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 .
Probability of getting red or green apple equal to .
Let us analyze problems in probability theory included in the collections for preparation for the Unified State Exam.
. The taxi company currently has 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 an odd number of 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 random experiment two dice are thrown. 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 the previous problem. 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 a problem from diagnostic work that many people 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.
To do this, we will compose all 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 arrange the corresponding three-digit numbers in ascending order:
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 us analyze one of the complex problems in probability theory.
To enter the institute for the specialty "Linguistics", applicant Z. must score at least 70 points on the Unified State Examination in each of three subjects - mathematics, Russian language and a foreign language. To enroll in the specialty "Commerce", you need to score at least 70 points in each of three subjects - mathematics, Russian language and social studies.
The probability that applicant Z. will receive at least 70 points in mathematics is 0.6, in Russian - 0.8, in 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.
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 specific event or another can happen.
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. Otherwise, events A and B are called dependent.
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?
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: .
