Probability theory formulas and examples of problem solving. Simple problems in probability theory
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.B random experiment two dice are thrown. 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 total number outcomes is 6·6 = 36. Therefore, the probability that a total of 8 points will appear 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.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: 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, 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: 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 total probability formula: 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. At the geometry exam, the student gets one question from the list of 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.Probability that 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.Agricultural company purchases chicken eggs in 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 conditional probability formula, the probabilities of these events are equal to 0.4·0.9 = 0.36 and 0.6·0.2 = 0.12, respectively. 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 firing automatic system makes 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, a football team needs 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 tall passengers. Passenger V. is tall. Find the probability that at check-in, if a seat is randomly selected, passenger B. will get 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. The taxi company has 50 available passenger 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 arrive in the US within a year warranty repair, is equal to 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", an applicant must score at least 70 points on the Unified State Examination in each of three subjects - mathematics, Russian language and a 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 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 at a random moment in time all three sellers are busy at the same time (assume 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.A bus runs daily from the district center 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. These events are incompatible, 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 test gives a positive result with a probability of 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. These are 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 the hour hand freezes, reaching the 10 o'clock position, but not reaching the 1 o'clock position. 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. Assuming that the choice of the further path is 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. 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.
Plan for a workshop for mathematics teachers at the educational institution of the city of Tula on the topic “Solving Unified State Examination tasks in mathematics from the sections: combinatorics, probability theory. Teaching Methodology"
Time spending: 12 00 ; 15 00
Location: MBOU "Lyceum No. 1", office. No. 8
I. Solving probability problems
1. Solving problems involving the classical determination of probability
We, as teachers, already know that the main types of problems in the Unified State Exam in probability theory are based on the classical definition of probability. Let's remember what is called the probability of an event?
Probability of the event is the ratio of the number of outcomes favorable to a given event to the total number of outcomes.
Our scientific and methodological association of mathematics teachers has developed general scheme solving probability problems. I would like to present it to your attention. By the way, we shared our work experience, and in the materials that we gave to your attention for joint discussion of problem solving, we gave this diagram. However, I want to voice it.
In our opinion, this scheme helps to quickly logically sort everything into pieces, and after that the problem can be solved much easier for both the teacher and the students.
So, I want to analyze in detail the following task.
I wanted to talk with you together to explain the methodology, how to convey to the guys such a solution, during which the children would understand this typical problem, and subsequently they would understand these problems themselves.
What is a random experiment in this problem? Now we need to isolate an elementary event in this experiment. What is this elementary event? Let's list them.
Questions about the task?
Dear colleagues, you, too, have obviously considered probability problems with dice. I think we need to analyze it, because it has its own nuances. Let's analyze this problem according to the scheme that we proposed to you. Since on each side of the cube there is a number from 1 to 6, then the elementary events are the numbers 1, 2, 3, 4, 5, 6. We found that the total number of elementary events is 6. Let us determine which elementary events favor the event. Only two events favor this event - 5 and 6 (since it follows from the condition that 5 and 6 points should fall out).
Explain that all elementary events are equally possible. What questions will there be about the task?
How do you know that a coin is symmetrical? Let's get this straight, sometimes certain phrases cause misunderstandings. Let's understand this problem conceptually. Let's figure out with you in the experiment that is described what the elementary outcomes could be. Do you all have any idea where is heads and where is tails? What are the possible dropout options? Are there other events? What is the total number of events? According to the problem, it is known that heads came up exactly once. This means that this eventelementary events from these four ORs and ROs are favorable; this cannot happen twice. We use the formula that calculates the probability of an event. As a reminder, answers in Part B must be either a whole number or a decimal.
Show on interactive whiteboard. We read the problem. What is the elementary outcome in this experiment? Clarify that the pair is ordered - that is, the number fell on the first die and on the second die. In any problem there are moments when you need to choose rational methods, forms and present the solution in the form of tables, diagrams, etc. In this problem it is convenient to use such a table. I'm already giving it to you ready-made solution, but during the solution it turns out that in this problem it is rational to use a solution in the form of a table. We explain what the table means. You can understand why the columns say 1, 2, 3, 4, 5, 6.
Let's draw a square. The lines correspond to the results of the first throw - there are six of them, because the die has six sides. So are the columns. In each cell we write the sum of the points drawn. We show the completed table. Let's color the cells where the sum is equal to eight (as this is required in the condition).
I believe that the next problem, after analyzing the previous ones, can be given to the children to solve on their own.
In the following problems there is no need to write down all the elementary outcomes. It is enough to simply count their number.
(No solution) I gave this problem to the guys to solve on their own. Algorithm for solving the problem
1. Define what a random experiment consists of and what is a random event.
2. Find the total number of elementary events.
3. Find the number of events favorable to the event specified in the problem statement.
4. Find the probability of an event using the formula.
Students can be asked a question: if 1000 batteries go on sale, and among them 6 are faulty, then the selected battery is determined by how? What is it in our task? Next I ask the question of finding what is being used as a number hereand I suggest you find itnumber. Next I ask, what is the event here? How many accumulators contribute to the event? Next, using the formula, we calculate this probability.
Here the guys can be offered a second solution. Let's discuss what this method could be?
1. What event can we consider now?
2. How to find probability of this event?
The guys need to be told about these formulas. They are as follows
The eighth problem can be offered to the children on their own, since it is similar to the sixth problem. It can be offered to them as independent work, or on a card near the board.
This problem can be solved in relation to the Olympiad, which is currently taking place. Despite the fact that different events are involved in the tasks, the tasks are typical.
2. The simplest rules and formulas for calculating probabilities (opposite events, sum of events, product of events)
This is a task from the Unified State Exam collection. We display the solution on the board. What questions should we ask students to understand this problem?
1. How many machines were there? If there are two machines, then there are already two events. I ask the children a question - what will the event be like?? What will be the second event?
2. is the probability of an event. We don't need to calculate it, since it is given in the condition. According to the conditions of the problem, the probability that “the coffee will run out in both machines” is 0.12. There was event A, there was event B. And a new event appears? I ask the children a question - which one? This is the event when both machines run out of coffee. In this case, in probability theory, this is a new event, which is called the intersection of two events A and B and is designated in this way.
Let's use the probability addition formula. The formula is as follows
We give it to you in the reference material and the guys can be given this formula. It allows you to find the probability of a sum of events. We were asked the probability of the opposite event, the probability of which is found using the formula.
Problem 13 uses the concept of a product of events, the formula for finding the probability of which is given in the appendix.
3. Problems using wood possible options
Based on the conditions of the problem, it is easy to draw up a diagram and find the indicated probabilities.
What theoretical material did you use to help students solve problems of this kind? Have you used a possible tree or other methods to solve such problems? Have you given the concept of graphs? In the fifth or sixth grade, children have such problems, the analysis of which gives the concept of graphs.
I would like to ask you, have you and your students considered using a tree of possible options when solving probability problems? The fact is that not only does the Unified State Exam have such tasks, but quite complex problems have appeared that we will now solve.
Let's discuss with you the methodology for solving such problems - if it coincides with my methodology, as I explain to the guys, then it will be easier for me to work with you, if not, then I will help you deal with this problem.
Let us discuss the events. What events in problem 17 can be isolated?
When constructing a tree on a plane, a point is designated, which is called the root of the tree. Next we begin to consider the eventsAnd. We will construct a segment (in probability theory it is called a branch). The condition states that the first factory produces 30% mobile phones this brand (which one? The one they produce), which means in this moment I ask the students, what is the probability that the first factory will produce phones of this brand, the ones they produce? Since the event is the release of a phone at the first factory, the probability of this event is 30% or 0.3. The remaining phones were produced at the second factory - we are building the second segment, and the probability of this event is 0.7.
Students are asked the question: what type of phone could be produced by the first factory? With or without defect. What is the probability that a phone produced by the first factory has a defect? The condition says that it is equal to 0.01. Question: What is the probability that the phone produced by the first factory does not have a defect? Since this event is opposite to the given one, its probability is equal.
You need to find the probability that the phone is defective. It could be from the first factory, or maybe from the second. Then we use the formula for adding probabilities and find that the entire probability is the sum of the probabilities that the phone with a defect is from the first factory, and that the phone with a defect is from the second factory. We will find the probability that the phone has a defect and was produced at the first factory using the product of probabilities formula, which is given in the appendix.
4. One of the most complex tasks from the Unified State Exam bank for probability
Let's look at, for example, No. 320199 from the FIPI Task Bank. This is one of the most difficult tasks in B6.
To enter the institute for the specialty "Linguistics", applicant Z. must score at least 70 points on the Unified State Examination in each of three subjects - mathematics, Russian language and a 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.
Note that the problem does not ask whether an applicant named Z. will study both linguistics and commerce at once and receive two diplomas. Here we need to find the probability that Z. will be able to enroll in at least one of these two specialties - that is, he will gain required amount points.
In order to enter at least one of the two specialties, Z. must score at least 70 points in mathematics. And in Russian. And also - social studies or foreign.
The probability for him to score 70 points in mathematics is 0.6.
The probability of scoring points in mathematics and Russian is equal.
Let's deal with foreign and social studies. The options that suit us are when the applicant has scored points in social studies, foreign studies, or both. The option is not suitable when he did not score any points in either language or “society”. This means that the probability of passing social studies or foreign language with at least 70 points is equal. As a result, the probability of passing mathematics, Russian and social studies or foreign is equal
This is the answer.
II . Solving combinatorial problems
1. Number of combinations and factorials
Let's briefly look at the theoretical material.
Expressionn ! reads as “en-factorial” and denotes the product of all natural numbers from 1 ton inclusive:n ! = 1 · 2 · 3 · ... ·n .
In addition, in mathematics, by definition, they believe that 0! = 1. Such an expression is rare, but still occurs in problems in probability theory.
Definition
Let there be objects (pencils, candies, whatever) from which you want to select exactly different objects. Then the number of options for such a choice is callednumber of combinations from elements by. This number is designated and calculated using a special formula.
Designation
What does this formula give us? In fact, almost no serious problem can be solved without it.
For a better understanding, let’s look at a few simple combinatorial problems:
Task
The bartender has 6 types of green tea. To perform a tea ceremony you need to submit green tea exactly 3 different varieties. In how many ways can the bartender fill an order?
Solution
Everything is simple here: there isn = 6 varieties to choose fromk = 3 varieties. The number of combinations can be found using the formula:
Answer
Substitute into the formula. We cannot solve all problems, but typical tasks We have written them out and they are presented to your attention.
Task
In a group of 20 students, you need to choose 2 representatives to speak at the conference. In how many ways can this be done?
Solution
Again, that's all we haven = 20 students, but you have to choosek = 2 students. Find the number of combinations:
Please note: the factors included in different factorials are marked in red. These multipliers can be painlessly reduced and thereby significantly reduce the overall amount of calculations.
Answer
190
Task
17 servers were delivered to the warehouse with various defects, which cost 2 times cheaper than normal servers. The director bought 14 such servers for the school, and used the money saved in the amount of 200,000 rubles to purchase other equipment. In how many ways can the director select defective servers?
Solution
The problem contains quite a lot of extra data that can be confusing. Most important facts: there is everythingn = 17 servers, and the director needsk = 14 servers. We count the number of combinations:
Multipliers that are being reduced are again indicated in red. In total, there were 680 combinations. In general, the director has plenty to choose from.
Answer
680
This task is tricky because there is extra data in this task. They confuse many students the right decision. There were 17 servers in total, and the director needed to choose 14. Substituting into the formula, we get 680 combinations.
2. Law of multiplication
Definition
Law of multiplication in combinatorics: the number of combinations (ways, combinations) in independent sets is multiplied.
In other words, let there beA ways to perform one action andB ways to perform another action. The path is also that these actions are independent, i.e. are not related to each other in any way. Then you can find the number of ways to perform the first and second actions using the formula:C = A · B .
Task
Petya has 4 coins of 1 ruble and 2 coins of 10 rubles. Petya, without looking, took from his pocket 1 coin with a face value of 1 ruble and another 1 coin with a face value of 10 rubles to buy a pen for 11 rubles. In how many ways can he choose these coins?
Solution
So, first Petya getsk = 1 coin fromn = 4 available coins with a face value of 1 ruble. The number of ways to do this isC 4 1 = ... = 4.
Then Petya reaches into his pocket again and takes outk = 1 coin fromn = 2 available coins with a face value of 10 rubles. Here the number of combinations is equal toC 2 1 = ... = 2.
Since these actions are independent, the total number of options is equal toC = 4 · 2 = 8.
Answer
Task
There are 8 white balls and 12 black balls in a basket. In how many ways can you get 2 white balls and 2 black balls from this basket?
Solution
Total in cartn = 8 white balls to choose fromk = 2 balls. It can be doneC 8 2 = ... = 28 different ways.
In addition, the cart containsn = 12 black balls, from which you must choose againk = 2 balls. The number of ways to do this isC 12 2 = ... = 66.
Since the choice of a white ball and the choice of a black ball are independent events, the total number of combinations is calculated according to the multiplication law:C = 28 · 66 = 1848. As you can see, there can be quite a lot of options.
Answer
1848
The law of multiplication shows how many ways a complex action can be performed that consists of two or more simple ones - provided that they are all independent.
3. Law of addition
If the law of multiplication operates with “isolated” events that do not depend on each other, then in the law of addition the opposite is true. It deals with mutually exclusive events that never happen at the same time.
For example, “Petya took 1 coin out of his pocket” and “Petya did not take out a single coin from his pocket” are mutually exclusive events, since it is impossible to take out one coin without taking out any.
Likewise, the events “Ball at random is white” and “Ball at random is black” are also mutually exclusive.
Definition
Law of addition in combinatorics: if two mutually exclusive actions can be performedA AndB methods accordingly, then these events can be combined. This will create a new event that you can executeX = A + B ways.
In other words, when combining mutually exclusive actions (events, options), the number of their combinations adds up.
We can say that the law of addition is a logical “OR” in combinatorics, when we are satisfied with any of the mutually exclusive options. Conversely, the law of multiplication is a logical “AND”, in which we are interested in the simultaneous execution of both the first and second actions.
Task
There are 9 black balls and 7 red balls in a basket. The boy takes out 2 balls of the same color. In how many ways can he do this?
Solution
If the balls are the same color, then there are few options: they are both either black or red. Obviously, these options are mutually exclusive.
In the first case, the boy has to choosek = 2 black balls fromn = 9 available. The number of ways to do this isC 9 2 = ... = 36.
Similarly, in the second case we choosek = 2 red balls fromn = 7 possible. The number of ways is equalC 7 2 = ... = 21.
All that remains is to find total ways. Since the options with black and red balls are mutually exclusive, according to the law of addition we have:X = 36 + 21 = 57.
Answer57
Task
The stall sells 15 roses and 18 tulips. A 9th grade student wants to buy 3 flowers for his classmate, and all the flowers must be the same. In how many ways can he make such a bouquet?
Solution
According to the condition, all flowers must be the same. This means we will buy either 3 roses or 3 tulips. Anyway,k = 3.
In the case of roses you will have to choose fromn = 15 options, so the number of combinations isC 15 3 = ... = 455. For tulipsn = 18, and the number of combinations isC 18 3 = ... = 816.
Since roses and tulips are mutually exclusive options, we work according to the law of addition. We get the total number of optionsX = 455 + 816 = 1271. This is the answer.
Answer
1271
Additional terms and restrictions
Very often, the text of the problem contains additional conditions that impose significant restrictions on the combinations of interest to us. Compare two sentences:
Set of 5 pens available different colors. In how many ways can you choose 3 pens to outline a drawing?
There is a set of 5 pens in different colors. In how many ways can you choose 3 pens for outlining a drawing if red must be among them?
In the first case, we have the right to take any colors we like - there are no additional restrictions. In the second case, everything is more complicated, since we are required to choose a red pen (it is assumed that it is in the original set).
Obviously, any restrictions sharply reduce the final number of options. Well, how can you find the number of combinations in this case? Just remember next rule:
Let there be a set ofn elements from which to choosek elements. When introducing additional restrictions on the numbern Andk decrease by the same amount.
In other words, if out of 5 pens you need to choose 3, and one of them should be red, then you will have to choose fromn = 5 − 1 = 4 elements eachk = 3 − 1 = 2 elements. So instead ofC 5 3 must be countedC 4 2 .
Now let's see how this rule works for specific examples:
Task
In a group of 20 students, including 2 excellent students, you must select 4 people to participate in the conference. In how many ways can these four be selected if excellent students must get to the conference?
Solution
So there is a group ofn = 20 students. But you just have to choosek = 4 of them. If there were no additional restrictions, then the number of options would be equal to the number of combinationsC 20 4 .
However, we were given additional condition: 2 excellent students must be among these four. So, according to the above rule, we reduce the numbersn Andk by 2. We have:
Answer
153
Task
Petya has 8 coins in his pocket, of which 6 are ruble coins and 2 are 10 ruble coins. Petya transfers some three coins to another pocket. In how many ways can Petya do this if it is known that both 10 ruble coins ended up in the other pocket?
Solution
So there isn = 8 coins. Petya shiftsk = 3 coins, 2 of which are ten-ruble coins. It turns out that out of 3 coins that will be transferred, 2 have already been fixed, so the numbersn Andk must be reduced by 2. We have:
Answer
III . Solving combined problems using formulas of combinatorics and probability theory
Task
Petya had 4 ruble coins and 2 ruble coins in his pocket. Petya, without looking, transferred some three coins to another pocket. Find the probability that both two-ruble coins are in the same pocket.
Solution
Suppose that both two-ruble coins actually ended up in the same pocket, then 2 options are possible: either Petya did not transfer them at all, or he transferred both at once.
In the first case, when two-ruble coins were not shifted, you will have to shift 3 ruble coins. Since there are 4 such coins in total, the number of ways to do this is equal to the number of combinations of 4 by 3:C 4 3 .
In the second case, when both two-ruble coins have been transferred, another ruble coin will have to be transferred. It must be chosen from 4 existing ones, and the number of ways to do this is equal to the number of combinations of 4 by 1:C 4 1 .
Now let's find the total number of ways to rearrange the coins. Since there are 4 + 2 = 6 coins in total, and you only need to choose 3 of them, the total number of options is equal to the number of combinations of 6 by 3:C 6 3 .
It remains to find the probability:
Answer
0,4
Show on the interactive whiteboard. Pay attention to the fact that, according to the conditions of the problem, Petya, without looking, put three coins in one pocket. In answering this question, we can assume that two two-ruble coins actually remained in one pocket. Refer to the formula for adding probabilities. Show the formula again.
Task
Petya had 2 coins of 5 rubles and 4 coins of 10 rubles in his pocket. Petya, without looking, transferred some 3 coins to another pocket. Find the probability that the five-ruble coins are now in different pockets.
Solution
To keep five-ruble coins in different pockets, you need to move only one of them. The number of ways to do this is equal to the number of combinations of 2 by 1:C 2 1 .
Since Petya shifted 3 coins in total, he will have to shift 2 more coins of 10 rubles each. Petya has 4 such coins, so the number of ways is equal to the number of combinations of 4 by 2:C 4 2 .
It remains to find how many options there are to transfer 3 coins out of 6 available. This quantity, as in the previous problem, is equal to the number of combinations of 6 by 3:C 6 3 .
We find the probability:
In the last step, we multiplied the number of ways to choose two-ruble coins and the number of ways to choose ten-ruble coins, since these events are independent.
Answer
0,6
So, coin problems have their own probability formula. It is so simple and important that it can be formulated as a theorem.
Theorem
Let the coin be tossedn once. Then the probability that heads will land exactlyk times, can be found using the formula:
WhereC n k - number of combinations ofn elements byk , which is calculated by the formula:
Thus, to solve the coin problem, you need two numbers: the number of tosses and the number of heads. Most often, these numbers are given directly in the text of the problem. Moreover, it does not matter what exactly you count: tails or heads. The answer will be the same.
At first glance, the theorem seems too cumbersome. But once you practice a little, you will no longer want to return to the standard algorithm described above.
The coin is tossed four times. Find the probability of getting heads exactly three times.
Solution
According to the problem, the total throws weren = 4. Required number of eagles:k = 3. Substituten Andk into the formula:
You can just as easily count the number of heads:k = 4 − 3 = 1. The answer will be the same.
Answer
0,25
Task [ Workbook“USE 2012 in mathematics. Problems B6"]
The coin is tossed three times. Find the probability that you will never get heads.
Solution
Writing out the numbers againn Andk . Since the coin is tossed 3 times,n = 3. And since there shouldn’t be heads,k = 0. It remains to substitute the numbersn Andk into the formula:
Let me remind you that 0! = 1 by definition. That's whyC 3 0 = 1.
Answer
0,125
Task [ Trial Unified State Exam in mathematics 2012. Irkutsk]
In a random experiment, a symmetrical coin is tossed 4 times. Find the probability that heads will appear more times than tails.
Solution
For there to be more heads than tails, they must appear either 3 times (then there will be 1 tails) or 4 times (then there will be no tails at all). Let's find the probability of each of these events.
Letp 1 - the probability that heads will appear 3 times. Thenn = 4, k = 3. We have:
Now let's findp 2 - the probability that heads will appear all 4 times. In this casen = 4, k = 4. We have:
To get the answer, all that remains is to add up the probabilitiesp 1 Andp 2 . Remember: you can only add probabilities for mutually exclusive events. We have:
p = p 1 + p 2 = 0,25 + 0,0625 = 0,3125
Answer
0,3125
In order to save your time when preparing with the guys for the Unified State Exam and State Examination, we have presented solutions to many more problems that you can choose and solve with the guys.
Materials from the State Examination Institute, Unified State Examination of various years, textbooks and websites.
IV. Reference material
In a shopping center, two identical machines sell coffee. The machines are serviced in the evenings after the center closes. It is known that the probability of the event “By evening the first machine will run out of coffee” is 0.25. The probability of the event “By evening the second machine will run out of coffee” is the same. The probability that both machines will run out of coffee by evening is 0.15. Find the probability that by evening there will be coffee left in both machines.
Solution.
Consider the events
A = coffee will run out in the first machine,
B = coffee will run out in the second machine.
A B = coffee will run out in both machines,
A + B = coffee will run out in at least one machine.
By condition P(A) = P(B) = 0.25; P(A·B) = 0.15.
Events A and B are joint, the probability of the sum of two joint events is equal to the sum of the probabilities of these events, reduced by the probability of their product:
P(A + B) = P(A) + P(B) − P(A B) = 0.25 + 0.25 − 0.15 = 0.35.
Therefore, the probability of the opposite event, that the coffee will remain in both machines, is 1 − 0.35 = 0.65.
Answer: 0.65.
Let's give another solution.
The probability that the coffee will remain in the first machine is 1 − 0.25 = 0.75. The probability that the coffee will remain in the second machine is 1 − 0.25 = 0.75. The probability that coffee will remain in the first or second machine is 1 − 0.15 = 0.85. Since P(A + B) = P(A) + P(B) − P(A B), we have: 0.85 = 0.75 + 0.75 − X, where does the required probability come from? X = 0,65.
Note.
Note that events A and B are not independent. Indeed, the probability of producing independent events would be equal to the product of the probabilities of these events: P(A·B) = 0.25·0.25 = 0.0625, however, according to the condition, this probability is equal to 0.15.
Elena Alexandrovna Popova 10.10.2018 09:57
I, associate professor, candidate pedagogical sciences, I consider it COMPLETELY STUPID AND RIDICULOUS TO INCLUDE TASKS ON DEPENDENT EVENTS FOR SCHOOLCHILDREN. Teachers DO NOT KNOW this section - I was invited to give lectures on TV at teacher training courses. This section does not and cannot be in the program. There is NO NEED to invent methods without justification. TASKS of this kind can simply be eliminated. Limit yourself to the CLASSICAL DEFINITION OF PROBABILITIES. Yes, and then study it first school books- see what the authors wrote about this. Look at Zubareva's 5th grade. She doesn’t even know the symbols and gives the probability as a percentage. After learning from such textbooks, students still believe that probability is a percentage. There are many interesting problems on the classical determination of probabilities. This is what schoolchildren need to ask. There is no limit to the indignation of university teachers at YOUR stupidity in introducing such tasks.
Lesson-lecture on the topic “probability theory”
Task No. 4 from the Unified State Exam 2016.
Profile level.
1 Group: tasks on using the classical probability formula.
- Exercise 1. The taxi company has 60 cars; 27 of them are black with yellow inscriptions on the sides, the rest are yellow with black inscriptions. Find the probability that a yellow car with black inscriptions will respond to a random call.
- Task 2. Misha, Oleg, Nastya and Galya cast lots as to who should start the game. Find the probability that Galya will not start the game.
- Task 3. On average, out of 1000 garden pumps sold, 7 leak. Find the probability that one pump randomly selected for control does not leak.
- Task 4. There are only 15 tickets in the collection of tickets for chemistry, 6 of them contain a question on the topic “Acids”. Find the probability that a student will get a question on the topic “Acids” on a randomly selected exam ticket.
- Task 5. 45 athletes are competing at the diving championship, including 4 divers from Spain and 9 divers from the USA. The order of performances is determined by drawing lots. Find the probability that a US jumper will be twenty-fourth.
- Task 6. The scientific conference is held over 3 days. A total of 40 reports are planned - 8 reports on the first day, the rest are distributed equally between the second and third 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?
- Exercise 1. Before the start of the first round of the tennis championship, participants are randomly divided into playing pairs using lots. In total, 26 tennis players are participating in the championship, including 9 participants from Russia, including Timofey Trubnikov. Find the probability that in the first round Timofey Trubnikov will play with any tennis player from Russia.
- Task 2. Before the start of the first round of the badminton championship, participants are randomly divided into playing pairs using lots. A total of 76 badminton players are participating in the championship, including 22 athletes from Russia, including Viktor Polyakov. Find the probability that in the first round Viktor Polyakov will play with any badminton player from Russia.
- Task 3. There are 16 students in the class, among them two friends - Oleg and Mikhail. The class is randomly divided into 4 equal groups. Find the probability that Oleg and Mikhail will be in the same group.
- Task 4. There are 33 students in the class, among them two friends - Andrey and Mikhail. Students are randomly divided into 3 equal groups. Find the probability that Andrey and Mikhail will be in the same group.
- Exercise 1: In a ceramic tableware factory, 20% of the plates produced are defective. During product quality control, 70% 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.
- Task 2. At a ceramic tableware factory, 30% of the plates produced are defective. During product quality control, 60% of defective plates are identified. The remaining plates are on sale. Find the probability that a plate randomly selected during purchase has a defect. Round your answer to the nearest hundredth.
- Task 3: 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 will be defective.
2 Group: finding the probability of the opposite event.
- Exercise 1. The probability of hitting the center of the target from a distance of 20 m for a professional shooter is 0.85. Find the probability of missing the center of the target.
- Task 2. When manufacturing bearings with a diameter of 67 mm, the probability that the diameter will differ from the specified one by less 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.
3 Group: Finding the probability of the occurrence of at least one of the incompatible events. Formula for adding probabilities.
- Exercise 1. Find the probability that when throwing a die you will get 5 or 6 points.
- Task 2. There are 30 balls in an urn: 10 red, 5 blue and 15 white. Find the probability of drawing a colored ball.
- Task 3. The shooter shoots at a target divided into 3 areas. The probability of hitting the first area is 0.45, the second is 0.35. Find the probability that the shooter will hit either the first or the second area with one shot.
- Task 4. A bus runs daily from the district center to the village. The probability that there will be fewer than 18 passengers on the bus on Monday is 0.95. The probability that there will be fewer than 12 passengers is 0.6. Find the probability that the number of passengers will be from 12 to 17.
- Task 5. 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.
- Task 6. The probability that student U. will correctly solve more than 9 problems during a biology test is 0.61. The probability that U. will correctly solve more than 8 problems is 0.73. Find the probability that U will solve exactly 9 problems correctly.
4 Group: The probability of simultaneous occurrence of independent events. Probability multiplication formula.
- Exercise 1. 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.
- Task 2. The room is illuminated by a lantern with three 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.
- Task 3. There are two sellers in the store. Each of them is busy with a client with probability 0.4. Find the probability that at a random moment in time both sellers are busy at the same time (assume that customers come in independently of each other).
- Task 4. There are three sellers in the store. Each of them is busy with a client with probability 0.2. Find the probability that at a random moment in time all three sellers are busy at the same time (assume that customers come in independently of each other).
- Task 5: Based on customer reviews, Mikhail Mikhailovich assessed the reliability of the two online stores. The probability that the desired product will be delivered from store A is 0.81. The probability that this product will be delivered from store B is 0.93. Mikhail Mikhailovich 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.
- Task 6: If grandmaster A. plays white, then he wins against grandmaster B. with probability 0.6. If A. plays black, then A. wins against B. with probability 0.4. 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.
5 Group: Problems involving the use of both formulas.
- Exercise 1: 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 test gives a positive result with a probability of 0.9. If the patient does not have hepatitis, the test may give a false positive result with a probability of 0.02. It is known that 66% 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.
- Task 2. Cowboy John has a 0.9 chance of hitting a fly on the wall if he fires a zeroed revolver. If John fires an unsighted 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.
Task 3:
In some areas, observations showed:
1. If a June morning is clear, then the probability of rain on that day is 0.1. 2. If a June morning is cloudy, then the probability of rain during the day is 0.4. 3. The probability that the morning in June will be cloudy is 0.3.
Find the probability that there will be no rain on a random day in June.
Task 4. 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 0.3, and with each subsequent shot it is 0.9. How many shots will be required to ensure that the probability of destroying the target is at least 0.96?