Attention to applicants! Several are discussed here Unified State Exam problems. The rest, more interesting, are in our free video. Watch and do!
We'll start with simple tasks and basic concepts of probability theory.
Random An event that cannot be accurately predicted in advance is called. It can either happen or not.
You won the lottery - a random event. You invited friends to celebrate your win, and on the way to you they got stuck in the elevator - also a random event. True, the master turned out to be nearby and freed the entire company in ten minutes - and this can also be considered a happy accident...
Our life is full of random events. About each of them we can say that it will happen with some probability. Most likely, you are intuitively familiar with this concept. Now we'll give mathematical definition probabilities.
Let's start from the very beginning simple example. You flip a coin. Heads or tails?
Such an action, which can lead to one of several results, is called in probability theory test.
Heads and tails - two possible outcome tests.
Heads will fall out in one case out of two possible. They say that probability that the coin will land on heads is .
Let's throw a dice. The die has six sides, so there are also six possible outcomes.
For example, you wished that three points would appear. This is one outcome out of six possible. In probability theory it will be called favorable outcome.
The probability of getting a three is equal (one favorable outcome out of six possible).
The probability of four is also
But the probability of a seven appearing is zero. After all, there is no edge with seven points on the cube.
The probability of an event is equal to the ratio of the number of favorable outcomes to the total number of outcomes.
Obviously, the probability cannot be greater than one.
Here's another example. There are apples in a bag, some of them are red, the rest are green. The apples do not differ in shape or size. You put your hand into the bag and take out an apple at random. The probability of drawing a red apple is equal to , and the probability of drawing a green apple is equal to .
The probability of getting a red or green apple is equal.
Let us analyze problems in probability theory included in the collections for preparation for the Unified State Exam.
. At the taxi company this moment free cars: red, yellow and green. One of the cars that happened to be closest to the customer responded to the call. Find the probability that a yellow taxi will come to her.
There are a total of cars, that is, one out of fifteen will come to the customer. There are nine yellow ones, which means that the probability of a yellow car arriving is equal to , that is.
. (Demo version) In the collection of tickets on biology of all tickets, in two of them there is a question about mushrooms. During the exam, the student receives one randomly selected ticket. Find the probability that this ticket will not contain a question about mushrooms.
Obviously, the probability of drawing a ticket without asking about mushrooms is equal to , that is.
. The Parents Committee purchased puzzles for graduation gifts for children. school year, including paintings by famous artists and images of animals. Gifts are distributed randomly. Find the probability that Vovochka will get a puzzle with an animal.
The problem is solved in a similar way.
Answer: .
. Athletes from Russia, from the USA, and the rest from China are participating in the gymnastics championship. The order in which the gymnasts perform is determined by lot. Find the probability that the last athlete to compete is from China.
Let's imagine that all the athletes simultaneously approached the hat and pulled out pieces of paper with numbers from it. Some of them will get number twenty. The probability that a Chinese athlete will pull it out is equal (since the athletes are from China). Answer: .
. The student was asked to name the number from to. What is the probability that he will name a number that is a multiple of five?
Every fifth a number from this set is divisible by . This means the probability is equal to .
A die is thrown. Find the probability of getting odd number points.
Odd numbers; - even. The probability of an odd number of points is .
Answer: .
. The coin is tossed three times. What is the probability of two heads and one tail?
Note that the problem can be formulated differently: three coins were thrown at the same time. This will not affect the decision.
How many possible outcomes do you think there are?
We toss a coin. This action has two possible outcomes: heads and tails.
Two coins - already four outcomes:
Three coins? That's right, outcomes, since .
Two heads and one tails appear three out of eight times.
Answer: .
. In a random experiment, two dice are rolled. Find the probability that the total will be points. Round the result to hundredths.
We throw the first die - six outcomes. And for each of them six more are possible - when we throw the second die.
We find that this action - throwing two dice - has a total of possible outcomes, since .
And now - favorable outcomes:
The probability of getting eight points is .
>. The shooter hits the target with probability. Find the probability that he hits the target four times in a row.
If the probability of a hit is equal, then the probability of a miss is . We reason in the same way as in previous task. The probability of two hits in a row is equal. And the probability of four hits in a row is equal.
Probability: brute force logic.
Here is the problem from diagnostic work, which many found difficult.
Petya had coins worth rubles and coins worth rubles in his pocket. Petya, without looking, transferred some coins to another pocket. Find the probability that the five-ruble coins are now in different pockets.
We know that the probability of an event is equal to the ratio of the number of favorable outcomes to the total number of outcomes. But how to calculate all these outcomes?
You can, of course, designate five-ruble coins with numbers, and ten-ruble coins with numbers - and then count how many ways you can select three elements from the set.
However, there is a simpler solution:
We code the coins with numbers: , (these are five-ruble coins), (these are ten-ruble coins). The problem condition can now be formulated as follows:
There are six chips with numbers from to . In how many ways can they be distributed equally into two pockets, so that the chips with numbers do not end up together?
Let's write down what we have in our first pocket.
Let's put everything together for this possible combinations from the set. A set of three chips will be a three-digit number. Obviously, in our conditions and are the same set of chips. In order not to miss anything or repeat ourselves, we have the appropriate three digit numbers Ascending:
All! We went through all possible combinations starting with . Let's continue:
Total possible outcomes.
We have a condition - chips with numbers should not be together. This means, for example, that the combination does not suit us - it means that both chips ended up not in the first, but in the second pocket. Outcomes that are favorable for us are those where there is either only , or only . Here they are:
134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256 – total favorable outcomes.
Then the required probability is equal to .
What tasks await you on the Unified State Examination in mathematics?
Let's look at one of them complex tasks according to probability theory.
To enter the institute for the specialty "Linguistics", applicant Z. must score at least 70 points on the Unified State Exam in each of three items- mathematics, Russian language and foreign language. To enroll in the specialty "Commerce", you need to score at least 70 points in each of three subjects - mathematics, Russian language and social studies.
The probability that applicant Z. will receive at least 70 points in mathematics is 0.6, in Russian - 0.8, in 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 0.6 0.8.
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 to
1 – 0,5 0,3.
As a result, the probability of passing mathematics, Russian and social studies or foreign is equal
0.6 0.8 (1 - 0.5 0.3) = 0.408. This is the answer.
Lesson-lecture on the topic “probability theory”
Task No. 4 from the Unified State Exam 2016.
Profile level.
1 Group: assignments for use classical formula probabilities.
- Exercise 1. The taxi company has 60 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 lettering 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. Scientific Conference takes place 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 at least one of the occurrences 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. From district center There is a daily bus 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: Probability of simultaneous attack 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 in random moment time, both sellers are busy at the same time (consider 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 analysis gives positive result with probability 0.9. If the patient does not have hepatitis, the test may give a false positive result with a probability of 0.02. 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?
Classic definition of probability
Random event – any event that may or may not occur as a result of some experience.
Probability of event R equal to the ratio of the number of favorable outcomes k to the number of possible outcomes n, i.e.
p=\frac(k)(n)
Formulas for addition and multiplication of probability theory
Event \bar(A) called opposite to event A, if event A did not occur.
Sum of probabilities of opposite events is equal to one, i.e.
P(\bar(A)) + P(A) =1
- The probability of an event cannot be greater than 1.
- If the probability of an event is 0, then it will not happen.
- If the probability of an event is 1, then it will happen.
Probability addition theorem:
“The probability of the sum of two incompatible events is equal to the sum of the probabilities of these events.”
P(A+B) = P(A) + P(B)
Probability amounts two joint events equal to the sum of the probabilities of these events without taking into account their joint occurrence:
P(A+B) = P(A) + P(B) - P(AB)
Probability multiplication theorem
“The probability of the occurrence of two events is equal to the product of the probabilities of one of them by the conditional probability of the other, calculated under the condition that the first took place.”
P(AB)=P(A)*P(B)
Events are called incompatible, if the appearance of one of them excludes the appearance of others. That is, only one thing can happen specific event, or another.
Events are called joint, if the occurrence of one of them does not exclude the occurrence of the other.
Two random events A and B are called independent, if the occurrence of one of them does not change the probability of the occurrence of the other. IN otherwise events A and B are called dependent.
IN mall 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. A lot of interesting tasks on classic definition 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.
Problems in probability theory with solutions
1. Combinatorics
Problem 1 . There are 30 students in the group. It is necessary to choose a headman, a deputy headman and a trade union organizer. How many ways are there to do this?
Solution. Any of the 30 students can be chosen as a headman, any of the remaining 29 students can be chosen as a deputy, and any of the remaining 28 students can be chosen as a trade union organizer, i.e. n1=30, n2=29, n3=28. According to the multiplication rule total number N ways to select a headman, his deputy and a trade union leader are equal to N=n1´n2´n3=30´29´28=24360.
Problem 2 . Two postmen must deliver 10 letters to 10 addresses. In how many ways can they distribute the work?
Solution. The first letter has n1=2 alternatives - either it is taken to the addressee by the first postman, or by the second. For the second letter there are also n2=2 alternatives, etc., i.e. n1=n2=…=n10=2. Therefore, by virtue of the multiplication rule, the total number of ways to distribute letters between two postmen is equal to
Problem 3. There are 100 parts in the box, of which 30 are 1st grade parts, 50 are 2nd grade, the rest are 3rd grade. How many ways are there to remove one grade 1 or grade 2 part from a box?
Solution. A part of the 1st grade can be extracted in n1=30 ways, a part of the 2nd grade can be extracted in n2=50 ways. According to the sum rule, there are N=n1+n2=30+50=80 ways to extract one part of the 1st or 2nd grade.
Problem 5 . The order of performance of the 7 participants in the competition is determined by lot. How many various options Is it possible to draw lots in this case?
Solution. Each variant of the draw differs only in the order of the participants in the competition, i.e. it is a permutation of 7 elements. Their number is equal
Problem 6 . 10 films are participating in the competition in 5 nominations. How many options for prize distribution are there if the following rules are established for all categories? various awards?
Solution. Each of the prize distribution options is a combination of 5 films out of 10, differing from other combinations both in composition and in their order. Since each film can receive awards in one or several categories, the same films can be repeated. Therefore, the number of such combinations is equal to the number of placements with repetitions of 10 elements of 5:
Problem 7 . 16 people participate in a chess tournament. How many games must be played in a tournament if one game must be played between any two participants?
Solution. Each game is played by two participants out of 16 and differs from the others only in the composition of the pairs of participants, i.e., it is a combination of 16 elements of 2. Their number is equal to 
Problem 8 . In the conditions of task 6, determine how many options for the distribution of prizes exist if for all nominations the same prizes?
Solution. If the same prizes are established for each nomination, then the order of films in a combination of 5 prizes does not matter, and the number of options is the number of combinations with repetitions of 10 elements of 5, determined by the formula
Task 9. The gardener must plant 6 trees within three days. In how many ways can he distribute his work over the days if he plants at least one tree a day?
Solution. Suppose a gardener plants trees in a row and can take various solutions regarding which tree to stop on the first day and which tree to stop on the second. Thus, one can imagine that the trees are separated by two partitions, each of which can stand in one of 5 places (between the trees). The partitions must be there one at a time, because otherwise not a single tree will be planted on some day. Thus, you need to select 2 elements out of 5 (no repetitions). Therefore, the number of ways .
Problem 10. How many four-digit numbers (possibly starting with zero) are there whose digits add up to 5?
Solution. Let's imagine the number 5 as a sum of consecutive ones, divided into groups by partitions (each group in total forms the next digit of the number). It is clear that 3 such partitions will be needed. There are 6 places for partitions (before all units, between them and after). Each place can be occupied by one or more partitions (in the latter case there are no units between them, and the corresponding sum is zero). Let's consider these places as elements of a set. Thus, you need to select 3 elements out of 6 (with repetitions). Therefore, the required number of numbers 
Problem 11 . In how many ways can a group of 25 students be divided into three subgroups A, B and C of 6, 9 and 10 people respectively?
Solution. Here n=25, k=3, n1=6, n2=9, n3=10..gif" width="160" height="41">
Problem 1 . There are 5 oranges and 4 apples in a box. 3 fruits are selected at random. What is the probability that all three fruits are oranges?
Solution. The elementary outcomes here are sets that include 3 fruits. Since the order of the fruits is indifferent, we will consider their choice to be unordered (and non-repetitive)..gif" width="21" height="25 src=">. The number of favorable outcomes is equal to the number of ways to choose 3 oranges from the available 5, i.e.. gif" width="161 height=83" height="83">.
Problem 2 . The teacher asks each of the three students to think of any number from 1 to 10. Assuming that each student’s choice of any given number is equally possible, find the probability that one of them will have the same number.
Solution. First, let's calculate the total number of outcomes. The first student chooses one of 10 numbers and has n1=10 possibilities, the second also has n2=10 possibilities, and finally, the third also has n3=10 possibilities. By virtue of the multiplication rule, the total number of ways is equal to: n= n1´n2´n3=103 = 1000, i.e. the entire space contains 1000 elementary outcomes. To calculate the probability of event A, it is convenient to move on to the opposite event, i.e., count the number of cases when all three students think different numbers. The first one still has m1=10 ways to choose a number. The second student now has only m2=9 possibilities, since he has to take care that his number does not coincide with the intended number of the first student. The third student is even more limited in his choice - he has only m3=8 possibilities. Therefore, the total number of combinations of conceived numbers in which there are no matches is m=10×9×8=720. There are 280 cases in which there are matches. Therefore, the desired probability is equal to P = 280/1000 = 0.28.
Problem 3 . Find the probability that in an 8-digit number exactly 4 digits are the same and the rest are different.
Solution. Event A=(an eight-digit number contains 4 same numbers). From the conditions of the problem it follows that the number contains five different digits, one of them is repeated. The number of ways to select it is equal to the number of ways to select one digit from 10 digits..gif" width="21" height="25 src="> . Then the number of favorable outcomes. The total number of ways to compose 8-digit numbers is |W|=108 The required probability is equal to
Problem 4 . Six clients randomly contact 5 firms. Find the probability that no one will contact at least one company.
Solution. Consider the opposite event https://pandia.ru/text/78/307/images/image020_10.gif" width="195" height="41">. The total number of ways to distribute 6 clients across 5 companies. Hence 
. Hence, .
Problem 5 . Let there be N balls in an urn, of which M are white and N–M are black. n balls are drawn from the urn. Find the probability that there will be exactly m white balls among them.
Solution. Since the order of the elements is unimportant here, the number of all possible sets of volume n of N elements is equal to the number of combinations of m white balls, n–m black ones, and, therefore, the required probability is equal to P(A) = https://pandia. ru/text/78/307/images/image031_2.gif" width="167" height="44">.
Problem 7 (meeting problem) . Two persons A and B agreed to meet at a certain place between 12 and 13 o'clock. The first person to arrive waits for the other person for 20 minutes and then leaves. What is the probability of a meeting between persons A and B, if the arrival of each of them can occur at random within the specified hour and the moments of arrival are independent?
Solution. Let us denote the moment of arrival of person A by x and person B by y. In order for the meeting to take place, it is necessary and sufficient that x-yô £20. Let's depict x and y as coordinates on a plane, choosing the minute as the scale unit. All possible outcomes are represented by the points of a square with a side of 60, and those favorable to the meeting are located in the shaded area. The desired probability is equal to the ratio of the area of the shaded figure (Fig. 2.1) to the area of the entire square: P(A) = (602–402)/602 = 5/9.
3. Basic formulas of probability theory
Problem 1 . There are 10 red and 5 blue buttons in the box. Two buttons are pulled out at random. What is the probability that the buttons will be the same color? ?
Solution. The event A=(buttons of the same color are taken out) can be represented as a sum , where the events and mean the choice of buttons red and of blue color respectively. The probability of pulling out two red buttons is equal, and the probability of pulling out two blue buttons https://pandia.ru/text/78/307/images/image034_2.gif" width="19 height=23" height="23">.gif" width="249" height="83">
Problem 2 . Among the company’s employees, 28% speak English, 30% speak German, 42% know French; English and German – 8%, English and French – 10%, German and French – 5%, all three languages – 3%. Find the probability that a randomly selected employee of the company: a) knows English or German; b) knows English, German or French; c) does not know any of the listed languages.
Solution. Let us denote by A, B and C the events that a randomly selected employee of the company speaks English, German or French, respectively. Obviously, the proportion of company employees who speak certain languages determines the probabilities of these events. We get:
a) P(AÈB)=P(A)+P(B) -P(AB)=0.28+0.3-0.08=0.5;
b) P(AÈBÈC)=P(A)+P(B)+P(C)-(P(AB)+P(AC)+P(BC))+P(ABC)=0.28+0, 3+0.42-
-(0,08+0,1+0,05)+0,03=0,8;
c) 1-P(AÈBÈC)=0.2.
Problem 3 . The family has two children. What is the probability that the eldest child is a boy if it is known that the family has children of both sexes?
Solution. Let A=(the eldest child is a boy), B=(the family has children of both sexes). Let us assume that the birth of a boy and the birth of a girl are equally probable events. If the birth of a boy is denoted by the letter M, and the birth of a girl by D, then the space of all elementary outcomes consists of four pairs: . In this space, only two outcomes (MD and DM) correspond to event B. Event AB means that the family has children of both sexes. The eldest child is a boy, therefore the second (youngest) child is a girl. This event AB corresponds to one outcome – MD. Thus, |AB|=1, |B|=2 and
Problem 4 . The master, having 10 parts, of which 3 are non-standard, checks the parts one by one until he comes across a standard one. What is the probability that he will check exactly two details?
Solution. Event A=(the master checked exactly two parts) means that during such a check the first part turned out to be non-standard, and the second was standard. This means, where =(the first part turned out to be non-standard) and =(the second part was standard). Obviously, the probability of event A1 is also equal to 
, since before taking the second part the master had 9 parts left, of which only 2 were non-standard and 7 were standard. By the multiplication theorem
Problem 5 . One box contains 3 white and 5 black balls, another box contains 6 white and 4 black balls. Find the probability that a white ball will be drawn from at least one box if one ball is drawn from each box.
Solution. The event A=(a white ball is taken out of at least one box) can be represented as a sum , where events mean the occurrence white ball from the first and second box, respectively..gif" width="91" height="23">..gif" width="20" height="23 src=">.gif" width="480" height="23 ">.
Problem 6 . Three examiners take an exam in a certain subject from a group of 30 people, with the first examining 6 students, the second - 3 students, and the third - 21 students (students are selected randomly from a list). The attitude of the three examiners towards those who are poorly prepared is different: the chances of such students passing the exam with the first teacher are 40%, with the second - only 10%, and with the third - 70%. Find the probability that a poorly prepared student will pass the exam .
Solution. Let us denote by the hypotheses that the poorly prepared student answered the first, second and third examiner, respectively. According to the conditions of the problem
, 
, 
.
Let event A=(poorly prepared student passed the exam). Then again, due to the conditions of the problem
, 
, 
.
According to the formula full probability we get:
Problem 7 . The company has three sources of supply of components - companies A, B, C. Company A accounts for 50% of the total supply, B - 30% and C - 20%. It is known from practice that among the parts supplied by company A, 10% are defective, by company B - 5% and by company C - 6%. What is the probability that a part taken at random will be suitable?
Solution. Let event G be the appearance of a suitable part. The probabilities of the hypotheses that the part was supplied by companies A, B, C are equal to P(A)=0.5, P(B)=0.3, P(C)=0.2, respectively. The conditional probabilities of the appearance of a good part are equal to P(G|A)=0.9, P(G|B)=0.95, P(G|C)=0.94 (as the probabilities of opposite events to the appearance of a defective part). Using the total probability formula we get:
P(G)=0.5×0.9+0.3×0.95+0.2×0.94=0.923.
Problem 8 (see task 6). Let it be known that the student did not pass the exam, i.e. received an “unsatisfactory” grade. Which of the three teachers was he most likely to answer? ?
Solution. The probability of getting a “failure” is equal to . You need to calculate conditional probabilities. Using Bayes' formulas we get:
https://pandia.ru/text/78/307/images/image059_0.gif" width="183" height="44 src=">, 
.
It follows that, most likely, the poorly prepared student took the exam to a third examiner.
4. Repeated independent tests. Bernoulli's theorem
Problem 1 . Dice thrown 6 times. Find the probability that a six will be rolled exactly 3 times.
Solution. Rolling a die six times can be thought of as a sequence independent tests with a probability of success (“sixes”) equal to 1/6 and a probability of failure of 5/6. We calculate the required probability using the formula 
.
Problem 2 . The coin is tossed 6 times. Find the probability that the coat of arms will appear no more than 2 times.
Solution. The required probability is equal to the sum of the probabilities of three events, consisting in the fact that the coat of arms will not appear even once, or once, or twice:
P(A) = P6(0) + P6(1) + P6(2) = https://pandia.ru/text/78/307/images/image063.gif" width="445 height=24" height= "24">.
Problem 4 . The coin is tossed 3 times. Find the most probable number of successes (coat of arms).
Solution. Possible values for the number of successes in the three trials under consideration are m = 0, 1, 2 or 3. Let Am be the event that the coat of arms appears m times in three coin tosses. Using Bernoulli's formula it is easy to find the probabilities of events Am (see table):
From this table it can be seen that the most probable values are the numbers 1 and 2 (their probabilities are 3/8). The same result can be obtained from Theorem 2. Indeed, n=3, p=1/2, q=1/2. Then
, i.e. .
Task 5. As a result of each visit of the insurance agent, the contract is concluded with probability 0.1. Find the most likely number of concluded contracts after 25 visits.
Solution. We have n=10, p=0.1, q=0.9. The inequality for the most probable number of successes takes the form: 25×0.1–0.9£m*£25×0.1+0.1 or 1.6£m*£2.6. This inequality has only one integer solution, namely, m*=2.
Problem 6 . It is known that the defect rate for a certain part is 0.5%. The inspector checks 1000 parts. What is the probability of finding exactly three defective parts? What is the probability of finding at least three defective parts?
Solution. We have 1000 Bernoulli tests with a probability of “success” p=0.005. Applying the Poisson approximation with λ=np=5, we obtain
2) P1000(m³3)=1-P1000(m<3)=1-»1-,
and P1000(3)"0.14; Р1000(m³3)»0.875.
Problem 7 . The probability of a purchase when a customer visits a store is p=0.75. Find the probability that with 100 visits the client will make a purchase exactly 80 times.
Solution. In this case, n=100, m=80, p=0.75, q=0.25. We find 
, and determine j(x)=0.2036, then the required probability is equal to Р100(80)= 
.
Task 8. The insurance company concluded 40,000 contracts. The probability of an insured event for each of them during the year is 2%. Find the probability that there will be no more than 870 such cases.
Solution. According to the task conditions, n=40000, p=0.02. We find np=800,. To calculate P(m £ 870) we use the Moivre-Laplace integral theorem:
P(0
We find from the table of values of the Laplace function:
P(0 Problem 9
.
The probability of an event occurring in each of 400 independent trials is 0.8. Find a positive number e such that, with probability 0.99, the absolute value of the deviation of the relative frequency of occurrence of an event from its probability does not exceed e. Solution. According to the conditions of the problem, p=0.8, n=400. We use a corollary from the Moivre-Laplace integral theorem: 5.
Discrete random variables Problem 1
.
In a set of 3 keys, only one key fits the door. The keys are searched until a suitable key is found. Construct a distribution law for random variable x – number of tested keys .
Solution. The number of keys tried could be 1, 2 or 3. If only one key was tried, this means that this first key immediately matched the door, and the probability of such an event is 1/3. So, Next, if there were 2 tested keys, i.e. x=2, this means that the first key did not work, but the second did. The probability of this event is 2/3×1/2=1/3..gif" width="100" height="21"> The result is the following distribution series: Problem 2
.
Construct the distribution function Fx(x) for the random variable x from Problem 1. Solution. The random variable x has three values 1, 2, 3, which divide the entire numerical axis into four intervals: . If x<1, то неравенство x£x невозможно (левее x нет значений случайной величины x) и значит, для такого x функция Fx(x)=0. If 1£x<2, то неравенство x£x возможно только если x=1, а вероятность такого события равна 1/3, поэтому для таких x функция распределения Fx(x)=1/3. If 2£x<3, неравенство x£x означает, что или x=1, или x=2, поэтому в этом случае вероятность P(x And finally, in the case of x³3 the inequality x£x holds for all values of the random variable x, so P(x So we got the following function: Problem 3.
The joint distribution law of random variables x and h is given using the table Calculate the particular laws of distribution of the component quantities x and h. Determine whether they are dependent..gif" width="423" height="23 src=">; https://pandia.ru/text/78/307/images/image086.gif" width="376" height="23 src=">. The partial distribution for h is obtained similarly: https://pandia.ru/text/78/307/images/image088.gif" width="229" height="23 src=">. The obtained probabilities can be written in the same table opposite the corresponding values of random variables: Now let’s answer the question about the independence of random variables x and h..gif" width="108" height="25 src="> in this cell. For example, in the cell for the values x=-1 and h=1 there is a probability of 1/ 16, and the product of the corresponding partial probabilities 1/4×1/4 is equal to 1/16, i.e. coincides with joint probability. This condition is also tested in the remaining five cells, and it turns out to be true in all. Therefore, the random variables x and h are independent. Note that if our condition were violated in at least one cell, then the quantities should be recognized as dependent. To calculate the probability Let's mark the cells for which the condition https://pandia.ru/text/78/307/images/image092.gif" width="574" height="23 src="> Problem 4
.
Let the random variable ξ have the following distribution law: Calculate expected value Mx, variance Dx and average standard deviation s. Solution. By definition, the mathematical expectation of x is equal to Standard deviation https://pandia.ru/text/78/307/images/image097.gif" width="51" height="21">. Solution. Let's use the formula Problem 6
.
For a pair of random variables from Problem 3, calculate the covariance cov(x, h). Solution. In the previous problem the mathematical expectation was already calculated .
It remains to calculate And .
Using the partial distribution laws obtained in solving Problem 3, we obtain ; ;
and that means which was to be expected due to the independence of the random variables. Task 7.
The random vector (x, h) takes values (0,0), (1,0), (–1,0), (0,1) and (0,–1) equally likely. Calculate the covariance of random variables x and h. Show that they are dependent. Solution. Since P(x=0)=3/5, P(x=1)=1/5, P(x=–1)=1/5; Р(h=0)=3/5, P(h=1)=1/5, P(h=–1)=1/5, then Мx=3/5´0+1/5´1+1 /5´(–1)=0 and Мh=0; М(xh)=0´0´1/5+1´0´1/5–1´0´1/5+0´1´1/5–0´1´1/5=0. We obtain cov(x, h)=М(xh)–МxМh=0, and the random variables are uncorrelated. However, they are dependent. Let x=1, then conditional probability event (h=0) is equal to Р(h=0|x=1)=1 and is not equal to the unconditional Р(h=0)=3/5, or the probability (ξ=0,η=0) is not equal to the product of probabilities: Р(x=0,h=0)=1/5¹Р(x=0)Р(h=0)=9/25. Therefore, x and h are dependent. Problem 8
. Random increments in stock prices of two companies on day x and h have joint distribution, given by the table: Find the correlation coefficient. Solution. First of all, we calculate Mxh=0.3-0.2-0.1+0.4=0.4. Next, we find the particular laws of distribution of x and h: We define Mx=0.5-0.5=0; Mh=0.6-0.4=0.2; Dx=1; Dh=1–0.22=0.96; cov(x, h)=0.4. We get Task 9.
Random increments in stock prices of two companies per day have variances Dx=1 and Dh=2, and their correlation coefficient r=0.7. Find the variance of the price increment of a portfolio of 5 shares of the first company and 3 shares of the second company. Solution. Using the properties of dispersion, covariance and the definition of the correlation coefficient, we obtain: Problem 10
.
The distribution of a two-dimensional random variable is given by the table: Find the conditional distribution and conditional expectation h at x=1. Solution. The conditional mathematical expectation is From the conditions of the problem we find the distribution of the components h and x (last column and last line tables).
. Hence, 
..gif" width="587" height="41">
. Namely, in each cell of the table we multiply the corresponding values and , multiply the result by the probability pij, and sum it all up across all cells of the table. As a result we get:
.