Events that happen in reality or in our imagination can be divided into 3 groups. These are certain events that are sure to happen, impossible events and random events. Probability theory studies random events, i.e. events that may or may not happen. This article will present in in brief probability theory formulas and examples of solving problems in probability theory that will be in task 4 of the Unified State Exam in mathematics (profile level).
Why do we need probability theory?
Historically, the need to study these problems arose in the 17th century in connection with the development and professionalization gambling and the emergence of casinos. This was a real phenomenon that required its own study and research.
Playing cards, dice, and roulette created situations where any of a finite number of equally possible events could occur. There was a need to give numerical estimates of the possibility of the occurrence of a particular event.
In the 20th century, it became clear that this seemingly frivolous science plays an important role in understanding the fundamental processes occurring in the microcosm. Was created modern theory probabilities.
Basic concepts of probability theory
The object of study of probability theory is events and their probabilities. If an event is complex, then it can be broken down into simple components, the probabilities of which are easy to find.
The sum of events A and B is called event C, which consists in the fact that either event A, or event B, or events A and B occurred simultaneously.
The product of events A and B is an event C, which means that both event A and event B occurred.
Events A and B are called incompatible if they cannot occur simultaneously.
An event A is called impossible if it cannot happen. Such an event is indicated by the symbol.
An event A is called certain if it is sure to happen. Such an event is indicated by the symbol.
Let each event A be associated with a number P(A). This number P(A) is called the probability of event A if the following conditions are met with this correspondence.
An important special case is the situation when there are equally probable elementary outcomes, and arbitrary of these outcomes form events A. In this case, the probability can be entered using the formula. Probability introduced in this way is called classical probability. It can be proven that in this case properties 1-4 are satisfied.
Probability theory problems that appear on the Unified State Examination in mathematics are mainly related to classical probability. Such tasks can be very simple. The probability theory problems in the demonstration versions are especially simple. It is easy to calculate the number of favorable outcomes; the number of all outcomes is written right in the condition.
We get the answer using the formula.
An example of a problem from the Unified State Examination in mathematics on determining probability
There are 20 pies on the table - 5 with cabbage, 7 with apples and 8 with rice. Marina wants to take the pie. What is the probability that she will take the rice cake?
Solution.
There are 20 equally probable elementary outcomes, that is, Marina can take any of the 20 pies. But we need to estimate the probability that Marina will take the rice pie, that is, where A is the choice of the rice pie. This means that the number of favorable outcomes (choices of pies with rice) is only 8. Then the probability will be determined by the formula:
Independent, Opposite and Arbitrary Events
However, in open jar More complex tasks began to be encountered. Therefore, let us draw the reader’s attention to other issues studied in probability theory.
Events A and B are said to be independent if the probability of each does not depend on whether the other event occurs.
Event B is that event A did not happen, i.e. event B is opposite to event A. The probability of the opposite event is equal to one minus the probability of the direct event, i.e. .
Probability addition and multiplication theorems, formulas
For arbitrary events A and B, the probability of the sum of these events is equal to the sum of their probabilities without the probability of their joint event, i.e. .
For independent events A and B, the probability of the occurrence of these events is equal to the product of their probabilities, i.e. in this case .
The last 2 statements are called the theorems of addition and multiplication of probabilities.
Counting the number of outcomes is not always so simple. In some cases it is necessary to use combinatorics formulas. The most important thing is to count the number of events that satisfy certain conditions. Sometimes these kinds of calculations can become independent tasks.
In how many ways can 6 students be seated in 6 free seats? The first student will take any of the 6 places. Each of these options corresponds to 5 ways for the second student to take a place. There are 4 free places left for the third student, 3 for the fourth, 2 for the fifth, and the sixth will take the only remaining place. To find the number of all options, you need to find the product, which is denoted by the symbol 6! and reads "six factorial".
In the general case, the answer to this question is given by the formula for the number of permutations of n elements. In our case.
Let us now consider another case with our students. In how many ways can 2 students be seated in 6 empty seats? The first student will take any of the 6 places. Each of these options corresponds to 5 ways for the second student to take a place. To find the number of all options, you need to find the product.
In general, the answer to this question is given by the formula for the number of placements of n elements over k elements
In our case .
And the last case in this series. In how many ways can you choose three students out of 6? The first student can be selected in 6 ways, the second - in 5 ways, the third - in four ways. But among these options, the same three students appear 6 times. To find the number of all options, you need to calculate the value: . In general, the answer to this question is given by the formula for the number of combinations of elements by element:
In our case .
Examples of solving problems from the Unified State Exam in mathematics to determine probability
Task 1. From the collection edited by. Yashchenko.
There are 30 pies on the plate: 3 with meat, 18 with cabbage and 9 with cherries. Sasha chooses one pie at random. Find the probability that he ends up with a cherry.
.
Answer: 0.3.
Task 2. From the collection edited by. Yashchenko.
In each batch of 1000 light bulbs, on average, 20 are defective. Find the probability that a light bulb taken at random from a batch will be working.
Solution: The number of working light bulbs is 1000-20=980. Then the probability that a light bulb taken at random from a batch will be working:
Answer: 0.98.
The probability that student U will solve more than 9 problems correctly during a math test is 0.67. 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.
If we imagine a number line and mark points 8 and 9 on it, then we will see that the condition “U. will solve exactly 9 problems correctly” is included in the condition “U. will solve more than 8 problems correctly”, but does not apply to the condition “U. will solve more than 9 problems correctly.”
However, the condition “U. will solve more than 9 problems correctly” is contained in the condition “U. will solve more than 8 problems correctly.” Thus, if we designate events: “U. will solve exactly 9 problems correctly" - through A, "U. will solve more than 8 problems correctly" - through B, "U. will correctly solve more than 9 problems” through C. That solution will look like this:
Answer: 0.06.
In a geometry exam, a student answers one question from a list of exam questions. The probability that this is a Trigonometry question is 0.2. The likelihood is that this is a question on the topic " External corners", is equal to 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.
Let's think about what events we have. We are given two incompatible events. That is, either the question will relate to the topic “Trigonometry” or to the topic “External angles”. According to the probability theorem, the probability of incompatible events is equal to the sum of the probabilities of each event, we must find the sum of the probabilities of these events, that is:
Answer: 0.35.
The room is illuminated by a lantern with three lamps. The probability of one lamp burning out within a year is 0.29. Find the probability that at least one lamp will not burn out during the year.
Let's consider possible events. We have three light bulbs, each of which may or may not burn out independently of any other light bulb. These are independent events.
Then we will indicate the options for such events. Let's use the following notations: - the light bulb is on, - the light bulb is burnt out. And immediately next we calculate the probability of the event. For example, the probability of an event in which three independent events “the light bulb is burned out”, “the light bulb is on”, “the light bulb is on” occurred: , where the probability of the event “the light bulb is on” is calculated as the probability of the event opposite to the event “the light bulb is not on”, namely: .
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 available passenger cars; 27 of them are black with yellow inscriptions on the sides, the rest are yellow color 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 identical glasses 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. Probability that new Electric kettle will last more than a year, is equal to 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 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 0.3, 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?
Probability theory on the Unified State Exam in mathematics can be presented both in the form of simple problems on the classical definition of probability, and in the form of quite complex ones on the application of the corresponding theorems.
In this part, we will consider problems for which it is sufficient to use the definition of probability. Sometimes here we will also use a formula to calculate the probability of the opposite event. Although you can do without this formula here, you will still need it when solving the following problems.
Theoretical part
Random is an event that may or may not occur (impossible to predict in advance) during an observation or test.
Let, when conducting a test (tossing a coin or dice, drawing exam card etc.) equally possible outcomes are possible. For example, when tossing a coin, the number of all outcomes is 2, since there cannot be any other outcomes other than heads or tails. When throwing a die, 6 outcomes are possible, since any number from 1 to 6 is equally possible to appear on the top face of the die. Let also some event A be favored by outcomes.
The probability of event A is the ratio of the number of outcomes favorable for this event to the total number of equally possible outcomes (this is the classical definition of probability). We write
For example, let event A consist of getting an odd number of points when throwing a die. There are a total of 6 possible outcomes: 1, 2, 3, 4, 5, 6 appearing on the top face of the cube. In this case, outcomes with 1, 3, 5 appearing are favorable for event A. Thus, 
.
Note that it always holds double inequality, therefore the probability of any event A lies on the interval, that is 
. If your answer has a probability greater than one, it means you made a mistake somewhere and the solution needs to be double-checked.
Events A and B are called opposite each other if any outcome is favorable for exactly one of them.
For example, when throwing a die, the event “an odd number is rolled” is the opposite of the event “an even number is rolled.”
The event opposite to event A is designated. From the definition of opposite events it follows 
, Means, 
.
Problems about selecting objects from a set
Task 1. There are 24 teams participating in the World Championship. Using lots, they need to be divided into four groups of six teams each. There are cards with group numbers mixed in the box:
1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4.
Team captains draw one card each. What is the probability that the Russian team will be in the third group?
The total number of outcomes is equal to the number of cards - there are 24 of them. There are 6 favorable outcomes (since number 3 is written on six cards). The required probability is equal to 
.
Answer: 0.25.
Task 2. There are 14 red, 9 yellow and 7 green balls in an urn. One ball is drawn at random from the urn. What is the probability that this ball will be yellow?
The total number of outcomes is equal to the number of balls: 14 + 9 + 7 = 30. The number of outcomes favorable for this event is 9. The required probability is equal to 
.
Task 3. There are 10 numbers on the phone keypad, from 0 to 9. What is the probability that a randomly pressed number will be even and greater than 5?
The outcome here is pressing a certain key, so there are a total of 10 equally possible outcomes. The specified event is favored by outcomes that mean pressing key 6 or 8. There are two such outcomes. The required probability is equal to .
Answer: 0.2.
Problem 4. What is the probability that a randomly selected natural number Is 4 to 23 divisible by three?
On the segment from 4 to 23 there are 23 – 4 + 1 = 20 natural numbers, which means there are a total of 20 possible outcomes. On this segment, the following numbers are multiples of three: 6, 9, 12, 15, 18, 21. There are 6 such numbers in total, so the event in question is favored by 6 outcomes. The required probability is equal to 
.
Answer: 0.3.
Task 5. Of the 20 tickets offered in the exam, the student can answer only 17. What is the probability that the student will not be able to answer the ticket chosen at random?
1st method.
Since a student can answer 17 tickets, he cannot answer 3 tickets. The probability of getting one of these tickets is by definition equal to .
2nd method.
Let us denote by A the event “the student can answer the ticket.” Then . The probability of the opposite event is =1 – 0.85 = 0.15.
Answer: 0.15.
Problem 6. 20 athletes are participating in the rhythmic gymnastics championship: 6 from Russia, 5 from Germany, the rest from France. The order in which the gymnasts perform is determined by lot. Find the probability that the athlete competing seventh is from France.
There are 20 athletes in total, everyone has an equal chance to compete seventh. Therefore, there are 20 equally probable outcomes. There are 20 – 6 – 5 = 9 athletes from France, so there are 9 favorable outcomes for the specified event. The required probability is equal to .
Answer: 0.45.
Task 7. The scientific conference is held over 5 days. A total of 50 reports are planned - the first three days have 12 reports each, 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 N.’s report will be scheduled for the last day of the conference?
First, let's find how many reports are scheduled for the last day. Presentations are scheduled for the first three days. There are still 50 – 36 = 14 reports left, which are distributed equally between the remaining two days, so there are reports scheduled on the last day.
We will consider the outcome to be the serial number of Professor N.’s report. There are 50 such equally possible outcomes. There are 7 outcomes that favor the specified event (the last 7 numbers in the list of reports). The required probability is equal to .
Answer: 0.14.
Problem 8. There are 10 seats on board the aircraft next to the emergency exits and 15 seats behind the partitions separating the cabins. The remaining seats are inconvenient for tall passengers. Passenger K. is tall. Find the probability that at check-in, if a seat is randomly selected, passenger K will get a comfortable seat if there are 200 seats on the plane.
The outcome in this task is the choice of location. There are a total of 200 equally possible outcomes. The event “the chosen place is convenient” is favored by 15 + 10 = 25 outcomes. The required probability is equal to .
Answer: 0.125.
Problem 9. Out of 1000 coffee grinders assembled at the plant, 7 were defective. An expert tests one coffee grinder chosen at random from these 1000. Find the probability that the coffee grinder being tested will be defective.
When choosing a coffee grinder at random, 1000 outcomes are possible; for event A “the selected coffee grinder is defective,” 7 outcomes are favorable. By definition of probability.
Answer: 0.007.
Problem 10. The plant produces refrigerators. On average, for every 100 quality refrigerators, there are 15 refrigerators with hidden defects. Find the probability that the purchased refrigerator will be of high quality. Round the result to hundredths.
This task is similar to the previous one. However, the formulation “for 100 high-quality refrigerators, there are 15 with defects” indicates to us that 15 defective pieces are not included in the 100 quality ones. Therefore, the total number of outcomes is 100 + 15 = 115 (equal to the total number of refrigerators), there are 100 favorable outcomes. The required probability is equal to . To calculate the approximate value of a fraction, it is convenient to use angle division. We get 0.869... which is 0.87.
Answer: 0.87.
Problem 11. Before the start of the first round of the tennis championship, participants are randomly divided into playing pairs using lots. In total, 16 tennis players are participating in the championship, including 7 participants from Russia, including Maxim Zaitsev. Find the probability that in the first round Maxim Zaitsev will play with any tennis player from Russia.
As in the previous task, you need to carefully read the condition and understand what is an outcome and what is a favorable outcome (for example, thoughtless application of the probability formula leads to an incorrect answer).
Here the outcome is the opponent of Maxim Zaitsev. Since there are 16 tennis players in total, and Maxim cannot play against himself, there are 16 – 1 = 15 equally probable outcomes. A favorable outcome is an opponent from Russia. There are 7 – 1 = 6 such favorable outcomes (we exclude Maxim himself from the number of Russians). The required probability is equal to .
Answer: 0.4.
Problem 12. The football section is attended by 33 people, among them two brothers - Anton and Dmitry. Those attending the section are randomly divided into three teams of 11 people each. Find the probability that Anton and Dmitry will be on the same team.
We will form teams, sequentially placing players in empty seats, starting with Anton and Dmitry. First, let's place Anton on a randomly selected place from the free 33. Now we place Dmitry on the free place (we will consider the choice of a place for him to be the outcome). There are 32 free places in total (Anton has already taken one), so there are 32 possible outcomes in total. There are 10 empty spots left on the same team as Anton, so the “Anton and Dmitry on the same team” event is favored by 10 outcomes. The probability of this event is 
.
Answer: 0.3125.
Problem 13. A mechanical watch with a twelve-hour dial broke down at some point and stopped running. Find the probability that the hour hand is frozen, reaching 11 o'clock, but not reaching 2 o'clock.
Conventionally, the dial can be divided into 12 sectors, located between the marks of adjacent numbers (between 12 and 1, 1 and 2, 2 and 3, ..., 11 and 12). We will consider the outcome to be the stop of the clock hand in one of the indicated sectors. There are a total of 12 equally possible outcomes. This event is favored by three outcomes (sectors between 11 and 12, 12 and 1, 1 and 2). The required probability is equal to 
.
Answer: 0.25.
Summarize
After studying the material on solving simple problems in probability theory, I recommend completing the tasks for independent decision which we publish on our Telegram channel. You can also check whether they are completed correctly by entering your answers in the form provided.
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Source “Preparation for the Unified State Exam. Mathematics. Probability Theory.” Edited by F.F. Lysenko, S.Yu. Kulabukhova
V-6-2014 (all 56 prototypes from the Unified State Exam bank)
Be able to build and study 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 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.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. 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.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. In total, 35% of eggs receive the highest category. 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 you start football match The referee tosses 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, for these countries there are 6 ways of 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, 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 a comfortable seat if there are 300 seats in total 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 yellow car with black inscriptions will respond to a random call. 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 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 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 the Magic Land 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.
