The need for operations on probabilities comes 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.

Probability addition is used when it is necessary to calculate the probability of a combination or a logical sum of random events.

Sum of events A And B designate 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 an event occurs during the observation A or event B, or at the same time 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.

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

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

Example 1 A box contains 30 balls of the same size: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball is taken without looking.

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

and events IN:

Events A And IN- mutually incompatible, since if one ball is taken, then balls cannot be taken different colors. Therefore, we use the addition of probabilities:

The theorem of addition of probabilities for several incompatible events. If the events make up the 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.

The probabilities of opposite events are usually denoted 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 dash is divided into 3 zones. The probability that a certain shooter will shoot at a 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 hits the target and the probability that the shooter misses the target.

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

Find the probability that the shooter misses the target:

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .

Addition of probabilities of mutually joint events

Two random events are said to be joint if the occurrence of one event does not preclude the occurrence of a second event in the same observation. For example, when throwing a dice, the event A is considered to be the occurrence of the number 4, and the event IN- dropping an even number. Since the number 4 is an even number, the two events are compatible. In practice, there are tasks for calculating the probabilities of the occurrence of one of the mutually joint events.

The theorem of addition of probabilities 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 is as follows:

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

Similarly:

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 the 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 as follows:

Example 3 In auto racing, when driving in the first car, the probability of winning, when driving in 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(first car wins) and IN(second car wins) - independent events. Find the probability that both cars win:

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

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .

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

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

Probability multiplication

Multiplication of probabilities is used when the probability of a logical product of events is to 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. The probability of the 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 fall out all three times.

Solution. The probability that the coat of arms will fall on the first toss of a coin, the second time, and the third time. Find the probability that the coat of arms will fall out all three times:

Solve problems for multiplying probabilities yourself, and then look at the solution

Example 6 There is a box with nine new tennis balls. Three balls are taken for the game, after the game they are put back. When choosing balls, they do not distinguish between played and unplayed balls. What is the probability that after three games there will be no unplayed balls in the box?

Example 7 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. 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 are of the same suit.

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

More complex tasks, in which you need to apply both addition and multiplication of probabilities, as well as calculate the product of several events - on the page "Various tasks for addition and multiplication of probabilities" .

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

Example 10 Cargoes are 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 0.90. Find the probability that the goods will be delivered by at least one of the three modes of transport.

Theorems of addition and multiplication of probabilities.
Dependent and independent events

The title looks scary, but it's actually very simple. In this lesson, we will get acquainted with the theorems of addition and multiplication of event probabilities, as well as analyze typical tasks that, along with task for the classical definition of probability will definitely meet or, more likely, have already met on your way. To effectively study the materials of this article, you need to know and understand the basic terms probability theory and be able to perform simple arithmetic operations. As you can see, very little is required, and therefore a fat plus in the asset is almost guaranteed. But on the other hand, I again warn against a superficial attitude to practical examples- there are also enough subtleties. Good luck:

The addition theorem for the probabilities of incompatible events: the probability of occurrence of one of the two incompatible events or (no matter what), is equal to the sum of the probabilities of these events:

A similar fact is also true for a larger number of incompatible events, for example, for three incompatible events and :

Dream theorem =) However, such a dream is also subject to proof, which can be found, for example, in study guide V.E. Gmurman.

Let's get acquainted with new, hitherto unseen concepts:

Dependent and independent events

Let's start with independent events. Events are independent if the probability of occurrence any of them does not depend from the appearance/non-appearance of other events of the considered set (in all possible combinations). ... But what is there to grind out common phrases:

The theorem of multiplication of probabilities of independent events: the probability of joint occurrence of independent events and is equal to the product of the probabilities of these events:

Let's return to the simplest example of the 1st lesson, in which two coins are tossed and the following events:

- heads will fall on the 1st coin;
- Heads on the 2nd coin.

Let's find the probability of the event (heads will appear on the 1st coin And Eagle will appear on the 2nd coin - remember how to read product of events!) . The probability of getting heads on one coin does not depend on the result of tossing another coin, therefore, the events and are independent.

Similarly:
is the probability that the 1st coin will land heads And on the 2nd tail;
is the probability that heads appear on the 1st coin And on the 2nd tail;
is the probability that the 1st coin will land on tails And on the 2nd eagle.

Note that events form full group and the sum of their probabilities is equal to one: .

The multiplication theorem obviously extends to a larger number of independent events, so, for example, if the events are independent, then the probability of their joint occurrence is: . Let's practice on concrete examples:

Task 3

Each of the three boxes contains 10 parts. In the first box there are 8 standard parts, in the second - 7, in the third - 9. One part is randomly removed from each box. Find the probability that all parts are standard.

Solution: the probability of extracting a standard or non-standard part from any box does not depend on which parts will be extracted from other boxes, so the problem is about independent events. Consider the following independent events:

– a standard part is removed from the 1st box;
– a standard part is removed from the 2nd box;
– A standard part has been removed from the 3rd drawer.

According to the classical definition:
are the corresponding probabilities.

Event we are interested in (Standard part will be taken from the 1st drawer And from the 2nd standard And from the 3rd standard) is expressed by the product.

According to the theorem of multiplication of probabilities of independent events:

is the probability that one standard part will be extracted from three boxes.

Answer: 0,504

After invigorating exercises with boxes, no less interesting urns await us:

Task 4

Three urns contain 6 white and 4 black balls. One ball is drawn at random from each urn. Find the probability that: a) all three balls will be white; b) all three balls will be the same color.

Based on the information received, guess how to deal with the “be” item ;-) An approximate sample solution is designed in an academic style with a detailed description of all events.

Dependent events. The event is called dependent if its probability depends from one or more events that have already happened. You don’t have to go far for examples - just go to the nearest store:

- tomorrow at 19.00 will be on sale fresh bread.

The probability of this event depends on many other events: whether fresh bread will be delivered tomorrow, whether it will be sold out before 7 pm or not, etc. Depending on various circumstances, this event can be both reliable and impossible. So the event is dependent.

Bread ... and, as the Romans demanded, circuses:

- at the exam, the student will get a simple ticket.

If you go not the very first, then the event will be dependent, since its probability will depend on which tickets the classmates have already drawn.

How to determine dependency/independence of events?

Sometimes this is directly stated in the condition of the problem, but most often you have to conduct an independent analysis. There is no unambiguous guideline here, and the fact of dependence or independence of events follows from natural logical reasoning.

In order not to throw everything in one heap, tasks for dependent events I will highlight the next lesson, but for now we will consider the most common bunch of theorems in practice:

Problems on addition theorems for inconsistent probabilities
and multiplying the probabilities of independent events

This tandem, according to my subjective assessment, works in about 80% of the tasks on the topic under consideration. A hit of hits and a real classic of probability theory:

Task 5

Two shooters fired one shot each at the target. The probability of hitting for the first shooter is 0.8, for the second - 0.6. Find the probability that:

a) only one shooter will hit the target;
b) at least one of the shooters will hit the target.

Solution: The hit/miss probability of one shooter is obviously independent of the other shooter's performance.

Consider the events:
– 1st shooter will hit the target;
- The 2nd shooter will hit the target.

By condition: .

Let's find the probabilities of opposite events - that the corresponding arrows will miss:

a) Consider the event: - only one shooter hits the target. This event consists of two incompatible outcomes:

1st shooter will hit And 2nd misses
or
1st will miss And 2nd will hit.

On the tongue event algebras this fact can be written as:

First, we use the theorem of addition of probabilities of incompatible events, then - the theorem of multiplication of probabilities of independent events:

is the probability that there will be only one hit.

b) Consider the event: - at least one of the shooters will hit the target.

First of all, LET'S THINK - what does the condition "AT LEAST ONE" mean? IN this case this means that either the 1st shooter will hit (the 2nd one will miss) or 2nd (1st misses) or both arrows at once - a total of 3 incompatible outcomes.

Method one: given the prepared probability of the previous item, it is convenient to represent the event as the sum of the following disjoint events:

one will get (an event consisting in turn of 2 incompatible outcomes) or
If both arrows hit, we denote this event by the letter .

Thus:

According to the theorem of multiplication of probabilities of independent events:
is the probability that the 1st shooter will hit And 2nd shooter will hit.

According to the theorem of addition of probabilities of incompatible events:
is the probability of at least one hit on the target.

Method two: consider the opposite event: – both shooters will miss.

According to the theorem of multiplication of probabilities of independent events:

As a result:

Special attention pay attention to the second method - in the general case it is more rational.

In addition, there is an alternative, third way of solving, based on the theorem of summing joint events, which was silent above.

! If you are reading the material for the first time, then in order to avoid confusion, it is better to skip the next paragraph.

Method three : the events are joint, which means that their sum expresses the event “at least one shooter hits the target” (see Fig. event algebra). By theorem of addition of probabilities of joint events and the theorem of multiplication of probabilities of independent events:

Let's check: events and (0, 1 and 2 hits respectively) form a complete group, so the sum of their probabilities must be equal to one:
, which was to be verified.

Answer:

With a thorough study of the theory of probability, you will come across dozens of tasks of a militaristic content, and, which is typical, after that you will not want to shoot anyone - the tasks are almost gift. Why not make the template even simpler? Let's shorten the entry:

Solution: according to the condition: , is the probability of hitting the corresponding shooters. Then their miss probabilities are:

a) According to the theorems of addition of probabilities of incompatible and multiplication of probabilities of independent events:
is the probability that only one shooter will hit the target.

b) According to the theorem of multiplication of probabilities of independent events:
is the probability that both shooters will miss.

Then: is the probability that at least one of the shooters will hit the target.

Answer:

In practice, you can use any design option. Of course, much more often they go the short way, but one should not forget the 1st method - although it is longer, it is more meaningful - it is clearer in it, what, why and why adds up and multiplies. In some cases, a hybrid style is appropriate when capital letters It is convenient to indicate only some events.

Similar tasks for independent solution:

Task 6

Two independently operating sensors are installed for fire alarm. The probabilities that the sensor will operate during a fire are 0.5 and 0.7 for the first and second sensors, respectively. Find the probability that in a fire:

a) both sensors will fail;
b) both sensors will work.
c) using addition theorem for the probabilities of events forming a complete group, find the probability that only one sensor will operate during a fire. Check the result by direct calculation of this probability (using addition and multiplication theorems).

Here, the independence of the operation of devices is directly spelled out in the condition, which, by the way, is an important clarification. The sample solution is designed in an academic style.

What if, in a similar problem, the same probabilities are given, for example, 0.9 and 0.9? You need to decide exactly the same! (which, in fact, has already been demonstrated in the example with two coins)

Task 7

The probability of hitting the target by the first shooter with one shot is 0.8. The probability that the target is not hit after the first and second shooters shoot one shot is 0.08. What is the probability of hitting the target by the second shooter with one shot?

And this is a small puzzle, which is framed in a short way. The condition can be reformulated more concisely, but I will not remake the original - in practice, I have to delve into more ornate fabrications.

Meet him - he is the one who cut an unmeasured amount of details for you =):

Task 8

A worker operates three machines. The probability that during the shift the first machine will require adjustment is 0.3, the second - 0.75, the third - 0.4. Find the probability that during the shift:

a) all machines will require adjustment;
b) only one machine will require adjustment;
c) at least one machine will require adjustment.

Solution: since the condition does not say anything about a single technological process, then the operation of each machine should be considered independent of the operation of other machines.

By analogy with Task No. 5, here you can enter into consideration events consisting in the fact that the corresponding machines will require adjustment during the shift, write down the probabilities , find the probabilities of opposite events, etc. But with three objects, I don’t really want to draw up the task like that - it will turn out long and tedious. Therefore, it is noticeably more profitable to use the “quick” style here:

By condition: - the probability that during the shift the corresponding machines will require tuning. Then the probabilities that they will not require attention are:

One of the readers found a cool typo here, I won’t even correct it =)

a) According to the theorem of multiplication of probabilities of independent events:
is the probability that during the shift all three machines will require adjustment.

b) The event "During the shift, only one machine will require adjustment" consists of three incompatible outcomes:

1) 1st machine will require attention And 2nd machine will not require And 3rd machine will not require
or:
2) 1st machine will not require attention And 2nd machine will require And 3rd machine will not require
or:
3) 1st machine will not require attention And 2nd machine will not require And 3rd machine will require.

According to the theorems of addition of probabilities of incompatible and multiplication of probabilities of independent events:

- the probability that during the shift only one machine will require adjustment.

I think by now it should be clear to you where the expression came from

c) Calculate the probability that the machines will not require adjustment, and then the probability of the opposite event:
– the fact that at least one machine will require adjustment.

Answer:

Item "ve" can also be solved through the sum , where is the probability that during the shift only two machines will require adjustment. This event, in turn, includes 3 incompatible outcomes, which are signed by analogy with the "be" item. Try to find the probability yourself to check the whole problem with the help of equality.

Task 9

Three guns fired a volley at the target. The probability of hitting with one shot only from the first gun is 0.7, from the second - 0.6, from the third - 0.8. Find the probability that: 1) at least one projectile hits the target; 2) only two projectiles will hit the target; 3) the target will be hit at least twice.

Solution and answer at the end of the lesson.

And again about coincidences: in the event that, by condition, two or even all values ​​of the initial probabilities coincide (for example, 0.7; 0.7 and 0.7), then exactly the same solution algorithm should be followed.

In conclusion of the article, we will analyze another common puzzle:

Task 10

The shooter hits the target with the same probability with each shot. What is this probability if the probability of at least one hit in three shots is 0.973.

Solution: denote by - the probability of hitting the target with each shot.
and through - the probability of a miss with each shot.

Let's write down the events:
- with 3 shots, the shooter will hit the target at least once;
- the shooter will miss 3 times.

According to the condition, then the probability of the opposite event:

On the other hand, according to the theorem of multiplication of probabilities of independent events:

Thus:

- the probability of a miss with each shot.

As a result:
is the probability of hitting each shot.

Answer: 0,7

Simple and elegant.

In the considered problem, additional questions can be raised about the probability of only one hit, only two hits, and the probability of three hits on the target. The solution scheme will be exactly the same as in the two previous examples:

However, the fundamental substantive difference is that there are repeated independent tests, which are performed sequentially, independently of each other and with the same probability of outcomes.

Task 2. Throw 4 dice. Find the probability of getting the same number of points on each of the rolled dice

Solution. There are 6 faces in total on each bone. The fallout of each face is equally probable. If the first dice rolled, say, 1, then the rest should be the same. The probability of any particular face falling out so that all 4 identical ones fall out is the product of the probabilities of the appearance of a particular face on all 4 dice. The result must be multiplied by the number of faces, since there are 6 different numbers. Let's denote the desired event - "one fell on the die", - , then the loss of four ones on all the cubes will be . To find a solution to the problem, you need to multiply the result by 6, because the events "two rolled on all dice", "three rolled on all dice" ... satisfy the condition of the problem. So the solution to the problem will be:

Task 3. A trainee student was taught to shoot a can with a gun. The probability of hitting a jar with one shot is 0.03. How many cartridges do you need to prepare so that with a probability of 0.94 a can would be knocked to the ground?

Solution. Write an equation to find the probability of an event occurring. To do this, use the Bernoulli formula, which is used if several repetitions of the same event are performed. If we assume that the can is knocked to the ground by the very first hit, then before that shots were fired (with a miss), i.e. all shots were fired. If the probability of hitting is , then the probability of missing is . The probability of a miss and 1 hit event can be written:

We substitute the known data into the last formula: and express from the resulting equation:

Let's take the logarithm of the last expression:

Where

The absolute value is used here because the probabilities can only be positive. . The number of shots cannot be integer, so finally

Task 4. A die is tossed 6 times. What is the probability of getting 6 different faces?

Solution. There are 6 faces in total on each bone. The fallout of each face is equally probable. Events occur sequentially, but it doesn't matter in what order. The probability of any particular face falling out is 1 (the die is thrown and one face will appear), therefore, the second time any number should appear, except for the one that fell out (probability), the third time - any, except for the first two (probability), etc. The probability of the desired event is:

Task 5. Homogeneous dice has the shape of a regular tetrahedron. The numbers 1, 2, 3, and 4 are marked on its faces. How many times do you need to toss a die in order to expect a 3 to roll at least in one case with a probability greater than 0.9?

Solution. There are 4 faces in total on the bone. Each face is equally likely to fall out, but it will have to be thrown several times, so we will be based on the use of the Bernoulli formula. Suppose that in the th test the required number appeared, therefore all previous times were different. In this case, the probability of the appearance of a particular face will be equal, since there are only 4 faces. The probability of the event "the required face did not appear and the required face appeared once" can be written:

We substitute the known data into the last formula: and express from the resulting equation.

Let's take the logarithm of the last expression:

Where

The absolute value is used here because the probabilities can only be positive. . The number of rolls cannot be non-integer, so round up to the nearest integer. By condition, the probability must be greater than 0.9, so the answer is >6.

Task 6. Two hunters shoot independently of each other at one target, and each of them makes one shot. The probability of hitting the target for the first hunter is 0.8, and for the second - 0.4. After shooting, one hole was found in the target. Find the probability that it belongs to the first shooter?

Solution. Let's try to use the Bayes formula. According to the Bayes formula, the numerator contains the probability of the required event occurring, and the denominator contains the total probability of possible outcomes, which will determine the appearance of one hole in the target, i.e. situations when one of the hunters hit, and the second missed. There were two hunters, so only 2 options are possible: "the first hit, the second missed" and "the first missed, the second hit." Both events cannot occur at the same time, so we are talking about the sum of the probabilities. The probability of the required event occurring is "first miss, second hit". The probability of the event "first hit, second missed" is equal to , and the probability of the second event "the first missed, the second hit" is equal to . Let's use the recommended formula:

Task 7. Three shots are fired at a duck flying not very high. The probabilities of hitting the first, second, and third shots are 0.1, respectively; 0.2 and 0.4. Determine the probability of at least two hits on the duck.

Solution. Since the shots are fired sequentially, one must consider the possibility of missing the first time, or the second, or the third. According to the condition of the problem, there must be at least two hits on the duck, which implies either 2 hits or 3. There can be three "2 hits" events: "hit, hit, miss"; "hit, miss, hit"; "miss, hit, hit", because it is not known in advance which shot was a miss. Thus, we have 4 events that cannot occur simultaneously, therefore we are talking about the sum of the probabilities of events, i.e. about the total probability formula. The probability of the event "hit, hit, hit" is equal to ; the probability of the event "hit, hit, miss" is ; the probability of the event "hit, miss, hit" is ; The probability of a miss, hit, hit event is . Now we calculate the desired probability:

Task 8. The laboratory assistant, performing chemical analyzes, uses the reagents standing in two refrigerators. In the first refrigerator, of all stored reagents, only 10% are expired, and in the second - 20%. Find the probability that any reagent taken by a laboratory assistant from any refrigerator will be fresh enough

Solution. Let us denote the event as A - a laboratory assistant takes out a sufficiently fresh reagent from any refrigerator. The laboratory assistant takes a reagent from any refrigerator, of which there are two according to the condition of the problem. Because the problem says nothing about refrigerators, then the choice of any of them is equiprobable, i.e. is equal to . The probability of the required event, therefore, consists in the simultaneous occurrence of two - "the choice of refrigerator, the choice of reagent." The probability of "taking a fresh reagent from the first refrigerator" is equal to ; the probability of "taking a fresh reagent from the second refrigerator" is equal to . The laboratory assistant takes a reagent only once, so both events in "take a fresh reagent from the first refrigerator" and "take a fresh reagent from the second refrigerator" cannot occur at the same time, so we are talking about the sum of probabilities. Let's use the total probability formula. Then the desired probability will be equal to:

Task 9. There are 5 boxes with ornamental stones malachite and marble. Two boxes contain 2 pieces of marble and 1 piece of malachite, one contains 10 pieces of malachite, and the others contain 3 pieces of marble and 1 piece of malachite. Find the probability that a piece taken at random from a box chosen by the craftsman is marble.

Solution. This is a task to use the total probability formula. The master selects an ornamental stone from any, "randomly selected" box. There are 5 boxes in total, it is assumed that they are the same, so the probability of choosing any box is . The probability of the required event, therefore, consists in the simultaneous occurrence of two - "the choice of the box and the choice of the marble." The probability of taking marble from the first box is ; the probability of taking marble from the second box is ; the probability of taking the marble from the third box is 0, because there is only malachite, the probability of taking marble from the fourth box is ;