve the problem “by action”, calculating the probability of surviving after a series of consecutive misses: P(1) = 0.6. P(2) = P(1) 0.4 = 0.24. P(3) = P(2) 0.4 = 0.096. P(4) = P(3) 0.4 = 0.0384; P(5) = P(4) 0.4 = 0.01536. The latter probability is less than 0.02, so five shots at the target is enough.
33.To advance to the next round of competition, football team you need to score at least 4 points in two games. If a team wins, it receives 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4. Solution : A team can get at least 4 points in two games in three ways: 3+1, 1+3, 3+3. These events are incompatible; the probability of their sum is equal to the sum of their probabilities. Each of these events is the product of two independent events - the result in the first and in the second game. From here we have:
34. In a certain city, out of 5,000 babies born, 2,512 are boys. Find the frequency of births of girls in this city. Round the result to the nearest thousand. Solution: 5000 – 2512 = 2488; 2488: 5000 = 0,4976 ≈0,498
35. On board the aircraft there are 12 seats next to the emergency exits and 18 seats behind the partitions separating the cabins. The remaining seats are inconvenient for the passenger tall. Passenger V. is tall. Find the probability that at registration at random selection passenger V. will get seats comfortable spot, if there are only 300 seats on the plane. Solution : There are 12 + 18 = 30 seats on the plane that are comfortable for passenger B, and there are 300 seats in total on the plane. Therefore, the probability that passenger B will get a comfortable seat is 30: 300 = 0.1. Answer: 0.1.
36. At an Olympiad at a university, participants are seated in three classrooms. In the first two there are 120 people each; the remaining ones are taken to a reserve auditorium in another building. When counting, it turned out that there were 250 participants in total. Find the probability that a randomly selected participant wrote the competition in a spare classroom. Solution: In total, 250 − 120 − 120 = 10 people were sent to the reserve audience. Therefore, the probability that a randomly selected participant wrote the Olympiad in a spare classroom is 10: 250 = 0.04. Answer: 0.04.
37. There are 26 people in the class, among them two twins - Andrey and Sergey. The class is randomly divided into two groups of 13 people each. Find the probability that Andrey and Sergey will be in the same group. Solution: Let one of the twins be in some group. Together with him, 12 people from the 25 remaining classmates will be in the group. The probability that the second twin will be among these 12 people is 12: 25 = 0.48.
38. A taxi company has 50 cars; 27 of them are black with yellow inscriptions on the sides, the rest are yellow with black inscriptions. Find the probability that a car will respond to a random call yellow color with black inscriptions. Solution: 23:50=0.46
39.There are 30 people in the group of tourists. They are dropped by helicopter into a hard-to-reach area in several stages, 6 people per flight. The order in which the helicopter transports tourists is random. Find the probability that tourist P. will take the first helicopter flight. Solution: There are 6 seats on the first flight, 30 seats in total. Then the probability that tourist P. will fly on the first helicopter flight is: 6:30 = 0.2
40. The probability that a new DVD player will be repaired under warranty within a year is 0.045. In a certain city, out of 1,000 DVD players sold during the year, 51 units were received by the warranty workshop. How different is the frequency of the “warranty repair” event from its probability in this city? Solution: The frequency (relative frequency) of the “warranty repair” event is 51: 1000 = 0.051. It differs from the predicted probability by 0.006.
41. When manufacturing bearings with a diameter of 67 mm, the probability that the diameter will differ from the specified one by no more than 0.01 mm is 0.965. Find the probability that a random bearing will have a diameter less than 66.99 mm or greater than 67.01 mm. Solution. According to the condition, the bearing diameter will lie in the range from 66.99 to 67.01 mm with a probability of 0.965. Therefore, the desired probability of the opposite event is 1 − 0.965 = 0.035.
42. The probability that student O. will correctly solve more than 11 problems on a biology test is 0.67. The probability that O. will correctly solve more than 10 problems is 0.74. Find the probability that O. will solve exactly 11 problems correctly. Solution: Consider the events A = “the student will solve 11 problems” and B = “the student will solve more than 11 problems.” Their sum is event A + B = “the student will solve more than 10 problems.” Events A and B are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: P(A + B) = P(A) + P(B). Then, using these problems, we get: 0.74 = P(A) + 0.67, whence P(A) = 0.74 − 0.67 = 0.07. Answer: 0.07.
43. To enter the institute for the specialty “Linguistics”, the applicant must score at least 70 points on the Unified State Exam in each of three items- mathematics, Russian language and foreign language. To enroll in the specialty "Commerce", you need to score at least 70 points in each of three subjects - mathematics, Russian language and social studies. The probability that applicant Z. will receive at least 70 points in mathematics is 0.6, in Russian - 0.8, in a foreign language - 0.7 and in social studies - 0.5. Find the probability that Z. will be able to enroll in at least one of the two mentioned specialties. Solution: In order to enroll anywhere, Z. needs to pass both Russian and mathematics with at least 70 points, and in addition to this, also pass a foreign language or social studies with at least 70 points. Let A, B, C and D - these are events in which Z. passes mathematics, Russian, foreign and social studies, respectively, with at least 70 points. Then since
For the probability of arrival we have:
44. At a ceramic tableware factory, 10% of the plates produced are defective. During product quality control, 80% of defective plates are identified. The remaining plates are on sale. Find the probability that a plate randomly selected upon purchase has no defects. Round your answer to the nearest hundredth. Solution : Let the factory produceplates. All quality plates and 20% of undetected defective plates will go on sale:plates. Because the quality ones, the probability of buying a high-quality plate is 0.9p:0.92p=0.978 Answer: 0.978.
45.There are three sellers in the store. Each of them is busy with a client with probability 0.3. Find the probability that in random moment time, all three sellers are busy at the same time (consider that customers come in independently of each other). Solution : The probability of a product of independent events is equal to the product of the probabilities of these events. Therefore, the probability that all three sellers are busy is equal
46.Based on customer reviews, Ivan Ivanovich assessed the reliability of two online stores. The probability that the desired product will be delivered from store A is 0.8. The probability that this product will be delivered from store B is 0.9. Ivan Ivanovich ordered goods from both stores at once. Assuming that online stores operate independently of each other, find the probability that no store will deliver the product. Solution: The probability that the first store will not deliver the goods is 1 − 0.9 = 0.1. The probability that the second store will not deliver the goods is 1 − 0.8 = 0.2. Since these events are independent, the probability of their occurrence (both stores will not deliver the goods) is equal to the product of the probabilities of these events: 0.1 · 0.2 = 0.02
47.From district center There is a daily bus to the village. The probability that there will be fewer than 20 passengers on the bus on Monday is 0.94. The probability that there will be fewer than 15 passengers is 0.56. Find the probability that the number of passengers will be between 15 and 19. Solution: Consider the events A = “there are less than 15 passengers on the bus” and B = “there are from 15 to 19 passengers on the bus.” Their sum is event A + B = “there are less than 20 passengers on the bus.” Events A and B are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: P(A + B) = P(A) + P(B). Then, using these problems, we obtain: 0.94 = 0.56 + P(B), whence P(B) = 0.94 − 0.56 = 0.38. Answer: 0.38.
48.Before the start of a volleyball match, team captains draw fair lots to determine which team will start the game with the ball. The “Stator” team takes turns playing with the “Rotor”, “Motor” and “Starter” teams. Find the probability that Stator will start only the first and last games. Solution. You need to find the probability of three events happening: “Stator” starts the first game, does not start the second game, and starts the third game. The probability of a product of independent events is equal to the product of the probabilities of these events. The probability of each of them is 0.5, from which we find: 0.5·0.5·0.5 = 0.125. Answer: 0.125.
49. IN Fairyland There are two types of weather: good and excellent, and the weather, once established in the morning, remains unchanged all day. It is known that with probability 0.8 the weather tomorrow will be the same as today. Today is July 3rd, the weather in the Magic Land is good. Find the probability that the weather will be great in Fairyland on July 6th. Solution. For the weather on July 4, 5 and 6, there are 4 options: ХХО, ХОО, ОХО, OOO (here X is good, O is excellent weather). Let's find the probabilities of such weather occurring: P(XXO) = 0.8·0.8·0.2 = 0.128; P(XOO) = 0.8 0.2 0.8 = 0.128; P(OXO) = 0.2 0.2 0.2 = 0.008; P(OOO) = 0.2 0.8 0.8 = 0.128. Specified events inconsistent, the probability of their sum is equal to the sum of the probabilities of these events: P(ХХО) + P(ХОО) + P(ХХО) + P(ООО) = 0.128 + 0.128 + 0.008 + 0.128 = 0.392.
50. All patients with suspected hepatitis undergo a blood test. If the test reveals hepatitis, the test result is called positive . In patients with hepatitis, the analysis gives positive result with probability 0.9. If the patient does not have hepatitis, the test may give a false positive result with a probability of 0.01. It is known that 5% of patients admitted with suspected hepatitis actually have hepatitis. Find the probability that a patient admitted to the clinic with suspected hepatitis will test positive. Solution . A patient’s analysis can be positive for two reasons: A) the patient has hepatitis, his analysis is correct; B) the patient does not have hepatitis, his analysis is false. This incompatible events, the probability of their sum is equal to the sum of the probabilities of these events. We have: p(A)=0.9 0.05=0.045; p(B)=0.01 0.95=0.0095; p(A+B)=P(A)+p(B)=0.045+0.0095=0.0545.
51. Misha had four candies in his pocket - “Grilyazh”, “Squirrel”, “Korovka” and “Swallow”, as well as the keys to the apartment. While taking out the keys, Misha accidentally dropped one piece of candy from his pocket. Find the probability that the “Grillage” candy was lost.
52.A mechanical watch with a twelve-hour dial broke down at some point and stopped running. Find the probability that hour hand froze, reaching the 10 mark, but not reaching the 1 hour mark. Solution: 3: 12=0.25
53. The probability that the battery is defective is 0.06. A buyer in a store chooses a random package containing two of these batteries. Find the probability that both batteries are good. Solution: The probability that the battery is good is 0.94. The probability of independent events occurring (both batteries will be good) is equal to the product of the probabilities of these events: 0.94·0.94 = 0.8836. Answer: 0.8836.
54. An automatic line produces batteries. The probability that a finished battery is faulty is 0.02. Before packaging, each battery goes through a control system. The probability that the system will reject a faulty battery is 0.99. The probability that the system will mistakenly reject a working battery is 0.01. Find the probability that a randomly selected manufactured battery will be rejected by the inspection system. Solution. A situation in which the battery will be rejected can arise as a result of the following events: A = the battery is really faulty and was correctly rejected, or B = the battery is working, but was mistakenly rejected. These are incompatible events, the probability of their sum is equal to the sum of the probabilities of these events. We have:
55.The picture shows a labyrinth. The spider crawls into the maze at the Entrance point. The spider cannot turn around and crawl back, so at each branch the spider chooses one of the paths along which it has not yet crawled. Believing that the choice further path purely random, determine with what probability the spider will come to the exit.
Solution.
At each of the four marked forks, the spider can choose either the path leading to exit D or another path with probability 0.5. This independent events, the probability of their product (the spider reaches exit D) is equal to the product of the probabilities of these events. Therefore, the probability of arriving at exit D is (0.5) 4 = 0,0625.
Problems in probability theory with solutions
1. Combinatorics
Problem 1 . There are 30 students in the group. It is necessary to choose a headman, a deputy headman and a 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 organizer, i.e. n1=30, n2=29, n3=28. According to the multiplication rule, the total number N of ways to select a headman, his deputy and a trade union leader is N=n1´n2´n3=30´29´28=24360.
Problem 2 . Two postmen must deliver 10 letters to 10 addresses. In how many ways can they distribute the work?
Solution. The first letter has n1=2 alternatives - either it is taken to the addressee by the first postman, or by the second. For the second letter there are also n2=2 alternatives, etc., i.e. n1=n2=…=n10=2. Therefore, by virtue of the multiplication rule, the total number of ways to distribute letters between two postmen is equal to
Problem 3. There are 100 parts in the box, of which 30 are 1st grade parts, 50 are 2nd grade, the rest are 3rd grade. How many ways are there to remove one grade 1 or grade 2 part from a box?
Solution. A part of the 1st grade can be extracted in n1=30 ways, a part of the 2nd grade can be extracted in 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 the 7 participants in the competition is determined by lot. How many various options Is it possible to draw lots in this case?
Solution. Each variant of the draw differs only in the order of the participants in the competition, i.e. it is a permutation of 7 elements. Their number is equal
Problem 6 . 10 films are participating in the competition in 5 nominations. How many options for prize distribution are there if the following rules are established for all categories? various awards?
Solution. Each of the prize distribution options is a combination of 5 films out of 10, differing from other combinations both in composition and in their order. Since each film can receive awards in one or several categories, 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 . 16 people participate in a chess tournament. How many games must be played in a tournament if one game must be played between any two participants?
Solution. Each game is played by two participants out of 16 and differs from the others only in the composition of the pairs of participants, i.e., it is a combination of 16 elements of 2. Their number is equal to 
Problem 8 . In the conditions of task 6, determine how many options for the distribution of prizes exist if for all nominations 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
Task 9. The gardener must plant 6 trees within three days. In how many ways can he distribute his work over the days if he plants at least one tree a day?
Solution. Suppose a gardener plants trees in a row and can take various solutions regarding which tree to stop on the first day and which tree to stop 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 must 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). Therefore, the number of ways .
Problem 10. How many four-digit numbers (possibly starting with zero) are there whose digits add up to 5?
Solution. Let's imagine the number 5 as a sum of consecutive ones, divided into groups by partitions (each group in total forms the next digit of the number). It is clear that 3 such partitions will be needed. There are 6 places for partitions (before all units, between them and after). Each place can be occupied by one or more partitions (in the latter case there are no units between them, and the corresponding sum is zero). Let's consider these places as elements of a set. Thus, you need to select 3 elements out of 6 (with repetitions). Therefore, the required number of numbers 
Problem 11 . In how many ways can a group of 25 students be divided 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. 3 fruits are selected at random. What is the probability that all three fruits are oranges?
Solution. The elementary outcomes here are sets that include 3 fruits. Since the order of the fruits is indifferent, we will consider their choice to be unordered (and non-repetitive)..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, i.e.. gif" width="161 height=83" height="83">.
Problem 2 . The teacher asks each of the three students to think of any number from 1 to 10. Assuming that each student’s choice of any given number is equally possible, find the probability that one of them will have the same number.
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 also has n3=10 possibilities. By virtue of the multiplication rule, the total number of ways is equal to: n= n1´n2´n3=103 = 1000, i.e. the entire space contains 1000 elementary outcomes. To calculate the probability of event A, it is convenient to move on to the opposite event, i.e., count the number of cases when all three students think 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 matches. Therefore, the desired probability is equal to P = 280/1000 = 0.28.
Problem 3 . Find the probability that in an 8-digit number exactly 4 digits are the same and the rest are different.
Solution. Event A=(an eight-digit number contains 4 same numbers). From the conditions of the problem it follows that the number contains five different digits, one of them is repeated. The number of ways to select 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. The total number of ways to compose 8-digit numbers is |W|=108 The required probability is equal to
Problem 4 . Six clients randomly contact 5 firms. Find the probability that no one will contact at least one company.
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 companies. Hence 
. Hence, .
Problem 5 . Let there be N balls in an urn, of which M are white and N–M are black. n balls are drawn from the urn. Find the probability that there will be 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 ones, and, therefore, the required probability is equal to P(A) = https://pandia. ru/text/78/307/images/image031_2.gif" width="167" height="44">.
Problem 7 (meeting problem) . Two persons A and B agreed to meet at certain place between 12 and 1 p.m. The first person to arrive waits for the other person for 20 minutes and then leaves. What is the probability of a meeting between persons A and B, if the arrival of each of them can occur at random within the specified hour and the moments of arrival are independent?
Solution. Let us denote the moment of arrival of person A by x and person B by y. In order for the meeting to take place, it is necessary and sufficient that x-yô £20. Let's depict x and y as coordinates on a plane, choosing the minute as the scale unit. All possible outcomes are represented by the points of a square with a side of 60, 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 entire 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 pulled out at random. What is the probability that the buttons will be the same color? ?
Solution. The event A=(buttons of the same color are taken out) can be represented as a sum , where the events and mean the choice of buttons red and of blue color 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">
Problem 2 . Among the company’s employees, 28% speak English, 30% speak German, 42% know 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 company: 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 the events that a randomly selected employee of the company speaks English, German or French, respectively. Obviously, the proportion of company employees who speak certain languages determines the probabilities 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 eldest child is a boy if it is known that the family has children of both sexes?
Solution. Let A=(the eldest child is a boy), B=(the family has children of both sexes). Let us 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 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 the family has children of both sexes. The eldest child is a boy, therefore the second (youngest) child is a girl. This event AB 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 probability that he will check exactly two details?
Solution. Event A=(the master checked exactly two parts) means that during such a check the first part turned out to be non-standard, and the second was standard. This means, where =(the first part turned out to be non-standard) and =(the second part was standard). Obviously, 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 were non-standard and 7 were standard. By the multiplication theorem
Problem 5 . One box contains 3 white and 5 black balls, another box contains 6 white and 4 black balls. Find the probability that a white ball will be drawn from at least one box if one ball is drawn from each box.
Solution. The event A=(a white ball is taken out of at least one box) can be represented as a sum , where events mean the occurrence white ball from the first and second box, 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 examining 6 students, the second - 3 students, and the third - 21 students (students are selected randomly from a list). The attitude of the three examiners towards those who are poorly prepared is different: the chances of such students passing the exam with the first teacher are 40%, with the second - only 10%, and 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 examiner, respectively. According to the conditions of the problem
, 
, 
.
Let event A=(poorly prepared student passed the exam). Then again, due to the conditions of the problem
, 
, 
.
Using the total probability formula we get:
Problem 7 . The company has three sources of supply of components - companies A, B, C. Company A accounts for 50% of the total supply, B - 30% and C - 20%. It is known from practice that among the parts supplied by company A, 10% are defective, by company B - 5% and by company C - 6%. What is the probability that a part taken at random will be suitable?
Solution. Let event G be the appearance of a suitable part. The probabilities of the hypotheses that the part was supplied by companies A, B, C are equal to P(A)=0.5, P(B)=0.3, P(C)=0.2, respectively. The conditional probabilities of the appearance of a suitable part are equal to 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 part). Using the total probability formula we get:
P(G)=0.5×0.9+0.3×0.95+0.2×0.94=0.923.
Problem 8 (see task 6). Let it be known that the student did not pass the exam, i.e. received an “unsatisfactory” grade. Which of the three teachers was he most likely to answer? ?
Solution. The probability of getting a “failure” is equal to . You need 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, the poorly prepared student took the exam to a third examiner.
4. Repeated independent tests. Bernoulli's theorem
Problem 1 . The die is thrown 6 times. Find the probability that a six will be rolled exactly 3 times.
Solution. Rolling a die six times can be thought of as a sequence independent tests with a probability of success (“sixes”) equal to 1/6 and a probability of failure of 5/6. We calculate the required probability using the formula 
.
Problem 2 . The coin is tossed 6 times. Find the probability that the coat of arms will appear no more than 2 times.
Solution. The required probability is equal to the sum of the probabilities of three events, consisting in the fact that the coat of arms will not appear even once, or 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 tossed 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 in three coin tosses. Using Bernoulli's formula it is easy to find the probabilities of events Am (see table):
From this table it can be seen that the most probable values are the 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 probability 0.1. Find the most likely number of concluded contracts 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 integer solution, namely, m*=2.
Problem 6 . It is known that the defect 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 tests with a 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 the client will make a purchase exactly 80 times.
Solution. IN in this case n=100, m=80, p=0.75, q=0.25. We find 
, and determine j(x)=0.2036, then the required probability is equal to Р100(80)= 
.
Task 8. The insurance company 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. According to the task conditions, n=40000, p=0.02. We 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 400 independent trials is 0.8. Find a positive number e such that, with probability 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. According to the conditions 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 set of 3 keys, only one key fits the door. The keys are searched until a suitable key is found. Construct a distribution law for random variable x – number of tested keys .
Solution. The number of keys tried could be 1, 2 or 3. If only one key was tried, this means that this first key immediately matched the door, and the probability of such an event is 1/3. So, Next, if there were 2 tested keys, i.e. x=2, this means that the first key did not work, but the second 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: Problem 2
.
Construct the distribution function Fx(x) for the random variable x from Problem 1. Solution. The random variable x has three values 1, 2, 3, which divide the entire numerical 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 of x³3 the inequality x£x holds for all values of the random variable x, so P(x So we got the following function: Problem 3.
The joint distribution law of random variables x and h is given using the table Calculate the particular laws of distribution of the component quantities x and h. Determine whether 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 similarly: 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 random variables: Now let’s answer the question about the independence of 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 of 1/ 16, and the product of the corresponding partial probabilities 1/4×1/4 is equal to 1/16, i.e. coincides with joint probability. This condition is also tested 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 Let's mark the cells for which the condition https://pandia.ru/text/78/307/images/image092.gif" width="574" height="23 src="> Problem 4
.
Let the random variable ξ have the following distribution law: Calculate expected value Mx, variance Dx and average standard deviation s. Solution. By definition, the mathematical expectation of x is equal to 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 problem the mathematical expectation was already calculated .
It remains to calculate And .
Using the partial distribution laws obtained in solving Problem 3, we obtain ; ;
and that means which was to be expected due to the independence of the random variables. Task 7.
The random vector (x, h) takes values (0,0), (1,0), (–1,0), (0,1) and (0,–1) equally likely. Calculate the covariance of random variables x and h. Show that they are dependent. Solution. Since P(x=0)=3/5, P(x=1)=1/5, P(x=–1)=1/5; Р(h=0)=3/5, P(h=1)=1/5, P(h=–1)=1/5, then Мx=3/5´0+1/5´1+1 /5´(–1)=0 and Мh=0; М(xh)=0´0´1/5+1´0´1/5–1´0´1/5+0´1´1/5–0´1´1/5=0. We obtain cov(x, h)=М(xh)–МxМh=0, and the random variables are uncorrelated. However, they are dependent. 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 probability P(h=0)=3/5, or the probability (ξ=0,η =0) is not equal to the product of probabilities: Р(x=0,h=0)=1/5¹Р(x=0)Р(h=0)=9/25. Therefore, x and h are dependent. Problem 8
. Random increments in stock prices of two companies on day x and h have 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 laws of distribution of x and h: We define 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 Task 9.
Random increments in stock prices of two companies per day have variances Dx=1 and Dh=2, and their correlation coefficient r=0.7. Find the variance of the price increment of a portfolio of 5 shares of the first company and 3 shares of the second company. Solution. Using the properties of dispersion, covariance and the definition 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 expectation h at x=1. Solution. The conditional mathematical expectation is From the conditions of the problem we find the distribution of the components h and x (last column and last line tables). Brought to date in open jar Unified State Examination problems in mathematics (mathege.ru), the solution of which is based on only one formula, which is classic definition probabilities. The easiest way to understand the formula is with examples. A comment. In problems in probability theory, something happens (in this case, our action of pulling out the ball) that can have a different result - an outcome. It should be noted that the result can be looked at in different ways. “We pulled out some kind of ball” is also a result. “We pulled out the blue ball” - the result. “We pulled out exactly this ball from all possible balls” - this least generalized view of the result is called an elementary outcome. It is the elementary outcomes that are meant in the formula for calculating the probability. Solution. Now let's calculate the probability of choosing the blue ball. For the same problem, let's calculate the probability of choosing a red ball. The probability of any event always lies between 0 and 1. We have reviewed classic example, illustrating the definition of probability. All similar Unified State Examination tasks According to probability theory, they are solved by using this formula. They differ slightly in the formulation of the problem of the probability theory of the Unified State Examination, where you need to calculate the probability of some event occurring on a certain day. ( , ) As in previous tasks you need to determine what is the elementary outcome, and then apply the same formula. Example 2. The conference lasts three days. On the first and second days there are 15 speakers, on the third day - 20. What is the probability that Professor M.’s report will fall on the third day if the order of reports is determined by drawing lots? What is the elementary outcome here? – Assigning a professor’s report one of all possible serial numbers for a performance. 15+15+20=50 people participate in the draw. Thus, Professor M.'s report may receive one of 50 issues. This means there are only 50 elementary outcomes. The drawing of lots here represents the establishment of a random correspondence between people and ordered places. In example 2, the establishment of correspondence was considered from the point of view of which of the places could be taken special person. You can approach the same situation from the other side: which of the people with what probability could get caught? specific place(prototypes , , , ): Example 3. The draw includes 5 Germans, 8 French and 3 Estonians. What is the probability that the first (/second/seventh/last – it doesn’t matter) will be a Frenchman. Number of elementary outcomes – number of all possible people, which could, by drawing lots, get to this place. 5+8+3=16 people. The prototype is slightly different. There are still problems about coins () and dice (), which are somewhat more creative. The solution to these problems can be found on the prototype pages. Here are a few examples of tossing a coin or dice. Example 4. When we toss a coin, what is the probability of landing on heads? Example 5. What if we toss a coin twice? What is the probability of getting heads both times? What is the probability that two coin tosses will result in tails? In such problems, another formula may be useful. So, you can find the probability of getting 5 heads out of 5 coin tosses. The same is true for dice. With one throw, there are 6 possible results. So, for two throws: 6 6 = 36, for three 6 6 6 = 216, etc. Example 6. We throw the dice. What is the probability that an even number will be rolled? Total outcomes: 6, according to the number of sides. Example 7. We throw two dice. What is the probability that the total will be 10? (round to the nearest hundredth) For one die there are 6 possible outcomes. This means that for two, according to the above rule, 6·6=36. Other types of B6 problems will be discussed in a future How to Solve article. Probability. Tasks profile Unified State Examination mathematics. Prepared by the teacher mathematicians MBOU"Lyceum No. 4" Ruzaevka Ovchinnikova T.V. Definition of probability Probability
events A are called number ratio m
favorable outcomes for this event total number
n
all equally possible incompatible events that can occur as a result of one test or observation: m
n
Let k
– the number of coin tosses, then the number of possible outcomes: n=2
k
.
Let k
– the number of dice rolls, then the number of possible outcomes: n=6
k
.
In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will appear exactly once. Solution.
There are only 4 options: O; o o; p p; p p; O .
Favorable 2: O; R And R; O .
The probability is 2/4 = 1/2 = 0,5
.
Answer: 0.5. In a random experiment, two dice are rolled. Find the probability that the total will be 8 points. Round the result to hundredths. Solution.
Dice are cubes with 6 sides. The first die can roll 1, 2, 3, 4, 5 or 6 points. Each scoring option corresponds to 6 scoring options on the second die. Those. total different options 6×6 = 36. The options (experiment outcomes) will be as follows: 1; 1 1; 2 1; 3 1; 4 1; 5 1; 6
2; 1 2; 2 2; 3 2; 4 2; 5 2; 6
etc. ........................... 6; 1 6; 2 6; 3 6; 4 6; 5 6; 6
Let's count the number of outcomes (options) in which the sum of the points of two dice is 8. 2; 6 3; 5; 4; 4 5; 3 6; 2.
There are 5 options in total. Let's find the probability: 5/36 = 0.138 ≈ 0.14. Answer: 0.14. There are only 55 tickets in the collection of tickets for biology, 11 of them contain a question on botany. Find the probability that a student will get a question on botany on a randomly selected exam ticket. Solution:
The probability that a student will get a question on botany on a randomly selected exam ticket is 11/55 = 1/5 = 0.2. Answer: 0.2. 20 athletes are participating 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 athlete competing first is from China. Solution.
A total of 20 athletes are participating, of which 20 – 8 – 7 = 5 athletes from China. The probability that the athlete competing first will be from China is 5/20 = 1/4 = 0.25. Answer: 0.25. The scientific conference is held over 5 days. A total of 75 reports are planned - the first three days contain 17 reports, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by drawing lots. What is the probability that Professor M.'s report will be scheduled for the last day of the conference? Solution:
On the last day of the conference it is planned (75 – 17 × 3) : 2 = 12 reports. The probability that Professor M.'s report will be scheduled for the last day of the conference is 12/75 = 4/25 = 0.16. Answer: 0.16. Before the start of the first round of the badminton championship, participants are randomly divided into playing pairs using lots. In total, 26 badminton players are participating in the championship, including 10 participants from Russia, including Ruslan Orlov. Find the probability that in the first round Ruslan Orlov will play with any badminton player from Russia? Solution:
It must be taken into account that Ruslan Orlov must play with some badminton player from Russia. And Ruslan Orlov himself is also from Russia. The probability that in the first round Ruslan Orlov will play with any badminton player from Russia is 9/25 = 36/100 = 0.36. Answer: 0.36. Dasha throws the dice twice. She got a total of 8 points. Find the probability that on the first roll you get 2 points. Solution.
A total of 8 points should appear on the two dice. This is possible if there are the following combinations: There are 5 options in total. Let's count the number of outcomes (options) in which 2 points were obtained on the first throw. This is option 1. Let's find the probability: 1/5 = 0.2. Answer: 0.2. There are 20 teams participating in the World Championship. Using lots, they need to be divided into five groups of four teams each. There are cards with group numbers mixed in the box: 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5.
Team captains draw one card each. What is the probability that the Russian team will be in the third group. Solution:
There are 20 teams in total, 5 groups. Each group has 4 teams. So, there are 20 total outcomes, the ones we need are 4, which means the probability of getting the desired outcome is 4/20 = 0.2. Answer: 0.2. Two factories produce the same glass for car headlights. The first factory produces 45% of these glasses, the second – 55%. The first factory produces 3% of defective glass, and the second – 1%. Find the probability that glass accidentally purchased in a store will be defective. Solution:
The probability that the glass was purchased at the first factory and is defective: R 1
= 0.45 · 0.03 = 0.0135. The probability that the glass was purchased from a second factory and is defective: R 2
=
0.55 · 0.01 = 0.0055. Therefore, according to the total probability formula, the probability that glass accidentally purchased in a store will be defective is equal to p = p 1
+ p 2
= 0,0135 + 0,0055 = 0,019.
Answer: 0.019. If grandmaster A. plays white, then he wins against grandmaster B. with probability 0.52. If A. plays black, then A. wins against B. with probability 0.3. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times. Solution:
The possibility of winning the first and second games does not depend on each other. The probability of a product of independent events is equal to the product of their probabilities: p = 0.52 · 0.3 = 0.156. Answer: 0.156. A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hits the targets the first three times and misses the last two times. Round the result to hundredths. Solution:
The result of each next shot does not depend on the previous ones. Therefore, the events “hit on the first shot,” “hit on the second shot,” etc. independent. The probability of each hit is 0.8. This means that the probability of a miss is 1 – 0.8 = 0.2. 1 shot: 0.8 2 shot: 0.8 3 shot: 0.8 4 shot: 0.2 5 shot: 0.2 Using the formula for multiplying the probabilities of independent events, we find that the desired probability is equal to: 0,8 ∙ 0,8 ∙ 0,8 ∙ 0,2 ∙ 0,2 = 0,02048 ≈ 0,02.
Answer: 0.02. There are two payment machines in the store. Each of them can be faulty with probability 0.05, regardless of the other machine. Find the probability that at least one machine is working. Solution:
Let's find the probability that both machines are faulty. These events are independent, the probability of their occurrence is equal to the product of the probabilities of these events: 0.05 · 0.05 = 0.0025. An event consisting in the fact that at least one machine is working, the opposite. Therefore, its probability is equal to 1 − 0,0025 = 0,9975.
Answer: 0.9975. Cowboy John has a 0.9 chance of hitting a fly on the wall if he fires a zeroed revolver. If John fires an unfired revolver, he hits the fly with probability 0.2. There are 10 revolvers on the table, only 4 of which have been shot. Cowboy John sees a fly on the wall, randomly grabs the first revolver he comes across and shoots the fly. Find the probability that John misses. Solution:
The probability that John will miss if he grabs a zeroed revolver is: 0.4 (1 − 0.9) = 0.04 The probability that John will miss if he grabs an unfired revolver is: 0.6 · (1 − 0.2) = 0.48 These events are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: 0,04 + 0,48 = 0,52.
Answer: 0.52. During artillery fire, the automatic system fires a shot at the target. If the target is not destroyed, the system fires a second shot. Shots are repeated until the target is destroyed. The probability of destroying a certain target with the first shot is 0.4, and with each subsequent shot it is 0.6. How many shots will be required to ensure that the probability of destroying the target is at least 0.98? Solution:
You can solve the problem “by action”, calculating the probability of surviving after a series of consecutive mistakes: P(1) = 0.6; P(2) = P(1) 0.4 = 0.24; P(3) = P(2) 0.4 = 0.096; P(4) = P(3) 0.4 = 0.0384; P(5) = P(4) 0.4 = 0.01536. The latter probability is less than 0.02, so five shots at the target is enough. Answer: 5. There are 26 people in the class, among them two twins - Andrey and Sergey. The class is randomly divided into two groups of 13 people each. Find the probability that Andrey and Sergey will be in the same group. Solution:
Let one of the twins be in some group. Together with him, 12 people from the 25 remaining classmates will be in the group. The probability that the second twin will be among these 12 people is P = 12: 25 = 0.48. Answer: 0.48. The picture shows a labyrinth. The spider crawls into the maze at the Entrance point. The spider cannot turn around and crawl back, so at each branch the spider chooses one of the paths along which it has not yet crawled. Assuming that the choice of the further path is purely random, determine with what probability the spider will come to exit D. Solution:
At each of the four marked forks, the spider can choose either the path leading to exit D or another path with probability 0.5. These are independent events, the probability of their occurrence (the spider reaches exit D) is equal to the product of the probabilities of these events. Therefore, the probability of arriving at exit D is (0.5) 4
= 0,0625.
. Hence, 
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. Namely, in each cell of the table we multiply the corresponding values and , multiply the result by the probability pij, and sum it all up across all cells of the table. As a result we get:
.
Example 1. There are 9 red balls and 3 blue balls in the basket. The balls differ only in color. We take out one of them at random (without looking). What is the probability that the ball chosen in this way will be blue?
Event A: “the selected ball turned out to be blue”
Total number of all possible outcomes: 9+3=12 (the number of all balls that we could draw)
Number of outcomes favorable for event A: 3 (the number of such outcomes in which event A occurred - that is, the number of blue balls)
P(A)=3/12=1/4=0.25
Answer: 0.25
The total number of possible outcomes will remain the same, 12. Number of favorable outcomes: 9. Probability sought: 9/12=3/4=0.75
Sometimes in everyday speech (but not in probability theory!) the probability of events is estimated as a percentage. The transition between math and conversational scores is accomplished by multiplying (or dividing) by 100%.
So,
Moreover, the probability is zero for events that cannot happen - incredible. For example, in our example this would be the probability of drawing a green ball from the basket. (The number of favorable outcomes is 0, P(A)=0/12=0, if calculated using the formula)
Probability 1 has events that are absolutely certain to happen, without options. For example, the probability that “the selected ball will be either red or blue” is for our task. (Number of favorable outcomes: 12, P(A)=12/12=1)
In place of the red and blue balls there may be apples and pears, boys and girls, learned and unlearned tickets, tickets containing and not containing a question on some topic (prototypes,), defective and high-quality bags or garden pumps (prototypes,) - the principle remains the same.
What are the favorable outcomes? - Those in which it turns out that the professor will speak on the third day. That is, the last 20 numbers.
According to the formula, probability P(A)= 20/50=2/5=4/10=0.4
Answer: 0.4
Favorable outcomes - French. 8 people.
Required probability: 8/16=1/2=0.5
Answer: 0.5
There are 2 outcomes – heads or tails. (it is believed that the coin never lands on its edge) A favorable outcome is tails, 1.
Probability 1/2=0.5
Answer: 0.5.
The main thing is to determine what elementary outcomes we will consider when tossing two coins. After tossing two coins, one of the following results can occur:
1) PP – both times it came up heads
2) PO – first time heads, second time heads
3) OP – heads the first time, tails the second time
4) OO – heads came up both times
There are no other options. This means that there are 4 elementary outcomes. Only the first one, 1, is favorable.
Probability: 1/4=0.25
Answer: 0.25
The number of elementary outcomes is the same, 4. Favorable outcomes are the second and third, 2.
Probability of getting one tail: 2/4=0.5
If during one toss of a coin possible options we have 2 results, then for two throws the results will be 2 2 = 2 2 = 4 (as in example 5), for three throws 2 2 2 = 2 3 = 8, for four: 2 2 2 2 =2 4 =16, ... for N throws possible results will be 2·2·...·2=2 N .
Total number of elementary outcomes: 2 5 =32.
Favorable outcomes: 1. (RRRRRR – heads all 5 times)
Probability: 1/32=0.03125
Favorable: 3 outcomes. (2, 4, 6)
Probability: 3/6=0.5
What outcomes will be favorable for the total to roll 10?
10 must be decomposed into the sum of two numbers from 1 to 6. This can be done in two ways: 10=6+4 and 10=5+5. This means that the following options are possible for the cubes:
(6 on the first and 4 on the second)
(4 on the first and 6 on the second)
(5 on the first and 5 on the second)
Total, 3 options. Required probability: 3/36=1/12=0.08
Answer: 0.08
