\(\blacktriangleright\) If the execution of the event \(C\) requires the execution of both simultaneous (which can occur simultaneously) events \(A\) and \(B\) (\(C=\(A\) and \( B\)\) ), then the probability of the event \(C\) is equal to the product of the probabilities of the events \(A\) and \(B\) .
Note that if the events are incompatible, then the probability of their simultaneous occurrence is \(0\) .
\(\blacktriangleright\) Each event can be represented as a circle. Then if the events are joint, then the circles must intersect. The probability of the event \(C\) is the probability of getting into both circles at the same time.
\(\blacktriangleright\) For example, when tossing a dice, find the probability \(C=\) (rolling the number \(6\) ).
The event \(C\) can be formulated as \(A=\) (an even number) and \(B=\) (a number divisible by three).
Then \(P\,(C)=P\,(A)\cdot P\,(B)=\dfrac12\cdot \dfrac13=\dfrac16\).
Task 1 #3092
Task level: Equal to the Unified State Examination
The store sells sneakers from two brands: Dike and Ananas. The probability that a randomly selected pair of sneakers is Dike is \(0.6\) . Each company can make a mistake in writing its name on sneakers. The probability that Dike spells the name wrong is \(0.05\) ; The probability that Ananas spells the name wrong is \(0.025\) . Find the probability that a randomly purchased pair of sneakers will have the correct spelling of the company name.
Event A: “the pair of sneakers will be with the correct name” is equal to the sum of the events B: “the pair of sneakers will be from Dike and with the correct name” and C: “the pair of sneakers will be from Ananas and with the correct name.”
The probability of event B is equal to the product of the probabilities of the events “sneakers will be made by Dike” and “the name of the company Dike spelled correctly”: \ Similarly for event C: \
Consequently, \
Answer: 0.96
Task 2 #166
Task level: Equal to the Unified State Examination
If Timur plays with white checkers, then he beats Vanya with a probability of 0.72. If Timur plays with black checkers, then he beats Vanya with a probability of 0.63. Timur and Vanya play two games, and in the second game they change the color of the checkers. Find the probability that Vanya wins both times.
Vanya wins with white with probability \(0.37\) , and with black with probability \(0.28\) . The events “from two games Vanya won with white”\(\ \) and “from two games Vanya won with black”\(\ \) are independent, then the probability of their simultaneous occurrence is equal to \
Answer: 0.1036
Task 3 #172
Task level: Equal to the Unified State Examination
The entrance to the museum is guarded by two guards. The probability that the oldest of them will forget the walkie-talkie is \(0,2\) , and the probability that the youngest of them will forget the walkie-talkie is \(0,1\) . What is the probability that they will not have any radios?
Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. Then the desired probability is equal to \
Answer: 0.02
Task 4 #167
Task level: Equal to the Unified State Examination
Jumping from a height of 1 meter, Kostya breaks his leg with a probability of \(0.05\) . Jumping from a height of 1 meter, Vanya breaks his leg with a probability \(0.01\) . Jumping from a height of 1 meter, Anton breaks his leg with a probability of \(0.01\) . Kostya, Vanya and Anton simultaneously jump from a height of 1 meter. What is the probability that only Kostya will break his leg? Round your answer to thousandths.
Events “When jumping from a height of 1 meter, Kostya broke his leg”\(,\ \) “When jumping from a height of 1 meter, Vanya did not break his leg”\(\ \) and “When jumping from a height of 1 meter, Anton did not break his leg”\( \ \) are independent, therefore, the probability of their simultaneous occurrence is equal to the product of their probabilities: \ After rounding, we finally get \(0,049\) .
Answer: 0.049
Task 5 #170
Task level: Equal to the Unified State Examination
Maxim and Vanya decided to go bowling. Maxim rightly estimated that on average he hits a strike once every eight throws. Vanya rightly estimated that on average he knocks out a strike once every five throws. Maxim and Vanya make exactly one throw each (regardless of the result). What is the probability that there will be no strikes among them?
Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. In this case, the probability that Maxim will not hit a strike is equal to \ The probability that Vanya won't hit a strike is \(1 - 0.2 = 0.8\) . Then the desired probability is equal to \[\dfrac(7)(8)\cdot 0.8 = 0.7.\]
Answer: 0.7
Task 6 #1646
Task level: Equal to the Unified State Examination
Anton and Kostya are playing table tennis. The probability that Kostya will hit the table with his signature blow is \(0.9\) . The probability that Anton will win the rally in which Kostya tried to deliver a signature blow is \(0,3\) . Kostya tried to hit the table with his signature blow. What is the probability that Kostya really hits with his signature blow and eventually wins this draw?
Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. At the same time, the probability that Anton will not win the rally in which Kostya tried to deliver his signature blow is \(1 - 0.3 = 0.7\) . Then the desired probability is equal to \
The theorem of multiplying the probabilities of two arbitrary events: the probability of the product of two arbitrary events is equal to the product of the probability of one of the events by the conditional probability of the other event, provided that the first has already happened:
P(AB)=P(A)P(B|A) = P(B)P(A|B). (10)
Proof (not rigorous): we will prove the multiplication theorem for the scheme of chances (equiprobable hypotheses). Let the possible outcomes of the experience be n chances. Assume that event A has m chances (shaded in Fig. 11); event B - k chances; simultaneously events A and B (AB) - l chances (in Fig. 11 they have light shading).
Figure 11
Obviously, m+k-l=n. According to the classical way of calculating the probabilities P(AB)=l/n; P(A)=m/n; P(B)=k/n. And the probability is P(B|A)=l/m, since it is known that one of the m chances of event A has occurred, and event B is favored by l similar chances. Substituting these expressions into theorem (10), we obtain the identity l/n=(m/n)(l/m). The theorem has been proven.
The probabilities multiplication theorem for three arbitrary events:
P(ABC)=|AB=D|=P(DC)=P(D)P(C|D)=P(AB)P(C|AB)=P(A)P(B|A)P( C|AB).(11)
By analogy, one can write probability multiplication theorems for more events.
Corollary 1. If event A does not depend on B, then event B does not depend on A either.
Proof. Because event A does not depend on B, then by the definition of independence of events P(A)=P(A|B)=P(A|). It is required to prove that P(B)=P(B|A).
By the multiplication theorem, P(AB)=P(A)P(B|A)=P(B)P(A|B), therefore, P(A)P(B|A)=P(B)P(A ). Assuming that P(A)>0, we divide both sides of the equality by P(A) and get: P(B)=P(B|A).
Corollary 1 implies that two events are independent if the occurrence of one of them does not change the probability of the occurrence of the other. In practice, events (phenomena) that are interconnected by a causal relationship are dependent.
Corollary 2. The probability of the product of two independent events is equal to the product of the probabilities of these events. Those. if events A and B are independent, then
P(AB)=P(A)P(B). (eleven)
The proof is obvious, since for independent events P(B|A)=P(B).
Identity (11), along with expressions (12) and (13), are necessary and sufficient conditions for the independence of two random events A and B.
P(A)=P(A|B); P(A)=P(A|); P(A|B)=P(A|); (12)
P(B)=P(B|A); P(B)=P(B|); P(B|A)=P(B|). (13)
The reliability of some system is increased by double redundancy (see Fig. 12). The probability of failure-free operation of the first subsystem (during some operating time) is 0.9, the second - 0.8. Determine the probability of failure of the system as a whole during a given operating time, if the failures of the subsystems are independent.
Figure 12 - Double redundant system
E: Reliability study of a doubly redundant control system;
A 1 =(non-failure operation (during some operating time) of the first subsystem); P(A1)=0.9;
A 2 =(failsafe operation of the second subsystem); P(A2)=0.8;
A=(failsafe operation of the system as a whole); P(A)=?
Solution. Let us express the event A in terms of the events A 1 and A 2 whose probabilities are known. Since the failure-free operation of at least one of its subsystems is sufficient for the failure-free operation of the system, it is obvious that A=A 1 A 2.
Applying the probability addition theorem, we get: P(A)=P(A 1 A 2)=P(A 1)+P(A 2)-P(A 1 A 2). The probability of the joint occurrence of events A 1 and A 2 is determined by the probability multiplication theorem: P(A 1 A 2)=P(A 1)P(A 2 |A 1). Considering that (by condition) the events A 1 and A 2 are independent, P(A 1 A 2)=P(A 1)P(A 2). Thus, the probability of failure-free operation of the system is P(A)=P(A 1 A 2)=P(A 1)+P(A 2)-P(A 1)P(A 2)=0.9+0, 8-0.90.8=0.98.
Answer: the probability of failure-free operation of the system during a given operating time is 0.98.
Comment. In example 20, another way of defining event A is possible through events A 1 and A 2: , i.e. A system failure is possible when both of its subsystems fail simultaneously. Applying the theorem of multiplication of probabilities of independent events, we obtain the following value of the probability of system failure: . Therefore, the probability of failure-free operation of the system during a given operating time is equal to.
Example 21 (paradox of independence)
E: Two coins are tossed.
A=(coat of arms loss on the first coin), P(A)=0.5;
B=(coat of arms loss on the second coin), P(B)=0.5;
C=(coat of arms appearing on only one of the coins), P(C)=0.5.
Events A, B and C are pairwise independent, since the conditions for the independence of two events (11)-(13) are satisfied:
P(A)=P(A|B)=0.5; P(B)=P(B|C)=0.5; P(C)=P(C|A)=0.5.
However, P(A|BC)=0P(A); P(A|C)=1P(A); P(B|AC)=0P(B); P(C|AB)=0P(C).
Comment. Pairwise independence of random events does not mean their independence in the aggregate.
Random events are collectively independent if the probability of occurrence of each of them does not change with the occurrence of any combination of other events. For random events A 1, A 2, ... A n, independent in the aggregate, the following probability multiplication theorem is valid (a necessary and sufficient condition for independence in the aggregate of n random events):
P(A 1 A 2 ... A n) \u003d P (A 1) P (A 2) ... P (A n). (fourteen)
For example 21, condition (14) is not satisfied: P(ABC)=0P(A)P(B)P(C)=0.50.50.5=0.125. Therefore, pairwise independent events A, B and C are mutually dependent.
Example 22
There are 12 transistors in the box, three of which are faulty. To assemble a two-stage amplifier, two transistors are randomly removed. What is the probability that the assembled amplifier will be faulty?
E: selection of two transistors from the box with 9 good and 3 bad transistors;
A=(faulty assembled amplifier); P(A)=?
Solution. Obviously, the assembled two-stage amplifier will be faulty if at least one of the two transistors selected for assembly is faulty. Therefore, we redefine event A as follows:
A=(at least one of the two selected transistors is faulty);
We define the following auxiliary random events:
A 01 = (only the first of the two selected transistors is faulty);
A 10 = (only the second of the two selected transistors is faulty);
A 00 = (both selected transistors are faulty);
It is obvious that A=A 01 A 10 A 00 (for the event A to occur, at least one of the events A 01 or A 10 or A 00 must occur), and the events A 01, A 10 and A 00 are incompatible (they cannot occur together) , so we find the probability of an event by the theorem of addition of probabilities of incompatible events:
P(A)=P(A 01 A 10 A 00)=P(A 01)+P(A 10)+P(A 00).
To determine the probabilities of events A 01, A 10 and A 00, we introduce auxiliary events:
B 1 =(the first selected transistor is defective);
B 2 =(Second selected transistor is defective).
It is obvious that A 01 =B 1 ; A 10 = B 2 ; A 00 = B 1 B 2 ; therefore, to determine the probabilities of events A 01, A 10 and A 00, we apply the probabilities multiplication theorem.
P(A 01)=P(B 1)=P(B 1)P(|B 1),
where P(B 1) is the probability that the first selected transistor will be faulty; P(|B 1) - the probability that the second selected transistor will be good, provided that the first selected transistor is faulty. Using the classical way of calculating probabilities, P(B 1)=3/12, and P(|B 1)=9/11 (because after choosing the first bad transistor, there are 11 transistors left in the box, 9 of which are good).
Thus, P(A 01)=P(B 1)=P(B 1)P(|B 1)=3/129/11=0.20(45). Similarly:
P(A 10)=P(B 2)=P()P(B 2 |)=9/123/11=0.20(45);
P(A 00)=P(B 1 B 2)=P(B 1)P(B 2 |B 1)=3/122/11=0.041(6).
Let us substitute the obtained values of the probabilities A 01, A 10 and A 00 into the expression for the probability of the event A:
P(A)=P(A 01 A 10 A 00)=P(A 01)+P(A 10)+P(A 00)=3/129/11+9/123/11+3/122/11 =0.45(45).
Answer: The probability that the assembled amplifier will be faulty is 0.4545.
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 subject to proof, which can be found, for example, in the textbook by 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 with specific 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 fresh bread will be on sale.
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 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 .
In this way:
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:
Pay special attention to the second method - in general 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 it is convenient to indicate only some events in capital letters.
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: 
In this way: 
- 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.
The study of probability theory begins with solving problems for addition and multiplication of probabilities. It is worth mentioning right away that when mastering this field of knowledge, a student may encounter a problem: if physical or chemical processes can be visualized and understood empirically, then the level of mathematical abstraction is very high, and understanding here comes only with experience.
However, the game is worth the candle, because the formulas - both those considered in this article and more complex ones - are used everywhere today and may well come in handy in work.
Origin
Oddly enough, the impetus for the development of this section of mathematics was ... gambling. Indeed, dice, coin toss, poker, roulette are typical examples that use addition and multiplication of probabilities. On the example of tasks in any textbook, this can be seen clearly. People were interested in learning how to increase their chances of winning, and I must say, some succeeded in this.
For example, already in the 21st century, one person, whose name we will not disclose, used this knowledge accumulated over the centuries to literally “cleanse” the casino, winning several tens of millions of dollars at roulette.
However, despite the increased interest in the subject, it was only by the 20th century that a theoretical base was developed that made the “theorver” complete. Today, in almost any science, one can find calculations using probabilistic methods.
Applicability
An important point when using the formulas for addition and multiplication of probabilities, conditional probability is the satisfiability of the central limit theorem. Otherwise, although it may not be realized by the student, all calculations, no matter how plausible they may seem, will be incorrect.
Yes, the highly motivated learner is tempted to use new knowledge at every opportunity. But in this case, one should slow down a little and strictly outline the scope of applicability.
Probability theory deals with random events, which in empirical terms are the results of experiments: we can roll a six-sided die, draw a card from a deck, predict the number of defective parts in a batch. However, in some questions it is categorically impossible to use formulas from this section of mathematics. We will discuss the features of considering the probabilities of an event, the theorems of addition and multiplication of events at the end of the article, but for now let's turn to examples.
Basic concepts
A random event is some process or result that may or may not appear as a result of an experiment. For example, we toss a sandwich - it can fall butter up or butter down. Either of the two outcomes will be random, and we do not know in advance which of them will take place.
When studying addition and multiplication of probabilities, we need two more concepts.
Joint events are such events, the occurrence of one of which does not exclude the occurrence of the other. Let's say two people shoot at a target at the same time. If one of them produces a successful one, it will not affect the ability of the second to hit the bull's-eye or miss.
Inconsistent events will be such events, the occurrence of which is simultaneously impossible. For example, by pulling out only one ball from the box, you cannot get both blue and red at once.
Designation
The concept of probability is denoted by the Latin capital letter P. Next, in parentheses, there are arguments denoting some events.
In the formulas of the addition theorem, conditional probability, multiplication theorem, you will see expressions in brackets, for example: A+B, AB or A|B. They will be calculated in various ways, and we will now turn to them.
Addition
Consider the cases in which the formulas for addition and multiplication of probabilities are used.
For incompatible events, the simplest addition formula is relevant: the probability of any of the random outcomes will be equal to the sum of the probabilities of each of these outcomes.
Suppose there is a box with 2 blue, 3 red and 5 yellow marbles. There are 10 items in total in the box. What is the percentage of the truth of the statement that we will draw a blue or red ball? It will be equal to 2/10 + 3/10, i.e. fifty percent.
In the case of incompatible events, the formula becomes more complicated, since an additional term is added. We will return to it in one paragraph, after considering one more formula.
Multiplication
Addition and multiplication of probabilities of independent events are used in different cases. If, according to the condition of the experiment, we are satisfied with either of the two possible outcomes, we will calculate the sum; if we want to get two certain outcomes one after the other, we will resort to using a different formula.
Returning to the example from the previous section, we want to draw the blue ball first and then the red one. The first number we know is 2/10. What happens next? There are 9 balls left, there are still the same number of red ones - three pieces. According to the calculations, you get 3/9 or 1/3. But what to do with two numbers now? The correct answer is to multiply to get 2/30.
Joint Events
Now we can again turn to the sum formula for joint events. Why are we digressing from the topic? To learn how probabilities are multiplied. Now we need this knowledge.
We already know what the first two terms will be (the same as in the addition formula considered earlier), but now we need to subtract the product of probabilities, which we just learned how to calculate. For clarity, we write the formula: P(A+B) = P(A) + P(B) - P(AB). It turns out that in one expression both addition and multiplication of probabilities are used.
Let's say we have to solve either of two problems in order to get credit. We can solve the first one with a probability of 0.3, and the second - 0.6. Solution: 0.3 + 0.6 - 0.18 = 0.72. Note that simply summing the numbers here will not be enough.
Conditional Probability
Finally, there is the concept of conditional probability, the arguments of which are indicated in brackets and separated by a vertical bar. The entry P(A|B) reads as follows: "probability of event A given event B".
Let's look at an example: a friend gives you some device, let it be a phone. It can be broken (20%) or good (80%). You are able to repair any device that falls into your hands with a probability of 0.4 or you are not able to do this (0.6). Finally, if the device is in working condition, you can get through to the right person with a probability of 0.7.
It's easy to see how the conditional probability works in this case: you can't get through to the person if the phone is broken, and if it's good, you don't need to fix it. Thus, in order to get any results at the "second level", you need to know which event was executed at the first.
calculations
Consider examples of solving problems for addition and multiplication of probabilities, using the data from the previous paragraph.
First, let's find the probability that you will repair the device given to you. To do this, firstly, it must be faulty, and secondly, you must cope with the repair. This is a typical multiplication problem: we get 0.2 * 0.4 = 0.08.
What is the probability that you will immediately get through to the right person? Easier than simple: 0.8 * 0.7 \u003d 0.56. In this case, you found that the phone is working and successfully made a call.
Finally, consider this scenario: you received a broken phone, fixed it, then dialed the number, and the person on the opposite end picked up the phone. Here, the multiplication of three components is already required: 0.2 * 0.4 * 0.7 \u003d 0.056.
But what if you have two non-working phones at once? How likely are you to fix at least one of them? on addition and multiplication of probabilities, since joint events are used. Solution: 0.4 + 0.4 - 0.4 * 0.4 = 0.8 - 0.16 = 0.64. Thus, if two broken devices fall into your hands, you will be able to fix it in 64% of cases.
Considerate use
As mentioned at the beginning of the article, the use of probability theory should be deliberate and conscious.
The larger the series of experiments, the closer the theoretically predicted value approaches the value obtained in practice. For example, we are tossing a coin. Theoretically, knowing about the existence of formulas for addition and multiplication of probabilities, we can predict how many times heads and tails will fall out if we conduct the experiment 10 times. We conducted an experiment, and by coincidence, the ratio of the sides that fell out was 3 to 7. But if you conduct a series of 100, 1000 or more attempts, it turns out that the distribution graph is getting closer and closer to the theoretical one: 44 to 56, 482 to 518, and so on.
Now imagine that this experiment is being conducted not with a coin, but with the production of some new chemical substance, the probability of which we do not know. We would run 10 experiments and, without getting a successful result, we could generalize: "the substance cannot be obtained." But who knows, if we made the eleventh attempt, would we have reached the goal or not?
Thus, if you are going into the unknown, into an unexplored area, the theory of probability may not be applicable. Each subsequent attempt in this case may succeed, and generalizations like "X does not exist" or "X is impossible" will be premature.
Final word
So, we have considered two types of addition, multiplication and conditional probabilities. With further study of this area, it is necessary to learn to distinguish situations when each specific formula is used. In addition, you need to understand whether probabilistic methods are generally applicable in solving your problem.
If you practice, after a while you will begin to carry out all the required operations exclusively in your mind. For those who are fond of card games, this skill can be considered extremely valuable - you will significantly increase your chances of winning by just calculating the probability of a particular card or suit falling out. However, the acquired knowledge can easily be applied in other areas of activity.
The product, or intersection, of events A and B is an event consisting in the simultaneous occurrence of events and A, and IN. Designation of the work AB or L and V.
For example, hitting the target twice is the product of two events, the answer to both questions of the ticket in the exam is the product of two events.
Events L and IN are called inconsistent if their product is an impossible event, i.e. LV = V.
For example, events L - the loss of the coat of arms and IN- loss of numbers during a single toss of a coin cannot occur simultaneously, their product is an impossible event, events L and B are incompatible.
The concepts of the sum and product of events have a clear geometric interpretation (Fig. 6.4).
Rice. 6.4. Geometric interpretation of the work (but) and amounts (b) two joint events
Let the event A be the set of points in the area L, and the event B be the set of points in the area B. The shaded area corresponds to the event LP in fig. 6 la and event A + B in Fig. 6.46.
For incompatible events A and B we have LP = V(Fig. 6.5a). The L + B event corresponds to the shaded area in Fig. 6.56.
Rice. 6.5. The geometric interpretation of the product ( but) and sums (b) two incompatible events
Developments BUT And BUT are called opposite if they are incompatible and in total constitute a reliable event, i.e.
A A = V; A+A=U.
For example, let's fire one shot at a target: event BUT- the shooter hit the target BUT- missed; coin tossed:
event BUT- eagle fall, BUT- loss of numbers; schoolchildren write a test: an event BUT- none
errors in control work, BUT- there are errors in the control work; student came to take the test: event BUT- passed
offset, BUT- did not submit a report.
There are boys and girls in the class, excellent students, good students and three students studying English and German. Let the event M be a boy, O an excellent student, A an English learner. Can a student who accidentally left the class be both a boy, and an excellent student, and an English learner? This will be the product or intersection of MOA events.
Example 6.15. Throw a dice - a cube made of a homogeneous material, the faces of which are numbered. Observe the number (number of points) falling on the top face. Let the event BUT - appearance of an odd number, event IN - appearance of a multiple of three. Find the outcomes that make up each of the events (?/, A, A + V U AB) and indicate their meaning.
Solution. Outcome - the appearance on the upper face of any of the numbers 1, 2, 3, 4, 5, 6. The set of all outcomes is the space of elementary events U= (1, 2, 3, 4, 5, 6). It is clear that the event A =(1, 3, 5), event B = {3, 6}.
Event BUT + B =(1, 3, 5, 6) - the appearance of either an odd number or a number that is a multiple of three. When listing the outcomes, it is taken into account that each outcome in the set can be contained only once.
Event AB =(3) - the appearance of both an odd number and a multiple of three.
Example 6.16. The homework of three students was checked. Let the event BUT ( - performance of the task by the i-th student, G = 1, 2, 3.
What is the meaning of the events: A = A t + A 2+ L 3, BUT And B \u003d A t A 2 A 3?
Solution. Event BUT = A x + A 2 + A 3 - completion of the task by at least one student, i.e. or any one student (or first, or second, or third), or any two, or all three.
Event A \u003d A x -A 2 -A 3- the task was not completed by any student - neither the first, nor the second, nor the third. Event B \u003d A ( A 2 A 3 - performance of the task by three students - and the first, and the second, and the third.
When considering the joint occurrence of several events, there may be cases when the occurrence of one of them affects the possibility of the occurrence of another. For example, if the day is sunny in autumn, then the weather is less likely to deteriorate (it starts to rain). If the sun is not visible, then it is more likely that it will rain.
Event L called event independent IN, if the probability of an event BUT does not change depending on whether the event occurred or not IN. Otherwise the event BUT is called event dependent IN. Two events A andIN are called independent if the probability of one of them does not depend on the occurrence or non-occurrence of the other, dependent - otherwise. Events are called pairwise independent if every two of them are independent of each other.
The probability multiplication theorem is formulated as follows. The probability of the product of two independent events is equal to the product of the probabilities of these events:
This theorem is valid for any finite number of events, as long as they are collectively independent, i.e. the probability of any of them does not depend on whether the other of these events occurred or not.
Example 6.17. The student takes three exams. The probability of passing the first exam is 0.9, the second - 0.65, the third - 0.35. Find the probability that he fails at least one exam.
Solution. Denote BUT event - the student did not pass at least one exam. Then P(A) = 1 - /-’(1/1), where BUT- the opposite event - the student passed all the exams. Since the passing of each exam does not depend on other exams, then P(A)= 1 - P(1/1) = = 1 - 0.9 0.65 0.35 = 0.7953.
Event Probability BUT, computed assuming an event occurs IN, called conditional probability developments BUT subject to the appearance IN and denoted R B (A) or P(A/B).
Theorem.The probability of occurrence of the product of two events is equal to the product of the probability of one of them by the conditional probability of the second, calculated under the condition that the first event occurred:
Example 6.18. The student draws one ticket out of 34 twice. What is the probability that he will pass the exam if he has prepared 30 tickets and the first time he takes out an unsuccessful ticket?
Solution. Let the event BUT consists in the fact that for the first time you got an unsuccessful ticket, an event IN- the second time a successful ticket is drawn. Then BUT?IN- the student will pass the exam (under the specified circumstances). Developments BUT And IN are dependent, since the probability of choosing a successful ticket on the second attempt depends on the outcome of the first choice. Therefore, we use formula (6.6):
Note that the probability obtained in the solution is “0.107. Why is the probability of passing the exam so small if 30 tickets out of 34 are learned and two attempts are given?!
Extended addition theorem is formulated as follows. The probability of the sum of two events is equal to the sum of the probabilities of these events without the probability of their joint occurrence (works):
Example 6.19. Two students solve a problem. The probability that the first student solves the problem (event BUT), equal to 0.9; the probability that the second student solves the problem (event IN), equals 0.8. What is the probability that the problem will be solved?
Solution. We are interested in the event C, which consists in the fact that the problem will be solved, i.e. the first, or the second student, or two students at the same time. Thus, the event of interest to pass C \u003d A +IN. Developments BUT And IN are joint, then the probability addition theorem is applicable for the case of joint events: P(A + IN) = P(A) + P(B) - P(AB). For our case P(A + B) = = 0.9 + 0.8 + 0.9 0.8 = 0.98 (events BUT And IN joint but independent).
Example 6.20. The student knows 20 questions out of 25. What is the probability of answering 3 questions out of 25?
Solution. Let's introduce the event A, - the student knows the answer to i-th proposed question, i= 1,2,3. Events L, L 2 , L 3 - dependent. That's why
When finding the probabilities of events, the classical definition of probability was used.