Problems on the classical determination of probability.
Examples of solutions

In the third lesson we will look at various problems involving the direct application of the classical definition of probability. To effectively study the materials in this article, I recommend that you familiarize yourself with the basic concepts probability theory And basics of combinatorics. The task of classically determining probability with a probability tending to one will be present in your independent/control work on terver, so let’s get ready for serious work. You may ask, what's so serious about this? ...just one primitive formula. I warn you against frivolity - thematic tasks are quite diverse, and many of them can easily confuse you. In this regard, in addition to working through the main lesson, try to study additional tasks on the topic that are in the piggy bank ready-made solutions for higher mathematics. Solution techniques are solution techniques, but “friends” still “need to be known by sight,” because even a rich imagination is limited and there are also enough standard tasks. Well, I’ll try to sort out as many of them as possible in good quality.

Let's remember the classics of the genre:

The probability of an event occurring in a certain test is equal to the ratio , where:

– total number of all equally possible, elementary outcomes of this test, which form full group of events;

- quantity elementary outcomes favorable to the event.

And immediately an immediate pit stop. Do you understand the underlined terms? This means clear, not intuitive understanding. If not, then it’s still better to return to the 1st article on probability theory and only after that move on.

Please do not skip the first examples - in them I will repeat one fundamentally important point, and also tell you how to correctly format a solution and in what ways this can be done:

Problem 1

An urn contains 15 white, 5 red and 10 black balls. 1 ball is drawn at random, find the probability that it will be: a) white, b) red, c) black.

Solution: The most important prerequisite for using the classical definition of probability is ability to count the total number of outcomes.

There are a total of 15 + 5 + 10 = 30 balls in the urn, and obviously the following facts are true:

– retrieving any ball is equally possible (equal opportunity outcomes), while the outcomes elementary and form full group of events (i.e., as a result of the test, one of the 30 balls will definitely be removed).

Thus, the total number of outcomes:

Consider the event: – a white ball will be drawn from the urn. This event is favored elementary outcomes, therefore, according to the classical definition:
– the probability that a white ball will be drawn from the urn.

Oddly enough, even in such a simple task one can make a serious inaccuracy, which I already focused on in the first article on probability theory. Where is the pitfall here? It is incorrect to argue here that “since half the balls are white, then the probability of drawing a white ball» . The classic definition of probability refers to ELEMENTARY outcomes, and the fraction must be written down!

With other points, similarly, consider the following events:

– a red ball will be drawn from the urn;
– a black ball will be drawn from the urn.

An event is favored by 5 elementary outcomes, and an event is favored by 10 elementary outcomes. So the corresponding probabilities are:

A typical check of many server tasks is carried out using theorems on the sum of probabilities of events forming a complete group. In our case, the events form a complete group, which means the sum of the corresponding probabilities must necessarily equal one: .

Let's check if this is true: that's what I wanted to make sure of.

Answer:

In principle, the answer can be written down in more detail, but personally, I’m used to putting only numbers there - for the reason that when you start “stamping out” problems in hundreds and thousands, you try to reduce the writing of the solution as much as possible. By the way, about brevity: in practice, the “high-speed” design option is common solutions:

Total: 15 + 5 + 10 = 30 balls in the urn. According to the classical definition:
– the probability that a white ball will be drawn from the urn;
– the probability that a red ball will be drawn from the urn;
– the probability that a black ball will be drawn from the urn.

Answer:

However, if there are several points in the condition, then it is often more convenient to formulate the solution in the first way, which takes a little more time, but at the same time “lays everything out on the shelves” and makes it easier to navigate the problem.

Let's warm up:

Problem 2

The store received 30 refrigerators, five of which have a manufacturing defect. One refrigerator is selected at random. What is the probability that it will be without a defect?

Select the appropriate design option and check the sample at the bottom of the page.

In the simplest examples, the number of common and the number of favorable outcomes lie on the surface, but in most cases you have to dig up the potatoes yourself. A canonical series of problems about a forgetful subscriber:

Problem 3

When dialing a phone number, the subscriber forgot the last two digits, but remembers that one of them is zero and the other is odd. Find the probability that he will dial the correct number.

Note : zero is an even number (divisible by 2 without a remainder)

Solution: First we find the total number of outcomes. By condition, the subscriber remembers that one of the digits is zero, and the other digit is odd. Here it is more rational not to be tricky with combinatorics and use method of direct listing of outcomes . That is, when making a solution, we simply write down all the combinations:
01, 03, 05, 07, 09
10, 30, 50, 70, 90

And we count them - in total: 10 outcomes.

There is only one favorable outcome: the correct number.

According to the classical definition:
– probability that the subscriber will dial the correct number

Answer: 0,1

Decimal fractions look quite appropriate in probability theory, but you can also adhere to the traditional Vyshmatov style, operating only with ordinary fractions.

Advanced task for independent solution:

Problem 4

The subscriber has forgotten the PIN code for his SIM card, but remembers that it contains three “fives”, and one of the numbers is either a “seven” or an “eight”. What is the probability of successful authorization on the first try?

Here you can also develop the idea of ​​​​the likelihood that the subscriber will face punishment in the form of a puk code, but, unfortunately, the reasoning will go beyond the scope of this lesson

The solution and answer are below.

Sometimes listing combinations turns out to be a very painstaking task. In particular, this is the case in the next, no less popular group of problems, where 2 dice are rolled (less often - larger quantities):

Problem 5

Find the probability that when throwing two dice the total number will be:

a) five points;
b) no more than four points;
c) from 3 to 9 points inclusive.

Solution: find the total number of outcomes:

Ways the side of the 1st die can fall out And in different ways the side of the 2nd cube can fall out; By rule for multiplying combinations, Total: possible combinations. In other words, each the face of the 1st cube can be ordered a couple with each the edge of the 2nd cube. Let us agree to write such a pair in the form , where is the number that appears on the 1st die, and is the number that appears on the 2nd die. For example:

– the first dice scored 3 points, the second dice scored 5 points, total points: 3 + 5 = 8;
– the first dice scored 6 points, the second dice scored 1 point, total points: 6 + 1 = 7;
– 2 points rolled on both dice, sum: 2 + 2 = 4.

Obviously, the smallest amount is given by a pair, and the largest by two “sixes”.

a) Consider the event: – when throwing two dice, 5 points will appear. Let's write down and count the number of outcomes that favor this event:

Total: 4 favorable outcomes. According to the classical definition:
– the desired probability.

b) Consider the event: – no more than 4 points will be rolled. That is, either 2, or 3, or 4 points. Again we list and count the favorable combinations, on the left I will write down the total number of points, and after the colon - the suitable pairs:

Total: 6 favorable combinations. Thus:
– the probability that no more than 4 points will be rolled.

c) Consider the event: – 3 to 9 points will roll, inclusive. Here you can take the straight road, but... for some reason you don’t want to. Yes, some pairs have already been listed in the previous paragraphs, but there is still a lot of work to be done.

What's the best way to proceed? In such cases, a roundabout path turns out to be rational. Let's consider opposite event: – 2 or 10 or 11 or 12 points will be rolled.

What's the point? The opposite event is favored by a significantly smaller number of couples:

Total: 7 favorable outcomes.

According to the classical definition:
– the probability that you will roll less than three or more than 9 points.

In addition to direct listing and counting of outcomes, various combinatorial formulas. And again an epic problem about the elevator:

Problem 7

3 people entered the elevator of a 20-story building on the first floor. And let's go. Find the probability that:

a) they will exit on different floors
b) two will exit on the same floor;
c) everyone will get off on the same floor.

Our exciting lesson has come to an end, and finally, I once again strongly recommend that if not solve, then at least figure out additional problems on the classical determination of probability. As I already noted, “hand padding” matters too!

Further along the course - Geometric definition of probability And Probability addition and multiplication theorems and... luck in the main thing!

Solutions and Answers:

Task 2: Solution: 30 – 5 = 25 refrigerators have no defect.

– the probability that a randomly selected refrigerator does not have a defect.
Answer :

Task 4: Solution: find the total number of outcomes:
ways you can select the place where the dubious number is located and on every Of these 4 places, 2 digits (seven or eight) can be located. According to the rule of multiplication of combinations, the total number of outcomes: .
Alternatively, the solution can simply list all the outcomes (fortunately there are few of them):
7555, 8555, 5755, 5855, 5575, 5585, 5557, 5558
There is only one favorable outcome (correct pin code).
Thus, according to the classical definition:
– probability that the subscriber logs in on the 1st attempt
Answer :

Task 6: Solution: find the total number of outcomes:
numbers on 2 dice can appear in different ways.

a) Consider the event: – when throwing two dice, the product of the points will be equal to seven. There are no favorable outcomes for a given event, according to the classical definition of probability:
, i.e. this event is impossible.

b) Consider the event: – when throwing two dice, the product of the points will be at least 20. The following outcomes are favorable for this event:

Total: 8
According to the classical definition:
– the desired probability.

c) Consider the opposite events:
– the product of points will be even;
– the product of points will be odd.
Let's list all the outcomes favorable to the event:

Total: 9 favorable outcomes.
According to the classical definition of probability:
Opposite events form a complete group, therefore:
– the desired probability.

Answer :

Problem 8: Solution: let's calculate the total number of outcomes: 10 coins can fall in different ways.
Another way: ways the 1st coin can fall And ways the 2nd coin can fall And … And ways the 10th coin can fall. According to the rule of multiplying combinations, 10 coins can fall ways.
a) Consider the event: – heads will appear on all coins. This event is favored by a single outcome, according to the classical definition of probability: .
b) Consider the event: – 9 coins will land heads, and one coin will land tails.
There are coins that can land on heads. According to the classical definition of probability: .
c) Consider the event: – heads will appear on half of the coins.
Exists unique combinations of five coins that can land heads. According to the classical definition of probability:
Answer :

When a coin is tossed, we can say that it will land heads up, or probability this is 1/2. Of course, this does not mean that if a coin is tossed 10 times, it will necessarily land on heads 5 times. If the coin is "fair" and if it is tossed many times, then heads will land very close half the time. Thus, there are two types of probabilities: experimental And theoretical .

Experimental and theoretical probability

If we flip a coin a large number of times - say 1000 - and count how many times it lands on heads, we can determine the probability that it lands on heads. If heads are thrown 503 times, we can calculate the probability of it landing:
503/1000, or 0.503.

This experimental definition of probability. This definition of probability comes from observation and study of data and is quite common and very useful. Here, for example, are some probabilities that were determined experimentally:

1. The probability that a woman will develop breast cancer is 1/11.

2. If you kiss someone who has a cold, then the probability that you will also get a cold is 0.07.

3. A person who has just been released from prison has an 80% chance of returning to prison.

If we consider tossing a coin and taking into account that it is just as likely that it will come up heads or tails, we can calculate the probability of getting heads: 1/2. This is a theoretical definition of probability. Here are some other probabilities that have been determined theoretically using mathematics:

1. If there are 30 people in a room, the probability that two of them have the same birthday (excluding year) is 0.706.

2. During a trip, you meet someone, and during the conversation you discover that you have a mutual friend. Typical reaction: “This can’t be!” In fact, this phrase is not suitable, because the probability of such an event is quite high - just over 22%.

Thus, experimental probabilities are determined through observation and data collection. Theoretical probabilities are determined through mathematical reasoning. Examples of experimental and theoretical probabilities, such as those discussed above, and especially those that we do not expect, lead us to the importance of studying probability. You may ask, "What is true probability?" In fact, there is no such thing. Probabilities within certain limits can be determined experimentally. They may or may not coincide with the probabilities that we obtain theoretically. There are situations in which it is much easier to determine one type of probability than another. For example, it would be sufficient to find the probability of catching a cold using theoretical probability.

Calculation of experimental probabilities

Let us first consider the experimental definition of probability. The basic principle we use to calculate such probabilities is as follows.

Principle P (experimental)

If in an experiment in which n observations are made, a situation or event E occurs m times in n observations, then the experimental probability of the event is said to be P (E) = m/n.

Example 1 Sociological survey. An experimental study was conducted to determine the number of left-handed people, right-handed people and people whose both hands are equally developed. The results are shown in the graph.

a) Determine the probability that the person is right-handed.

b) Determine the probability that the person is left-handed.

c) Determine the probability that a person is equally fluent in both hands.

d) Most Professional Bowling Association tournaments are limited to 120 players. Based on the data from this experiment, how many players could be left-handed?

Solution

a)The number of people who are right-handed is 82, the number of left-handers is 17, and the number of those who are equally fluent in both hands is 1. The total number of observations is 100. Thus, the probability that a person is right-handed is P
P = 82/100, or 0.82, or 82%.

b) The probability that a person is left-handed is P, where
P = 17/100, or 0.17, or 17%.

c) The probability that a person is equally fluent in both hands is P, where
P = 1/100, or 0.01, or 1%.

d) 120 bowlers, and from (b) we can expect that 17% are left-handed. From here
17% of 120 = 0.17.120 = 20.4,
that is, we can expect about 20 players to be left-handed.

Example 2 Quality control . It is very important for a manufacturer to keep the quality of its products at a high level. In fact, companies hire quality control inspectors to ensure this process. The goal is to produce the minimum possible number of defective products. But since the company produces thousands of products every day, it cannot afford to test every product to determine whether it is defective or not. To find out what percentage of products are defective, the company tests far fewer products.
The USDA requires that 80% of the seeds sold by growers must germinate. To determine the quality of the seeds that an agricultural company produces, 500 seeds from those that were produced are planted. After this, it was calculated that 417 seeds sprouted.

a) What is the probability that the seed will germinate?

b) Do the seeds meet government standards?

Solution a) We know that out of 500 seeds that were planted, 417 sprouted. Probability of seed germination P, and
P = 417/500 = 0.834, or 83.4%.

b) Since the percentage of seeds germinated has exceeded 80% as required, the seeds meet government standards.

Example 3 Television ratings. According to statistics, there are 105,500,000 households with televisions in the United States. Every week, information about viewing programs is collected and processed. In one week, 7,815,000 households tuned in to the hit comedy series "Everybody Loves Raymond" on CBS and 8,302,000 households tuned in to the hit series "Law & Order" on NBC (Source: Nielsen Media Research). What is the probability that one household's TV is tuned to "Everybody Loves Raymond" during a given week? to "Law & Order"?

Solution The probability that the TV in one household is tuned to "Everybody Loves Raymond" is P, and
P = 7,815,000/105,500,000 ≈ 0.074 ≈ 7.4%.
The chance that the household's TV was tuned to Law & Order is P, and
P = 8,302,000/105,500,000 ≈ 0.079 ≈ 7.9%.
These percentages are called ratings.

Theoretical probability

Suppose we are conducting an experiment, such as throwing a coin or darts, drawing a card from a deck, or testing products for quality on an assembly line. Each possible result of such an experiment is called Exodus . The set of all possible outcomes is called outcome space . Event it is a set of outcomes, that is, a subset of the space of outcomes.

Example 4 Throwing darts. Suppose that in a dart throwing experiment, a dart hits a target. Find each of the following:

b) Outcome space

Solution
a) The outcomes are: hitting black (B), hitting red (R) and hitting white (B).

b) The space of outcomes is (hitting black, hitting red, hitting white), which can be written simply as (H, K, B).

Example 5 Throwing dice. A die is a cube with six sides, each with one to six dots on it.


Suppose we are throwing a die. Find
a) Outcomes
b) Outcome space

Solution
a) Outcomes: 1, 2, 3, 4, 5, 6.
b) Outcome space (1, 2, 3, 4, 5, 6).

We denote the probability that an event E occurs as P(E). For example, “the coin will land on heads” can be denoted by H. Then P(H) represents the probability that the coin will land on heads. When all outcomes of an experiment have the same probability of occurring, they are said to be equally likely. To see the differences between events that are equally likely and events that are not, consider the target shown below.

For target A, the events of hitting black, red and white are equally probable, since the black, red and white sectors are the same. However, for target B, the zones with these colors are not the same, that is, hitting them is not equally probable.

Principle P (Theoretical)

If an event E can happen in m ways out of n possible equally probable outcomes from the outcome space S, then theoretical probability events, P(E) is
P(E) = m/n.

Example 6 What is the probability of rolling a die to get a 3?

Solution There are 6 equally probable outcomes on a dice and there is only one possibility of rolling the number 3. Then the probability P will be P(3) = 1/6.

Example 7 What is the probability of rolling an even number on a die?

Solution The event is the throwing of an even number. This can happen in 3 ways (if you roll a 2, 4 or 6). The number of equally probable outcomes is 6. Then the probability P(even) = 3/6, or 1/2.

We will use a number of examples involving a standard 52 card deck. This deck consists of the cards shown in the figure below.

Example 8 What is the probability of drawing an Ace from a well-shuffled deck of cards?

Solution There are 52 outcomes (the number of cards in the deck), they are equally likely (if the deck is well shuffled), and there are 4 ways to draw an Ace, so according to the P principle, the probability
P(draw an ace) = 4/52, or 1/13.

Example 9 Suppose we choose, without looking, one ball from a bag with 3 red balls and 4 green balls. What is the probability of choosing a red ball?

Solution There are 7 equally probable outcomes of drawing any ball, and since the number of ways to draw a red ball is 3, we get
P(red ball selection) = 3/7.

The following statements are results from Principle P.

Properties of Probability

a) If event E cannot happen, then P(E) = 0.
b) If event E is certain to happen then P(E) = 1.
c) The probability that event E will occur is a number from 0 to 1: 0 ≤ P(E) ≤ 1.

For example, in a coin toss, the event that the coin lands on its edge has zero probability. The probability that a coin is either heads or tails has a probability of 1.

Example 10 Let's assume that 2 cards are drawn from a 52-card deck. What is the probability that both of them are peaks?

Solution The number n of ways to draw 2 cards from a well-shuffled deck of 52 cards is 52 C 2 . Since 13 of the 52 cards are spades, the number of ways m to draw 2 spades is 13 C 2 . Then,
P(pulling 2 peaks) = m/n = 13 C 2 / 52 C 2 = 78/1326 = 1/17.

Example 11 Suppose 3 people are randomly selected from a group of 6 men and 4 women. What is the probability that 1 man and 2 women will be selected?

Solution The number of ways to select three people from a group of 10 people is 10 C 3. One man can be chosen in 6 C 1 ways, and 2 women can be chosen in 4 C 2 ways. According to the fundamental principle of counting, the number of ways to choose 1 man and 2 women is 6 C 1. 4 C 2 . Then, the probability that 1 man and 2 women will be selected is
P = 6 C 1 . 4 C 2 / 10 C 3 = 3/10.

Example 12 Throwing dice. What is the probability of rolling a total of 8 on two dice?

Solution Each dice has 6 possible outcomes. The outcomes are doubled, meaning there are 6.6 or 36 possible ways in which the numbers on the two dice can appear. (It’s better if the cubes are different, say one is red and the other is blue - this will help visualize the result.)

The pairs of numbers that add up to 8 are shown in the figure below. There are 5 possible ways to obtain a sum equal to 8, hence the probability is 5/36.

Events that happen in reality or in our imagination can be divided into 3 groups. These are certain events that will definitely happen, impossible events and random events. Probability theory studies random events, i.e. events that may or may not happen. This article will briefly present the theory of probability formulas and examples of solving problems in probability theory, which 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 of 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. The modern theory of probability was created.

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, more complex tasks began to be found in the open task bank. 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 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. 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 probability that this is a question on External Angles 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.

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 right next to it we will 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: .

Every person encounters the concept of probability every day. People calculate the chance of catching the bus, the probability that they will receive a salary today, and they come up with various combinations for winning the lottery. The theory of probability in computer programs and artificial intelligence is seriously affected, and it is also closely intertwined with financial exchanges and the like. There are elementary examples of how to find probability.

The classic case is with a coin. It is thrown up, and there are two different options for its landing: falling on the obverse and falling on the reverse. The possibility of falling on an edge is excluded in advance, that is, there are two probable outcomes. Since there are only two of them, and they happen with the same frequency, the probability of getting, for example, heads is 1/2. This is the basic law of how to find probability in mathematics.

Where did this 1/2 come from? The principle is that the probability of one (1) event out of two (2) possible events is calculated. Their ratio is resolved by the division operation, which gives 1/2. Similarly, you can calculate the probability of a certain number falling out on a dice. As you know, the surface of a cube has 6 faces, therefore any number from 1 to 6 can appear - six different options. How to find the probability of rolling, for example, a four?

Four can only come out in the only way (1) out of six in every possible way, therefore, the probability will be equal to 1: 6 = 1/6. One-sixth can be converted to a decimal by dividing on a calculator: 1/6 = 0.6(6) . By multiplying the value by 100 and adding the “%” sign, you can get an estimate of the probability of the event as a percentage. It is extremely important to know that the probability of an event is estimated as a number from 0 to 1, which in percentage ranges from 0% to 100%.

All other probability values ​​are absurd. A specific example should be considered: a random card is drawn from a classic deck of cards (36 cards). What is the probability that the card will be red and its number will be odd? A red odd card can only be a seven or nine of diamonds or hearts. There are 4 such cards in total. This means that the probability of such a card appearing is 4 / 36 = 1 / 9 = 0.1(1). The probability should be calculated as a percentage, this is equal to 1.1%.

Very often in problems you should use the complex probability formula. For example, there are 10 balls in an urn, 3 black and 7 white. What is the probability that two balls drawn at random in a row will be black? This problem should be solved as two separate ones. First, you should calculate the probability of drawing a black ball from all of them. There are 3 such balls, and there are 10 of them in total, which means the probability will be equal to 3/10. Next, we need to move on to the second part of the problem, where probability theory allows us to harmonize the results.

After extraction, there will already be 9 balls left in the urn, 2 of which will be black. In this case, the chance of getting a black ball is 2/9. Next, you should multiply the obtained probabilities for the final result: 3/10 * 2/9 = 6/90 = 1/15 = 0.6(6), which is approximately equal to 6.7%. This means that the probability of this event is quite low.

The need to act on probabilities occurs when the probabilities of some events are known, and it is necessary to calculate the probabilities of other events that are associated with these events.

Addition of probabilities is used when you need to calculate the probability of a combination or logical sum of random events.

Sum of events A And B denote A + B or A ∪ B. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B– an event that occurs if and only if the event occurred during observation A or event B, or simultaneously A And B.

If events A And B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.

Probability addition theorem. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:

For example, while hunting, two shots are fired. Event A– hitting a duck with the first shot, event IN– hit from the second shot, event ( A+ IN) – a hit from the first or second shot or from two shots. So, if two events A And IN– incompatible events, then A+ IN– the occurrence of at least one of these events or two events.

Example 1. There are 30 balls of the same size in a box: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball will be picked up without looking.

Solution. Let us assume that the event A- “the red ball is taken”, and the event IN- “The blue ball was taken.” Then the event is “a colored (not white) ball is taken.” Let's find the probability of the event A:

and events IN:

Events A And IN– mutually incompatible, since if one ball is taken, then it is impossible to take balls of different colors. Therefore, we use the addition of probabilities:

The theorem for adding probabilities for several incompatible events. If events constitute a complete set of events, then the sum of their probabilities is equal to 1:

The sum of the probabilities of opposite events is also equal to 1:

Opposite events form a complete set of events, and the probability of a complete set of events is 1.

Probabilities of opposite events are usually indicated in small letters p And q. In particular,

from which the following formulas for the probability of opposite events follow:

Example 2. The target in the shooting range is divided into 3 zones. The probability that a certain shooter will shoot at the target in the first zone is 0.15, in the second zone – 0.23, in the third zone – 0.17. Find the probability that the shooter will hit the target and the probability that the shooter will miss the target.

Solution: Find the probability that the shooter will hit the target:

Let's find the probability that the shooter will miss the target:

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".

Addition of probabilities of mutually simultaneous events

Two random events are called joint if the occurrence of one event does not exclude the occurrence of a second event in the same observation. For example, when throwing a die the event A The number 4 is considered to be rolled out, and the event IN– rolling an even number. Since 4 is an even number, the two events are compatible. In practice, there are problems of calculating the probabilities of the occurrence of one of the mutually simultaneous events.

Probability addition theorem for joint events. The probability that one of the joint events will occur is equal to the sum of the probabilities of these events, from which the probability of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events has the following form:

Since events A And IN compatible, event A+ IN occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:

Event A will occur if one of two incompatible events occurs: or AB. However, the probability of the occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:

Likewise:

Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:

When using formula (8), it should be taken into account that events A And IN can be:

• mutually independent;
• mutually dependent.

Probability formula for mutually independent events:

Probability formula for mutually dependent events:

If events A And IN are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is:

Example 3. In auto racing, when you drive the first car, you have a better chance of winning, and when you drive the second car. Find:

• the probability that both cars will win;
• the probability that at least one car will win;

1) The probability that the first car will win does not depend on the result of the second car, so the events A(the first car wins) and IN(the second car will win) – independent events. Let's find the probability that both cars win:

2) Find the probability that one of the two cars will win:

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".

Solve the addition of probabilities problem yourself, and then look at the solution

Example 4. Two coins are tossed. Event A- loss of the coat of arms on the first coin. Event B- loss of the coat of arms on the second coin. Find the probability of an event C = A + B .

Multiplying Probabilities

Probability multiplication is used when the probability of a logical product of events must be calculated.

In this case, random events must be independent. Two events are said to be mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.

Probability multiplication theorem for independent events. Probability of simultaneous occurrence of two independent events A And IN is equal to the product of the probabilities of these events and is calculated by the formula:

Example 5. The coin is tossed three times in a row. Find the probability that the coat of arms will appear all three times.

Solution. The probability that the coat of arms will appear on the first toss of a coin, the second time, and the third time. Let's find the probability that the coat of arms will appear all three times:

Solve probability multiplication problems on your own and then look at the solution

Example 6. There is a box of nine new tennis balls. To play, three balls are taken, and after the game they are put back. When choosing balls, played balls are not distinguished from unplayed balls. What is the probability that after three games there will be no unplayed balls left in the box?

Example 7. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. Find the probability that the letters will form the word "end".

Example 8. From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards will be of different suits.

Example 9. The same task as in example 8, but each card after being removed is returned to the deck.

More complex problems, in which you need to use both addition and multiplication of probabilities, as well as calculate the product of several events, can be found on the page "Various problems involving addition and multiplication of probabilities".

The probability that at least one of the mutually independent events will occur can be calculated by subtracting from 1 the product of the probabilities of opposite events, that is, using the formula:

Example 10. Cargo is delivered by three modes of transport: river, rail and road transport. The probability that the cargo will be delivered by river transport is 0.82, by rail 0.87, by road transport 0.90. Find the probability that the cargo will be delivered by at least one of the three modes of transport.