Mathematics and us. Basic formulas of probability theory I will solve the Unified State Exam theory of probability profile level

Unified State Exam 2016. Mathematics. Probability theory. Workbook.

M.: 2016. - 64 p.

The mathematics workbook of the "Unified State Exam 2016. Mathematics" series is aimed at preparing high school students for successfully passing the unified state exam in mathematics in 2016 at the basic and specialized levels. The workbook presents tasks for one position of the control measuring materials of the Unified State Exam-2016. At various stages of training, the manual will help to provide a leveled approach to organizing repetition, to monitor and self-monitor knowledge on the topic “Probability Theory”. The workbook is focused on one academic year, but if necessary, it will allow you to quickly fill in the gaps in the graduate’s knowledge. The notebook is intended for high school students, mathematics teachers, and parents.

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CONTENT
From the editor of series 3
Introduction 4
Diagnostic work 1 6
Solutions to diagnostic work problems 1 10
Training work 1 (for task D1.1) 22
Training work 2 (for tasks D1.2, D.1.4) 24
Training work 3 (for tasks D1.3, D1.5) 26
Training work 4 (for tasks D1.1-D1.5) 28
Training work 5 (for problems D1.6-D1.9) 30
Training work 6 (for tasks D1.6-D1.9) 32
Training work 7 (for tasks D1.6-D1.9) 34
Training work 8 (for problems D1.10-D1.14) 36
Training work 9 (for problems D1.10-D1.14) 39
Training work 10 (for problems D1.10-D1.14) 41
Training work 11 (for problems D1.15-D.18) 43
Training work 12 (for problems D1.15-D.18) 45
Diagnostic work 2 47
Diagnostic work 3 51
Diagnostic work 4 54
References 57
Replies 58

This manual is intended to prepare for the task on probability theory of the unified state exam (task 4 of the profile level and task 10 basic level in the 2016 version).
The manual consists of diagnostic work D1 with analysis of decisions, ten training works and three additional diagnostic works D2-D4, intended for intermediate control. At the end of the collection, answers to all problems are given.
Due to the fact that the tasks of the first part of the Unified State Examination in mathematics are formed using an open bank, the probability problems will also not be a surprise for the exam participants.
Probability theory is one of the most important applied branches of mathematics. Many phenomena in the world around us can only be described using probability theory. It is taught in schools in many countries, but in Russia it was returned to school in the 2004 standard and remains a new section for now.
Students and teachers still experience certain difficulties in studying probability theory and statistics due to the lack of deep teaching traditions and the small number of educational materials. Therefore, in 2016, the Unified State Exam will include only the simplest problems in probability theory.

At a ceramic tile factory, 5% of the tiles produced have a defect. During product quality control, only 40% of defective tiles are detected. The remaining tiles are sent for sale. Find the probability that a tile chosen at random upon purchase will have no defects. Round your answer to the nearest hundredth.

Solution

During product quality control, 40% of defective tiles are identified, which account for 5% of the tiles produced, and they do not go on sale. This means that 0.4 · 5% = 2% of the tiles produced do not go on sale. The rest of the tiles produced - 100% - 2% = 98% - go on sale.

100% - 95% of produced tiles are free from defects. The probability that the purchased tile does not have a defect is 95%: 98% = \frac(95)(98)\approx 0.97

Answer

Condition

The probability that the battery is not charged is 0.15. A customer in a store purchases a random package that contains two of these batteries. Find the probability that both batteries in this package will be charged.

Solution

The probability that the battery is charged is 1-0.15 = 0.85. Let’s find the probability of the event “both batteries are charged.” Let us denote by A and B the events “the first battery is charged” and “the second battery is charged”. We got P(A) = P(B) = 0.85. The event “both batteries are charged” is the intersection of events A \cap B, its probability is equal to P(A\cap B) = P(A)\cdot P(B) = 0.85\cdot 0.85 = 0,7225.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The probability that a new washing machine will be repaired under warranty within a year is 0.065. In a certain city, 1,200 washing machines were sold during the year, of which 72 were handed over to a warranty workshop. Determine how different the relative frequency of the occurrence of the “warranty repair” event is from its probability in this city?

Solution

The frequency of the event “the washing machine will be repaired under warranty within a year” is equal to \frac(72)(1200) = 0.06. It differs from probability by 0.065-0.06=0.005.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The probability that the pen is defective is 0.05. A customer in a store purchases a random package that contains two pens. Find the probability that both pens in this package will be good.

Solution

The probability that the handle is working is 1-0.05 = 0.95. Let's find the probability of the event “both handles are working.” Let us denote by A and B the events “the first handle is working” and “the second handle is working”. We got P(A) = P(B) = 0.95. The event “both handles are working” is the intersection of events A\cap B, its probability is equal to P(A\cap B) = P(A)\cdot P(B) = 0.95\cdot 0.95 = 0,9025.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The picture shows a labyrinth. The beetle crawls into the maze at the “Entrance” point. The beetle cannot turn around and crawl in the opposite direction, so at each fork it chooses one of the paths it has not been on yet. With what probability is the beetle coming to exit D if the choice further path is random.

Solution

Let's place arrows at intersections in the directions in which the beetle can move (see figure).

At each intersection we will choose one direction out of two possible ones and assume that when it gets to the intersection the beetle will move in the direction we have chosen.

In order for the beetle to reach exit D, it is necessary that at each intersection the direction indicated by the solid red line is chosen. In total, the choice of direction is made 4 times, each time regardless of the previous choice. The probability that the solid red arrow is selected each time is \frac12\cdot\frac12\cdot\frac12\cdot\frac12= 0,5^4= 0,0625.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

There are 16 athletes in the section, among them two friends - Olya and Masha. Athletes are randomly assigned to 4 equal groups. Find the probability that Olya and Masha will end up in the same group.

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 two events occurring is equal to the product of the probabilities of one of them and the conditional probability of the other, calculated under the condition that the first occurred.”

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.