The probability is higher or greater. Probability theory

If you are interested in the title question, you are probably a student or schoolboy who is faced with a new subject. Probability theory problems are now being solved by fifth-grade students at advanced schools, high school students before the Unified State Exam, and students of literally all specialties - from geographers to mathematicians. What kind of object is this, and how to approach it?

Probability. What is this?

Probability theory, as the name suggests, deals with probabilities. We are surrounded by many things and phenomena about which, no matter how developed science is, it is impossible to make accurate predictions.

We don't know which card we'll draw from the deck at random or how many days it will rain in May, but having some additional information, we can make forecasts and calculate probabilities these random events.

Thus, we are faced with the basic concept random event- a phenomenon whose behavior cannot be predicted, an experiment whose result cannot be calculated in advance, etc. It is the probabilities of events that are calculated in typical problems.

Probability- this is some, strictly speaking, function that takes values ​​from 0 to 1 and characterizes a given random event. 0 - the event is practically impossible, 1 - the event is almost certain, 0.5 (or "50 to 50") - with equal probability the event will happen or not.

Algorithm for solving probability problems

You can learn more about the basics of probability theory, for example, in the online textbook.

Now let's not beat around the bush, and let's formulate diagram, which should be used to solve standard learning objectives to calculate the probability of a random event, and then below Let's illustrate with examples its application.

• Carefully read the task and understand what exactly is happening (what is being pulled out of which box, what was lying where, how many devices are working, etc.)
• Find the main question of a problem like "calculate the likelihood that, that..." and write this ellipsis in the form of an event, the probability of which must be found.
• The event is recorded. Now we need to understand which “scheme” of probability theory the problem belongs to in order to correctly select formulas for the solution. Reply to test questions type:
  • there is one test (for example, throwing two dice) or several (for example, checking 10 devices);
  • if there are several tests, are the results of one dependent on the others (dependence or independence of events);
  • an event occurs in a single situation or a task speaks of several possible hypotheses(for example, the ball is taken from any of the three boxes, or from a specific one).
  The more experience you have in solving problems, the easier it will be to determine which formulas are appropriate.
• A formula (or several) has been selected for the solution. We write down all the task data and substitute it into this formula.
• Voila, the probability has been found.

How to solve problems: classical probability

Example 1. In a group of 30 students, on the test, 6 students received a “5”, 10 students received a “4”, 9 students received a “3”, the rest received a “2”. Find the probability that 3 students called to the board received a “2” on the test.

We begin the solution according to the points described above.

• In the problem we're talking about about selecting 3 students from a group who meet certain conditions.
• Enter the main event $X$ = (All 3 students called to the board received a “2” on the test).
• Since only one test occurs in the task and it is associated with selection/choice under a certain condition, we are talking about the classical definition of probability. Let us write the formula: $P=m/n$, where $m$ is the number of outcomes favorable to the occurrence of event $X$, and $n$ is the number of all equally possible elementary outcomes.
• Now we need to find the values ​​of $m$ and $n$ for this problem. First, let's find the number of all possible outcomes - the number of ways to choose 3 students out of 30. Since the order of choice does not matter, this is the number of combinations of 30 by 3: $$n=C_(30)^3=\frac(30{3!27!}=\frac{28\cdot 29 \cdot 30}{1\cdot 2 \cdot 3}=4060.$$ Найдем число способов вызвать только студентов, получивших "2". Всего таких студентов было $30-6-10-9=5$ человек, поэтому $$m=C_{5}^3=\frac{5!}{3!2!}=\frac{4 \cdot 5}{1\cdot 2}=10.$$!}
• We get the probability: $$P(X)=\frac(m)(n)=\frac(10)(4060)=0.002.$$ Problem solved.

No time to decide? Find a solved problem

Ready-made solutions to problems for any section of probability theory, more than 10,000 examples! Find your task.

At When assessing the probability of the occurrence of any random event, it is very important to have a good understanding of whether the probability () of the occurrence of the event we are interested in depends on how other events develop.

In the case of the classical scheme, when all outcomes are equally probable, we can already estimate the probability values ​​of the individual event of interest to us independently. We can do this even if the event is a complex collection of several elementary outcomes. What if several random events occur simultaneously or sequentially? How does this affect the likelihood of the event we are interested in happening?

If I roll a die several times and want a six to come up, and I keep getting unlucky, does that mean I should increase my bet because, according to probability theory, I'm about to get lucky? Alas, probability theory does not state anything like this. No dice, no cards, no coins can't remember what they showed us in last time. It doesn’t matter to them at all whether it’s the first time or the tenth time I’m testing my luck today. Every time I repeat the roll, I know only one thing: and this time the probability of getting a six is ​​again one sixth. Of course, this does not mean that the number I need will never come up. This only means that my loss after the first throw and after any other throw are independent events.

Events A and B are called independent, if the implementation of one of them does not in any way affect the probability of another event. For example, the probabilities of hitting a target with the first of two weapons do not depend on whether the target was hit by the other weapon, so the events “the first weapon hit the target” and “the second weapon hit the target” are independent.

If two events A and B are independent, and the probability of each of them is known, then the probability of the simultaneous occurrence of both event A and event B (denoted AB) can be calculated using the following theorem.

Probability multiplication theorem for independent events

Example.The probabilities of hitting the target when firing the first and second guns are respectively equal: p 1 =0.7; p 2 =0.8. Find the probability of a hit with one salvo by both guns simultaneously.

Solution: as we have already seen, events A (hit by the first gun) and B (hit by the second gun) are independent, i.e. P(AB)=P(A)*P(B)=p 1 *p 2 =0.56.

What happens to our estimates if the initial events are not independent? Let's change the previous example a little.

Example.Two shooters shoot at targets at a competition, and if one of them shoots accurately, the opponent begins to get nervous and his results worsen. How to turn this everyday situation into math problem and outline ways to solve it? It is intuitively clear that we need to somehow separate the two options developments, essentially create two scenarios, two different tasks. In the first case, if the opponent missed, the scenario will be favorable for the nervous athlete and his accuracy will be higher. In the second case, if the opponent took his chance decently, the probability of hitting the target for the second athlete decreases.

For separation possible scenarios(they are often called hypotheses) for the development of events, we will often use the “probability tree” diagram. This diagram is similar in meaning to the decision tree that you have probably already dealt with. Each branch represents a separate scenario for the development of events, only now it has eigenvalue so-called conditional probabilities (q 1, q 2, q 1 -1, q 2 -1).

This scheme is very convenient for analyzing sequential random events.

It remains to clarify one more important question: where do the initial values ​​of the probabilities in real situations ? After all, probability theory doesn’t work with just coins and dice? Usually these estimates are taken from statistics, and when statistical information is not available, we conduct our own research. And we often have to start not with collecting data, but with the question of what information we actually need.

Example.Let's say we need to estimate in a city with a population of one hundred thousand inhabitants the market volume for a new product that is not an essential item, for example, for a balm for the care of colored hair. Let's consider the "probability tree" diagram. In this case, we need to approximately estimate the probability value on each “branch”. So, our estimates of market capacity:

1) of all city residents, 50% are women,

2) of all women, only 30% dye their hair often,

3) of them, only 10% use balms for colored hair,

4) of them, only 10% can muster the courage to try a new product,

5) 70% of them usually buy everything not from us, but from our competitors.

Solution: According to the law of multiplication of probabilities, we determine the probability of the event we are interested in A = (a city resident buys this new balm from us) = 0.00045.

Let's multiply this probability value by the number of city residents. As a result, we have only 45 potential customers, and considering that one bottle of this product lasts for several months, the trade is not very lively.

And yet there is some benefit from our assessments.

Firstly, we can compare forecasts of different business ideas; they will have different “forks” in the diagrams, and, of course, the probability values ​​will also be different.

Secondly, as we have already said, random variable It is not called random because it does not depend on anything at all. Just her exact the meaning is not known in advance. We know that the average number of buyers can be increased (for example, by advertising a new product). So it makes sense to focus our efforts on those “forks” where the probability distribution does not suit us particularly, on those factors that we are able to influence.

Let's look at one more quantitative example research on purchasing behavior.

Example. On average, 10,000 people visit the food market per day. Probability that a market visitor enters a pavilion dairy products, is equal to 1/2. It is known that this pavilion sells an average of 500 kg of various products per day.

Can we say that the average purchase in the pavilion weighs only 100 g?

Discussion. Of course not. It is clear that not everyone who entered the pavilion ended up buying something there.

As shown in the diagram, to answer the question about the average weight of a purchase, we must find an answer to the question, what is the probability that a person entering the pavilion will buy something there. If we do not have such data at our disposal, but we need it, we will have to obtain it ourselves by observing the visitors to the pavilion for some time. Let's say our observations showed that only a fifth of pavilion visitors buy something.

Once we have obtained these estimates, the task becomes simple. Of the 10,000 people who come to the market, 5,000 will go to the dairy products pavilion; only 1,000 will purchase. Average weight purchase is equal to 500 grams. It is interesting to note that to construct full picture happening, the logic of conditional “branching” must be defined at each stage of our reasoning as clearly as if we were working with a “concrete” situation, and not with probabilities.

Self-test tasks

1. Let it be electrical circuit, consisting of n series-connected elements, each of which operates independently of the others.

The probability p of failure of each element is known. Determine the probability of proper operation of the entire section of the circuit (event A).

2. The student knows 20 out of 25 exam questions. Find the probability that the student knows the three questions given to him by the examiner.

3. Production consists of four successive stages, at each of which equipment operates, for which the probabilities of failure over the next month are equal to p 1, p 2, p 3 and p 4, respectively. Find the probability that there will be no production stoppages due to equipment failure in a month.

Let's talk about problems in which the phrase “at least one” appears. Surely you have encountered such tasks at home and tests, and now you will learn how to solve them. First I'll talk about general rule, and then consider a special case and write out formulas and examples for each.

General methodology and examples

General technique to solve problems in which the phrase “at least one” occurs is as follows:

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. This 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 things will happen is mutually exclusive joint events, 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, That A+ IN– the occurrence of at least one of these events or two events.

Example 1. There are 30 balls in a box same sizes: 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,

what follows following formulas probabilities of opposite events:

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:

For more complex problems, in which you need to use both addition and multiplication of probabilities, see 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 dice event A The number 4 is considered to be rolled out, and the event IN– loss even number. Since 4 is an even number, these 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 is subtracted general offensive both events, that is, the product of 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 occurs possible events: 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 may 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:

For more complex problems, in which you need to use both addition and multiplication of probabilities, see 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.

At the same time 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 Are there any 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. Probability that the cargo will be delivered river transport, is 0.82, by rail 0.87, by motor transport 0.90. Find the probability that the cargo will be delivered by at least one of three types transport.

It is clear that each event has a varying degree of possibility of its occurrence (its implementation). In order to quantitatively compare events with each other according to the degree of their possibility, obviously, it is necessary to associate with each event certain number, which is greater the more possible the event. This number is called the probability of an event.

Probability of event– is a numerical measure of degree objective possibility the occurrence of this event.

Consider a stochastic experiment and a random event A observed in this experiment. Let's repeat this experiment n times and let m(A) be the number of experiments in which event A occurred.

Relation (1.1)

called relative frequency events A in the series of experiments performed.

It is easy to verify the validity of the properties:

if A and B are inconsistent (AB= ), then ν(A+B) = ν(A) + ν(B) (1.2)

Relative frequency is determined only after a series of experiments and, generally speaking, can vary from series to series. However, experience shows that in many cases, as the number of experiments increases, the relative frequency approaches a certain number. This fact of stability of the relative frequency has been repeatedly verified and can be considered experimentally established.

Example 1.19.. If you throw one coin, no one can predict which side it will land on top. But if you throw two tons of coins, then everyone will say that about one ton will fall up with the coat of arms, that is, the relative frequency of the coat of arms falling out is approximately 0.5.

If, with an increase in the number of experiments, the relative frequency of the event ν(A) tends to a certain fixed number, then it is said that event A is statistically stable, and this number is called the probability of event A.

Probability of the event A some fixed number P(A) is called, to which the relative frequency ν(A) of this event tends as the number of experiments increases, that is,

This definition is called statistical definition probabilities .

Let's consider a certain stochastic experiment and let its space elementary events consists of a finite or infinite (but countable) set of elementary events ω 1, ω 2, …, ω i, …. Let us assume that each elementary event ω i is assigned a certain number - р i , characterizing the degree of possibility of the occurrence of a given elementary event and satisfying the following properties:

This number p i is called probability of an elementary eventωi.

Let now A be a random event observed in this experiment, and let it correspond to a certain set

In this setting probability of an event A call the sum of the probabilities of elementary events favoring A(included in the corresponding set A):

(1.4)

The probability introduced in this way has the same properties as the relative frequency, namely:

And if AB = (A and B are incompatible),

then P(A+B) = P(A) + P(B)

Indeed, according to (1.4)

In the last relation we took advantage of the fact that not a single elementary event can favor two incompatible events at the same time.

We especially note that probability theory does not indicate methods for determining p i; they must be sought for practical reasons or obtained from a corresponding statistical experiment.

As an example, consider classic scheme probability theory. To do this, consider a stochastic experiment, the space of elementary events of which consists of a finite (n) number of elements. Let us additionally assume that all these elementary events are equally possible, that is, the probabilities of elementary events are equal to p(ω i)=p i =p. It follows that

Example 1.20. When throwing a symmetrical coin, getting heads and tails are equally possible, their probabilities are equal to 0.5.

Example 1.21. When throwing a symmetrical die, all faces are equally possible, their probabilities are equal to 1/6.

Now let event A be favored by m elementary events, they are usually called outcomes favorable to event A. Then

Received classical definition of probability: the probability P(A) of event A is equal to the ratio of the number of outcomes favorable to event A to the total number of outcomes

Example 1.22. The urn contains m white balls and n black balls. What is the probability of getting it out? white ball?

Solution. The total number of elementary events is m+n. They are all equally probable. Favorable event A of which m. Hence, .

The following properties follow from the definition of probability:

Property 1. Probability reliable event equal to one.

Indeed, if the event is reliable, then every elementary outcome of the test favors the event. In this case t=p, hence,

P(A)=m/n=n/n=1.(1.6)

Property 2. The probability of an impossible event is zero.

Indeed, if an event is impossible, then none of the elementary outcomes of the test favor the event. In this case T= 0, therefore, P(A)=m/n=0/n=0. (1.7)

Property 3.There is a probability of a random event positive number, enclosed between zero and one.

Indeed, a random event is favored only by some of the total number elementary test outcomes. That is, 0≤m≤n, which means 0≤m/n≤1, therefore, the probability of any event satisfies the double inequality 0≤ P(A) ≤1. (1.8)

Comparing the definitions of probability (1.5) and relative frequency (1.1), we conclude: definition of probability does not require testing to be carried out in reality; the definition of relative frequency assumes that tests were actually carried out. In other words, the probability is calculated before the experiment, and the relative frequency - after the experiment.

However, calculating the probability requires having preliminary information about the number or probabilities of favorable this event elementary outcomes. In the absence of such preliminary information, empirical data are used to determine the probability, that is, the relative frequency of the event is determined based on the results of a stochastic experiment.

Example 1.23. Technical control department discovered 3 non-standard parts in a batch of 80 randomly selected parts. Relative frequency of occurrence of non-standard parts r(A)= 3/80.

Example 1.24. According to the purpose.produced 24 shot, and 19 hits were recorded. Relative target hit rate. r(A)=19/24.

Long-term observations have shown that if experiments are carried out under identical conditions, in each of which the number of tests is sufficiently large, then the relative frequency exhibits the property of stability. This property is what's in various experiences the relative frequency changes little (the less, the more tests are performed), fluctuating around a certain constant number. It turned out that this constant number can be taken as an approximate probability value.

The relationship between relative frequency and probability will be described in more detail and more precisely below. Now let us illustrate the property of stability with examples.

Example 1.25. According to Swedish statistics, the relative frequency of births of girls for 1935 by month is characterized by the following numbers (the numbers are arranged in order of months, starting with January): 0,486; 0,489; 0,490; 0.471; 0,478; 0,482; 0.462; 0,484; 0,485; 0,491; 0,482; 0,473

The relative frequency fluctuates around the number 0.481, which can be taken as approximate value probability of having girls.

Note that statistical data from different countries give approximately the same relative frequency value.

Example 1.26. Coin tossing experiments were carried out many times, in which the number of appearances of the “coat of arms” was counted. The results of several experiments are shown in the table.