Probability addition and multiplication theorems.
Dependent and independent events
The title looks scary, but in reality everything is very simple. On this lesson we will get acquainted with the theorems of addition and multiplication of event probabilities, and also analyze typical problems that, along with problem on the classical determination of probability will definitely meet or, more likely, have already met on your way. For effective learning materials of this article you need to know and understand the basic terms probability theory and be able to do the simplest 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 plenty of subtleties. Good luck:
Theorem for adding probabilities of incompatible events: probability of occurrence of one of two incompatible events or (no matter what), is equal to the sum of the probabilities of these events:
A similar fact is true for a larger number of incompatible events, for example, for three incompatible events and:
The theorem is a dream =) However, such a dream is subject to proof, which can be found, for example, in textbook V.E. Gmurman.
Let's get acquainted with new, hitherto unknown 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 on the appearance/non-appearance of other events of the set under consideration (in all possible combinations). ...But why bother trying out general phrases:
Theorem for multiplying the 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 appear on the 1st coin;
– heads will appear on the 2nd coin.
Let's find the probability of the event (heads will appear on the 1st coin And an eagle will appear on the 2nd coin - remember how to read product of events!)
. The probability of heads on one coin does not depend in any way on the result of throwing another coin, therefore, the events are independent. 
Likewise: 
– the probability that the 1st coin will land heads And on the 2nd tails; 
– the probability that heads will appear on the 1st coin And on the 2nd tails; 
– the probability that the 1st coin will show heads And on the 2nd eagle.
Notice that the 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, for example, if the events are independent, then the probability of their joint occurrence is equal to: . Let's practice on specific examples:
Problem 3
Each of the three boxes contains 10 parts. The first box contains 8 standard parts, the second – 7, the third – 9. One part is randomly removed from each box. Find the probability that all parts will be standard.
Solution: The probability of drawing a standard or non-standard part from any box does not depend on what parts are taken from other boxes, so the problem deals with independent events. Consider the following independent events:
– a standard part is removed from the 1st box;
– a standard part was removed from the 2nd box;
– a standard part is removed from the 3rd box.
According to the classical definition:
are the corresponding probabilities.
Event of interest to us (a standard part will be removed from the 1st box And from 2nd standard And from 3rd standard) is expressed by the product.
According to the theorem of multiplication of probabilities of independent events:
– the probability that one standard part will be removed from three boxes.
Answer: 0,504
After invigorating exercises with boxes, no less interesting urns await us:
Problem 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” point ;-) Approximate sample solutions are designed in an academic style with a detailed list of all events.
Dependent Events. The event is called dependent , if its probability depends from one or more events that have already occurred. You don’t have to go far for examples - just go to the nearest store:
– will be on sale tomorrow at 19.00 fresh bread.
The likelihood 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 either reliable or impossible. So the event is dependent.
Bread... and, as the Romans demanded, circuses:
– at the exam, the student will receive a simple ticket.
If you are not the very first, then the event will be dependent, since its probability will depend on what tickets have already been drawn by classmates.
How to determine the dependence/independence of events?
Sometimes this is directly stated in the problem statement, 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 lump everything into one pile, tasks for dependent events I will highlight the following lesson, but for now we will consider the most common set of theorems in practice:
Problems on addition theorems for incompatible probabilities
and multiplying the probabilities of independent events
This tandem, according to my subjective assessment, works in approximately 80% of tasks on the topic under consideration. Hit of hits and a real classic of probability theory:
Problem 5
Two shooters each fired one shot at the target. The probability of a hit 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: One shooter's hit/miss rate is obviously independent of the other shooter's performance.
Let's consider the events:
– 1st shooter will hit the target;
– The 2nd shooter will hit the target.
By condition: .
Let's find the probabilities opposing events– that the corresponding arrows will miss: 
a) Consider the event: – only one shooter will hit the target. This event consists of two incompatible outcomes:
1st shooter will hit And 2nd one will miss
or
1st one will miss And The 2nd one will hit.
On the tongue event algebras this fact will be written by the following formula: 
First, we use the theorem for adding the probabilities of incompatible events, then the theorem for multiplying the probabilities of independent events:
– the probability that there will be only one hit.
b) Consider the event: – at least one of the shooters hits the target.
First of all, LET’S THINK – what does the condition “AT LEAST ONE” mean? IN in this case this means that either the 1st shooter will hit (the 2nd will miss) or 2nd (1st will miss) or both shooters at once - a total of 3 incompatible outcomes.
Method one: taking into account the ready probability of the previous point, it is convenient to represent the event as the sum of the following incompatible events:
someone will get there (an event consisting in turn of 2 incompatible outcomes) or
If both arrows hit, we denote this event with the letter .
Thus:
According to the theorem of multiplication of probabilities of independent events:
– probability that the 1st shooter will hit And The 2nd shooter will hit.
According to the theorem of addition of probabilities of incompatible events:
– 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 general, it is more rational.
In addition, there is an alternative, third way of solving it, based on the theorem of addition of joint events, which was not mentioned above.
! If you are getting acquainted with 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 compatible, which means their sum expresses the event “at least one shooter will hit the target” (see. algebra of events). By the theorem for adding 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 equal one:
, which was what needed to be checked.
Answer: 
With a thorough study of probability theory, you will come across dozens of problems with a militaristic content, and, characteristically, after this you will not want to shoot anyone - the problems are almost a gift. Why not simplify the template as well? Let's shorten the entry:
Solution: by condition: , – probability of hitting the corresponding shooters. Then the probabilities of their miss: 
a) According to the theorems of addition of probabilities of incompatible and multiplication of probabilities of independent events:
– the probability that only one shooter will hit the target.
b) According to the theorem of multiplication of probabilities of independent events:
– the probability that both shooters will miss.
Then: – 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 take the short route, but we must not forget the 1st method - although it is longer, it is more meaningful - it is clearer, what, why and why adds and multiplies. In some cases, a hybrid style is appropriate, when it is convenient to use capital letters to indicate only some events.
Similar tasks for independent decision:
Problem 6
To signal a fire, two independently operating sensors are installed. The probabilities that the sensor will operate in the event of a fire are 0.5 and 0.7, respectively, for the first and second sensors. Find the probability that in a fire:
a) both sensors will fail;
b) both sensors will work.
c) Using the theorem for adding the probabilities of events forming a complete group, find the probability that in a fire only one sensor will work. Check the result by directly calculating this probability (using addition and multiplication theorems).
Here, the independence of the operation of the devices is directly stated 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)
Problem 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 fire one shot each is 0.08. What is the probability of the second shooter hitting the target with one shot?
And this is a small puzzle, which is designed in a short way. The condition can be reformulated more succinctly, but I will not redo the original - in practice, I have to delve into more ornate fabrications.
Meet him - he is the one who has planned an enormous amount of details for you =):
Problem 8
A worker operates three machines. The probability that during a 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 Problem No. 5, here you can enter into consideration the events that the corresponding machines will require adjustments during the shift, write down the probabilities, find the probabilities of opposite events, etc. But with three objects, I don’t really want to format the task like this anymore – it will turn out long and tedious. Therefore, it is noticeably more profitable to use the “fast” style here:
According to the 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:
– the probability that during the shift all three machines will require adjustments.
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 won't require And 3rd machine won't require
or:
2) 1st machine won't require attention And 2nd machine will require And 3rd machine won't require
or:
3) 1st machine won't require attention And 2nd machine won't 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 a shift only one machine will require adjustment.
I think by now you should understand where the expression comes from
c) Let’s calculate the probability that the machines will not require adjustment, and then the probability of the opposite event:
– that at least one machine will require adjustment.
Answer:
Point “ve” can also be solved through the sum , where is the probability that during a shift only two machines will require adjustment. This event, in turn, includes 3 incompatible outcomes, which are described by analogy with the “be” point. Try to find the probability yourself to check the whole problem using equality.
Problem 9
A salvo was fired from three guns at the target. The probability of a hit with one shot from only 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 will hit the target; 2) only two shells will hit the target; 3) the target will be hit at least twice.
The solution and answer are at the end of the lesson.
And again about coincidences: if, according to the 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.
To conclude the article, let’s look at another common puzzle:
Problem 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 with three shots is 0.973.
Solution: let us denote by – the probability of hitting the target with each shot.
and through - the probability of a miss with each shot.
And let’s write down the events:
– with 3 shots the shooter will hit the target at least once;
– the shooter will miss 3 times.
By 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:
– the probability of a hit with each shot.
Answer: 0,7
Simple and elegant.
In the problem considered, additional questions can be asked 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 here there are repeated independent tests, which are performed sequentially, independently of each other and with the same probability of outcomes.
Educational institution "Belarusian State
agricultural Academy"
Department of Higher Mathematics
ADDITION AND MULTIPLICATION OF PROBABILITIES. REPEATED INDEPENDENT TESTS
Lecture for students of the Faculty of Land Management
correspondence courses
Gorki, 2012
Addition and multiplication of probabilities. Repeated
independent tests
Addition of probabilities
The sum of two joint events A And IN called event WITH, consisting in the occurrence of at least one of the events A or IN. Similarly, the sum of several joint events is an event consisting of the occurrence of at least one of these events.
The sum of two incompatible events A And IN called event WITH consisting of an occurrence or event A, or events IN. Similarly, the sum of several incompatible events is an event consisting of the occurrence of any one of these events.
The theorem for adding the probabilities of incompatible events is valid: the probability of the sum of two incompatible events is equal to the sum of the probabilities of these events , i.e. . This theorem can be extended to any finite number of incompatible events.
From this theorem it follows:
the sum of the probabilities of events forming a complete group is equal to one;
the sum of the probabilities of opposite events is equal to one, i.e. 
.
Example 1 . The box contains 2 white, 3 red and 5 blue balls. The balls are mixed and one is drawn at random. What is the probability that the ball will be colored?
Solution . Let's denote the events:
A=(colored ball drawn);
B=(white ball drawn);
C=(red ball drawn);
D=(blue ball drawn).
Then A= C+ D. Since events C, D are inconsistent, then we will use the theorem for adding the probabilities of incompatible events: .
Example 2 . The urn contains 4 white balls and 6 black ones. 3 balls are drawn at random from the urn. What is the probability that they are all the same color?
Solution . Let's denote the events:
A=(balls of the same color are drawn);
B=(white balls are taken out);
C=(black balls are taken out).
Because A=
B+
C and events IN And WITH are inconsistent, then by the theorem of addition of probabilities of incompatible events 
. Probability of event IN equal to 
, Where 
4,
. Let's substitute k And n into the formula and we get 
Similarly, we find the probability of the event WITH:
, Where 
,
, i.e. 
. Then 
.
Example 3 . From a deck of 36 cards, 4 cards are drawn at random. Find the probability that there will be at least three aces among them.
Solution . Let's denote the events:
A=(among the cards taken out there are at least three aces);
B=(among the cards taken out are three aces);
C=(among the cards taken out are four aces).
Because A=
B+
C, and events IN And WITH are incompatible, then 
. Let's find the probabilities of events IN And WITH:
,
. Therefore, the probability that among the drawn cards there are at least three aces is equal to
0.0022.
Multiplying Probabilities
The work
two events A And IN called event WITH, consisting in the joint occurrence of these events: 
. This definition applies to any finite number of events.
The two events are called independent
, if the probability of one of them occurring does not depend on whether the other event occurred or not. Events 
,
,
… ,
are called collectively independent
, if the probability of the occurrence of each of them does not depend on whether other events occurred or did not occur.
Example 4 . Two shooters shoot at a target. Let's denote the events:
A=(the first shooter hit the target);
B=(the second shooter hit the target).
Obviously, the probability of the first shooter hitting the target does not depend on whether the second shooter hit or missed, and vice versa. Therefore, events A And IN independent.
The theorem for multiplying the probabilities of independent events is valid: the probability of the product of two independent events is equal to the product of the probabilities of these events : .
This theorem is also valid for n collectively independent events: .
Example 5 . Two shooters shoot at the same target. The probability of hitting the first shooter is 0.9, and the second is 0.7. Both shooters fire one shot at a time. Determine the probability that there will be two hits on the target.
Solution . Let's denote the events:
A
B
C=(both shooters will hit the target).
Because 
, and events A And IN are independent, then 
, i.e..
Events A And IN are called dependent
, if the probability of one of them occurring depends on whether another event occurred or not. Probability of an event occurring A provided that the event IN it's already arrived, it's called conditional probability
and is designated 
or 
.
Example 6 . The urn contains 4 white and 7 black balls. Balls are drawn from the urn. Let's denote the events:
A=(white ball drawn) ;
B=(black ball drawn).
Before starting to remove balls from the urn 
. One ball was taken from the urn and it turned out to be black. Then the probability of the event A after the event IN there will be another, equal 
. This means that the probability of an event A depends on the event IN, i.e. these events will be dependent.
The theorem for multiplying the probabilities of dependent events is valid: the probability of two dependent events occurring is equal to the product of the probability of one of them and the conditional probability of the other, calculated under the assumption that the first event has already occurred, i.e. or.
Example 7 . The urn contains 4 white balls and 8 red balls. Two balls are drawn sequentially from it at random. Find the probability that both balls are black.
Solution . Let's denote the events:
A=(black ball drawn first);
B=(the second black ball is drawn).
Events A And IN dependent because 
, A 
. Then 
.
Example 8 . Three shooters shoot at the target independently of each other. The probability of hitting the target for the first shooter is 0.5, for the second – 0.6 and for the third – 0.8. Find the probability that there will be two hits on the target if each shooter fires one shot.
Solution . Let's denote the events:
A=(there will be two hits on the target);
B=(the first shooter will hit the target);
C=(the second shooter will hit the target);
D=(the third shooter will hit the target);
=(the first shooter will not hit the target);
=(the second shooter will not hit the target);
=(the third shooter will not hit the target).
According to the example 
,
,
,
,
,
. Since, using the theorem for adding the probabilities of incompatible events and the theorem for multiplying the probabilities of independent events, we obtain:
Let events 
form a complete group of events of some test, and the events A can only occur with one of these events. If the probabilities and conditional probabilities of the event are known A, then the probability of event A is calculated by the formula:
or 
. This formula is called total probability formula
, and events 
hypotheses
.
Example 9
. The assembly line receives 700 parts from the first machine and 300 parts 
from the second. The first machine produces 0.5% scrap, and the second - 0.7%. Find the probability that the part taken will be defective.
Solution . Let's denote the events:
A=(the part taken will be defective);
=(the part was made on the first machine);
=(the part is made on the second machine).
The probability that the part is made on the first machine is equal to 
. For the second machine 
. According to the condition, the probability of receiving a defective part made on the first machine is equal to 
. For the second machine this probability is equal to 
. Then the probability that the taken part will be defective is calculated using the total probability formula 
If it is known that some event occurred as a result of the test A, then the probability that this event occurred with the hypothesis 
, is equal 
, Where 
- total probability of an event A. This formula is called Bayes formula
and allows you to calculate the probabilities of events 
after it became known that the event A has already arrived.
Example 10 . The same type of car parts are produced at two factories and delivered to the store. The first plant produces 80% of the total number of parts, and the second - 20%. The products of the first plant contain 90% of standard parts, and the second - 95%. The buyer bought one part and it turned out to be standard. Find the probability that this part was manufactured at the second plant.
Solution . Let's denote the events:
A=(standard part purchased);
=(the part was manufactured at the first plant);
=(the part was manufactured at the second plant).
According to the example 
,
,
And 
. Let's calculate full probability events A: 0.91. We calculate the probability that the part was manufactured at the second plant using the Bayes formula: 
.
Tasks for independent work
The probability of hitting the target for the first shooter is 0.8, for the second – 0.7 and for the third – 0.9. The shooters fired one shot each. Find the probability that there are at least two hits on the target.
The repair shop received 15 tractors. It is known that 6 of them need to replace the engine, and the rest need to replace individual components. Three tractors are selected at random. Find the probability that engine replacement is necessary for no more than two selected tractors.
The reinforced concrete plant produces panels, 80% of which are of the highest quality. Find the probability that out of three randomly selected panels, at least two will be of the highest grade.
Three workers are assembling bearings. The probability that the bearing assembled by the first worker is of the highest quality is 0.7, by the second – 0.8 and by the third – 0.6. For control, one bearing was taken at random from those assembled by each worker. Find the probability that at least two of them will be of the highest quality.
The probability of winning the first lottery ticket is 0.2, the second is 0.3 and the third is 0.25. There is one ticket for each issue. Find the probability that at least two tickets will win.
The accountant performs calculations using three reference books. The probability that the data he is interested in is in the first directory is 0.6, in the second - 0.7 and in the third - 0.8. Find the probability that the data the accountant is interested in is contained in no more than two directories.
Three machines produce parts. The first machine produces a part of the highest quality with probability 0.9, the second with probability 0.7 and the third with probability 0.6. One part is taken at random from each machine. Find the probability that at least two of them are of the highest quality.
The same type of parts are processed on two machines. The probability of producing a non-standard part for the first machine is 0.03, for the second – 0.02. The processed parts are stored in one place. Among them, 67% are from the first machine, and the rest are from the second. The part taken at random turned out to be standard. Find the probability that it was made on the first machine.
The workshop received two boxes of the same type of capacitors. The first box contained 20 capacitors, of which 2 were faulty. The second box contains 10 capacitors, of which 3 are faulty. The capacitors were placed in one box. Find the probability that a capacitor taken at random from a box will be in good condition.
Three machines produce the same type of parts, which are supplied to a common conveyor. Among all the parts, 20% are from the first machine, 30% from the second and 505 from the third. The probability of producing a standard part on the first machine is 0.8, on the second – 0.6 and on the third – 0.7. The part taken turned out to be standard. Find the probability that this part was made on the third machine.
The assembler receives 40% of the parts from the factory for assembly A, and the rest - from the factory IN. The likelihood that the part is from the factory A– superior quality, equal to 0.8, and from the factory IN– 0.9. The assembler took one part at random and it turned out to be of poor quality. Find the probability that this part is from the factory IN.
10 students from the first group and 8 from the second were allocated to participate in student sports competitions. The probability that a student from the first group will be included in the academy team is 0.8, and from the second - 0.7. A randomly selected student was included in the team. Find the probability that he is from the first group.
Bernoulli's formula
The tests are called independent
, if for each of them the event A occurs with the same probability 
, independent of whether this event appeared or did not appear in other trials. Probability of the opposite event 
in this case equals 
.
Example 11
. Dice being thrown n once. Let's denote the event A=(rolling three points). Probability of an event occurring A in each trial is equal and does not depend on whether this event occurred or did not occur in other trials. Therefore, these tests are independent. Probability of the opposite event 
(not rolling three points) is equal to 
.
The probability that in n independent trials, in each of which the probability of the event occurring A equal to p, the event will occur exactly k times (it doesn’t matter in what order), calculated by the formula 
, Where 
. This formula is called Bernoulli's formula
and it is convenient if the number of tests n is not too large.
Example 12 . The proportion of fruits infected with the disease in latent form is 25%. 6 fruits are randomly selected. Find the probability that among those selected there will be: a) exactly 3 infected fruits; b) no more than two infected fruits.
Solution . According to the example conditions.
a) According to Bernoulli’s formula, the probability that among six selected fruits exactly three will be infected is equal to
0.132.
b) Let us denote the event A=(no more than two fruits will be infected). Then . According to Bernoulli's formula:
0.297.
Hence, 
0.178+0.356+0.297=0.831.
Laplace's and Poisson's theorems
Bernoulli's formula is used to find the probability that an event A will come k once every n independent trials and in each trial the probability of an event A is constant. At large values n calculations using Bernoulli's formula become laborious. In this case, to calculate the probability of an event A It would be better to use a different formula.
Local Laplace theorem . Let the probability p occurrence of an event A in each trial is constant and different from zero and one. Then the probability that the event A will come exactly k times with enough large number n tests, calculated by the formula
, Where 
, and the function values 
are given in the table.
Main properties of the function 
are:
Function 
defined and continuous in the interval 
.
Function 
is positive, i.e. 
>0.
Function 
even, i.e. 
.
Since the function 
is even, then the table shows its values only for positive values X.
Example 13 . The germination rate of wheat seeds is 80%. 100 seeds are selected for the experiment. Find the probability that exactly 90 of the selected seeds will sprout.
Solution
. According to the example n=100,
k=90,
p=0.8,
q=1-0.8=0.2. Then 
. Using the table we find the value of the function 
:
. The probability that exactly 90 of the selected seeds will sprout is equal to 
0.0044.
When solving practical problems, it becomes necessary to find the probability of an event occurring A at n independent tests no less 
once and no more 
once. This problem is solved using Laplace's integral theorem
: Let the probability p occurrence of an event A in each n independent tests is constant and different from zero and one. Then the probability that the event will occur is at least 
once and no more 
times with a sufficiently large number of tests, is calculated by the formula
Where 
,
.
Function 
called Laplace function
and is not expressed through elementary functions. The values of this function are given in special tables.
Main properties of the function 
are:
.
Function 
increases in the interval 
.
at 
.
Function 
odd, i.e. 
.
Example 14 . The company produces products, 13% of which are not of the highest quality. Determine the probability that in an untested batch of 150 units of the highest quality product there will be no less than 125 and no more than 135.
Solution
. Let's denote . Let's calculate 
,
Addition and multiplication of probabilities. This article will focus on solving problems in probability theory. Previously, we have already analyzed some of the simplest tasks; to solve them, it is enough to know and understand the formula (I advise you to repeat it).
There are some problems that are a little more complicated; to solve them you need to know and understand: the rule of adding probabilities, the rule of multiplying probabilities, the concepts of dependent and independent events, opposite events, compatible and incompatible events. Don't be scared by the definitions, it's simple)).In this article we will consider just such tasks.
A little important and simple theory:
incompatible , if the appearance of one of them excludes the appearance of others. That is, only one specific event or another can happen.
A classic example: when throwing a dice, only a one can come up, or only a two, or only a three, etc. Each of these events is incompatible with the others, and the occurrence of one of them excludes the occurrence of the other (in one trial). It’s the same with a coin—when heads come up, it eliminates the possibility of tails coming up.
This also applies to more complex combinations. For example, two lighting lamps are on. Each of them may or may not burn out over time. There are options:
- The first burns out and the second burns out
- The first burns out and the second does not burn out
- The first one does not burn out and the second one burns out
- The first one does not burn out and the second one burns out.
All these 4 options for events are incompatible - they simply cannot happen together and none of them with any other...
Definition: Events are called joint, if the appearance of one of them does not exclude the appearance of the other.
Example: a queen will be taken from the deck of cards and a spades card will be taken from the deck of cards. Two events are considered. These events are not mutually exclusive - you can draw the queen of spades and thus both events will occur.
About the sum of probabilities
The sum of two events A and B is called the event A+B, which consists in the fact that either event A or event B will occur, or both at the same time.
If there are incompatible events A and B, then the probability of the sum of these events is equal to the sum of the probabilities of the events:
Dice example:
We throw the dice. What is the probability of rolling a number less than four?
Numbers less than four are 1,2,3. We know that the probability of getting a one is 1/6, a two is 1/6, and a three is 1/6. These are incompatible events. We can apply the addition rule. The probability of rolling a number less than four is:
Indeed, if we proceed from the concept of classical probability: then the number of possible outcomes is 6 (the number of all sides of the cube), the number of favorable outcomes is 3 (the appearance of a one, two or three). The desired probability is 3 to 6 or 3/6 = 0.5.
*The probability of the sum of two joint events is equal to the sum of the probabilities of these events without taking into account their joint occurrence: P(A+B)=P(A)+P(B) -P(AB)
About multiplying probabilities
Let two incompatible events A and B occur, their probabilities are respectively equal to P(A) and P(B). The product of two events A and B is an event A B, which consists in the fact that these events will occur together, that is, both event A and event B will occur. The probability of such an event is equal to the product of the probabilities of events A and B.Calculated by the formula:
As you have already noticed, the logical connective “AND” means multiplication.
Example with the same die:We throw the dice twice. What is the probability of rolling two sixes?
The probability of rolling a six the first time is 1/6. The second time is also equal to 1/6. The probability of rolling a six the first time and the second time is equal to the product of the probabilities:
Speaking in simple language: when in one trial some event occurs, AND then another (others) occur, then the probability that they will occur together is equal to the product of the probabilities of these events.
We solved problems with dice, but we used only logical reasoning and did not use the product formula. In the tasks considered below, you cannot do without formulas; or rather, with them it will be easier and faster to get the result.
It is worth mentioning one more nuance. When reasoning in solving problems, the concept of SIMULTANEOUSNESS of events is used. Events occur SIMULTANEOUSLY - this does not mean that they occur in one second (at one point in time). This means that they occur over a certain period of time (within one test).
For example:
Two lamps burn out within a year (it can be said - simultaneously within a year)
Two machines break down within a month (one might say simultaneously within a month)
The dice are rolled three times (points appear at the same time, this means on one trial)
The biathlete fires five shots. Events (shots) occur during one trial.
Events A and B are INDEPENDENT if the probability of either of them does not depend on the occurrence or non-occurrence of the other event.
Let's consider the tasks:
Two factories produce the same glass for car headlights. The first factory produces 35% of these glasses, the second – 65%. The first factory produces 4% of defective glass, and the second – 2%. Find the probability that glass accidentally purchased in a store will be defective.
The first factory produces 0.35 products (glass). The probability of buying defective glass from the first factory is 0.04.
The second factory produces 0.65 glasses. The probability of buying defective glass from the second factory is 0.02.
The probability that the glass was purchased at the first factory and that it turns out to be defective is 0.35∙0.04 = 0.0140.
The probability that the glass was purchased at the second factory and that it turns out to be defective is 0.65∙0.02 = 0.0130.
Buying defective glass in a store implies that it (the defective glass) was purchased EITHER from the first factory OR from the second. These are incompatible events, that is, we add up the resulting probabilities:
0,0140 + 0,0130 = 0,027
Answer: 0.027
If grandmaster A. plays white, then he wins against grandmaster B. with probability 0.62. If A. plays black, then A. wins against B. with probability 0.2. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times.
The possibility of winning the first and second games does not depend on each other. It is said that a grandmaster must win both times, that is, win the first time AND at the same time win the second time. In the case when independent events must occur together, the probabilities of these events are multiplied, that is, the multiplication rule is used.
The probability of the occurrence of these events will be equal to 0.62∙0.2 = 0.124.
Answer: 0.124
In a geometry exam, a student gets one question from the list exam questions. The probability that this is an inscribed circle question is 0.3. The probability that this is a Parallelogram question is 0.25. There are no questions that simultaneously relate to these two topics. Find the probability that a student will get a question on one of these two topics in the exam.
That is, it is necessary to find the probability that the student will get a question EITHER on the topic “Inscribed Circle” OR on the topic “Parallelogram”. In this case, the probabilities are summed up, since these are incompatible events and any of these events can happen: 0.3 + 0.25 = 0.55.
*Incompatible events are events that cannot happen at the same time.
Answer: 0.55
A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.9. Find the probability that the biathlete hits the targets the first four times and misses the last one. Round the result to hundredths.
Since the biathlete hits the target with probability 0.9, he misses with probability 1 – 0.9 = 0.1
*Miss and hit are events that cannot occur simultaneously with one shot; the sum of the probabilities of these events is equal to 1.
We are talking about the occurrence of several (independent) events. If an event occurs and at the same time another (subsequent) event occurs at the same time (test), then the probabilities of these events are multiplied.
The probability of a product of independent events is equal to the product of their probabilities.
Thus, the probability of the event “hit, hit, hit, hit, missed” is 0.9∙0.9∙0.9∙0.9∙0.1 = 0.06561.
Round to the nearest hundredth, we get 0.07
Answer: 0.07
There are two payment machines in the store. Each of them can be faulty with probability 0.07, regardless of the other machine. Find the probability that at least one machine is working.
Let's find the probability that both machines are faulty.
These events are independent, which means the probability will be equal to the product of the probabilities of these events: 0.07∙0.07 = 0.0049.
This means that the probability that both machines or one of them is working will be equal to 1 – 0.0049 = 0.9951.
*Both are operational and one of them is fully operational – meets the “at least one” condition.
One can present the probabilities of all (independent) events to be tested:
1. “faulty-faulty” 0.07∙0.07 = 0.0049
2. “defective-defective” 0.93∙0.07 = 0.0651
3. “defective-defective” 0.07∙0.93 = 0.0651
4. “defective-defective” 0.93∙0.93 = 0.8649
To determine the probability that at least one machine is working, it is necessary to add the probabilities of independent events 2,3 and 4: A reliable event an event that is certain to occur as a result of an experience is called. The event is called impossible, if it never occurs as a result of experience.
For example, if one ball is drawn at random from a box containing only red and green balls, then the appearance of a white one among the drawn balls is an impossible event. The appearance of the red and the appearance of the green balls form a complete group of events.
Definition: The events are called equally possible , unless there is reason to believe that one of them is more likely to appear as a result of experience.
In the example above, the appearance of red and green balls are equally likely events if there are the same number of red and green balls in the box. If there are more red balls in the box than green ones, then the appearance of a green ball is a less probable event than the appearance of a red one.
In we will look at more problems where the sum and product of the probabilities of events are used, don’t miss it!
That's all. I wish you success!
Sincerely, Alexander Krutitskikh.
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- Petrov, why weren’t you at school yesterday?!
“My mother washed my pants yesterday.”
- So what?
- And I walked past the house and saw that yours were hanging. I thought you wouldn't come.
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Basic Concepts
Events are called incompatible if the occurrence of one of them excludes the occurrence of other events in the same trial. Otherwise they are called joint.
A complete group is a set of events, the combination of which is a reliable event.
The only two possible events that form a complete group are called opposite.
Events are called dependent if the probability of the occurrence of one of them depends on the occurrence or non-occurrence of other events.
Events are called independent if the probability of one of them does not depend on the occurrence or non-occurrence of others.
Theorem for adding probabilities of incompatible events
P(A+B)=P(A)+P(B),
where A, B are incompatible events.
Theorem for adding probabilities of joint events
P(A+B)=P(A)+P(B)-P(AB), where A and B are joint events.
Theorem for multiplying the probabilities of independent events
,
where A and B are independent events.
Theorem for multiplying probabilities of dependent events
P(AB)=P(A)P A (B),
where P A (B) is the probability of the occurrence of event B, provided that event A has occurred; A and B are dependent events.
Task 1.
The shooter fires two shots at the target. The probability of hitting each shot is 0.8. Compose a complete group of events and find their probabilities. Solution.
Test - Two shots are fired at a target.
Event A- missed both times.
Event IN- hit once.
Event WITH- hit both times. 
.
Control: P(A) +P(B) +P(C) = 1.
Task 2.
According to meteorologists' forecast, P(rain)=0.4; P(wind)=0.7; R(rain and wind)=0.2. What is the probability that it will rain or wind? Solution. By the theorem of addition of probabilities and due to the compatibility of the proposed events, we have:
P(rain or wind or both)=P(rain) +P(wind) –P(rain and wind)=0.4+0.7-0.2=0.9.
Task 3.
At the departure station there are 8 orders for goods to be sent: five for domestic shipments and three for export. What is the probability that two orders chosen at random will be for domestic consumption? Solution. Event A– the first order taken at random is within the country. Event IN– the second is also intended for domestic consumption. We need to find the probability. Then, by the theorem on the multiplication of probabilities of dependent events, we have 
Task 4.
From a batch of products, the merchandiser randomly selects the highest grade products. The probability that the selected item will be of the highest quality is 0.8; first grade – 0.7; second grade – 0.5. Find the probability that out of three randomly selected products there will be:
a) only two premium grades;
b) everyone is different. Solution. Let the event be a product of the highest quality; event - first-class product; the event is a second-class product.
According to the conditions of the problem; ; Events are independent.
a) Event A– only two top-grade products will look like this then
b) Event IN– all three products are different - let’s put it this way: 
, Then .
Task 5.
The probabilities of hitting the target when firing from three guns are as follows: p1= 0,8; p2=0,7; p3=0.9. Find the probability of at least one hit (event A) with one salvo from all guns. Solution. The probability of each gun hitting the target does not depend on the results of firing from other guns, therefore the events under consideration (hit by the first gun), (hit by the second gun) and (hit by the third gun) are independent in the aggregate.
The probabilities of events opposite to events (i.e., the probability of misses) are respectively equal to: 
Required probability
Task 6.
The printing house has 4 printing machines. For each machine, the probability that it is currently running is 0.9. Find the probability that at least one machine is currently working (event A).
Solution. The events “the machine is working” and “the machine is not working” (at the moment) are opposite, therefore the sum of their probabilities is equal to one:
Hence the probability that the machine is currently not working is equal to
The required probability. Problem 7. In the reading room there are 6 textbooks on probability theory, three of which are bound. The librarian took two textbooks at random. Find the probability that both textbooks will be bound.
Solution. Consider the following events:
A1 - the first bound textbook taken;
A2 is the second bound textbook taken.
An event consisting in the fact that both taken textbooks are bound. Events A1 and A2 are dependent, since the probability of the occurrence of event A2 depends on the occurrence of event A1. To solve this problem, we use the theorem for multiplying the probabilities of dependent events: .
The probability of the occurrence of event A1 p(A1) in accordance with the classical definition of probability:
P(A1)=m/n=3/6=0.5.
The probability of the occurrence of event A2 is determined by the conditional probability of the occurrence of event A2 subject to the occurrence of event A1, i.e. (A2)==0.4.
Then the desired probability of the event occurring:
P(A)=0.5*0.4=0.2.
The need to act on probabilities occurs when the probabilities of some events are known, and it is necessary to calculate the probabilities of other events that are associated with these events.
Addition of probabilities is used when you need to calculate the probability of a combination or logical sum of random events.
Sum of events A And B denote A + B or A ∪ B. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B– an event that occurs if and only if the event occurred during observation A or event B, or simultaneously A And B.
If events A And B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.
Probability addition theorem. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:
For example, while hunting, two shots are fired. Event A– hitting a duck with the first shot, event IN– hit from the second shot, event ( A+ IN) – a hit from the first or second shot or from two shots. So, if two events A And IN– incompatible events, then A+ IN– the occurrence of at least one of these events or two events.
Example 1. There are 30 balls of the same size in a box: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball will be picked up without looking.
Solution. Let us assume that the event A- “the red ball is taken”, and the event IN- “The blue ball was taken.” Then the event is “a colored (not white) ball is taken.” Let's find the probability of the event A:
and events IN:
Events A And IN– mutually incompatible, since if one ball is taken, then the balls cannot be taken different colors. Therefore, we use the addition of probabilities:
The theorem for adding probabilities for several incompatible events. If events constitute a complete set of events, then the sum of their probabilities is equal to 1:
The sum of the probabilities of opposite events is also equal to 1:
Opposite events form a complete set of events, and the probability of a complete set of events is 1.
Probabilities of opposite events are usually indicated in small letters p And q. In particular,
from which the following formulas for the probability of opposite events follow:
Example 2. The target in the shooting range is divided into 3 zones. The probability that a certain shooter will shoot at the target in the first zone is 0.15, in the second zone – 0.23, in the third zone – 0.17. Find the probability that the shooter will hit the target and the probability that the shooter will miss the target.
Solution: Find the probability that the shooter will hit the target:
Let's find the probability that the shooter will miss the target:
More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".
Addition of probabilities of mutually simultaneous events
Two random events are called joint if the occurrence of one event does not exclude the occurrence of a second event in the same observation. For example, when throwing a die the event A The number 4 is considered to be rolled out, and the event IN– rolling an even number. Since the number 4 is 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 of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events has the following form:
Since events A And IN compatible, event A+ IN occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:
Event A will occur if one of two incompatible events occurs: or AB. However, the probability of the occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:
Likewise:
Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:
When using formula (8), it should be taken into account that events A And IN can be:
- mutually independent;
- mutually dependent.
Probability formula for mutually independent events:
Probability formula for mutually dependent events:
If events A And IN are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is:
Example 3. In auto racing, when you drive the first car, you have a better chance of winning, and when you drive the second car. Find:
- the probability that both cars will win;
- the probability that at least one car will win;
1) The probability that the first car will win does not depend on the result of the second car, so the events A(the first car wins) and IN(the second car will win) – independent events. Let's find the probability that both cars win:
2) Find the probability that one of the two cars will win:
More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".
Solve the addition of probabilities problem yourself, and then look at the solution
Example 4. Two coins are tossed. Event A- loss of the coat of arms on the first coin. Event B- loss of the coat of arms on the second coin. Find the probability of an event C = A + B .
Multiplying Probabilities
Probability multiplication is used when the probability of a logical product of events must be calculated.
Wherein random events must be independent. Two events are said to be mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.
Probability multiplication theorem for independent events. Probability of simultaneous occurrence of two independent events A And IN is equal to the product of the probabilities of these events and is calculated by the formula:
Example 5. The coin is tossed three times in a row. Find the probability that the coat of arms will appear all three times.
Solution. The probability that the coat of arms will appear on the first toss of a coin, the second time, and the third time. Let's find the probability that the coat of arms will appear all three times:
Solve probability multiplication problems on your own and then look at the solution
Example 6. There is a box of nine new tennis balls. To play, three balls are taken, and after the game they are put back. When choosing balls, played balls are not distinguished from unplayed balls. What is the probability that after three games there will be no unplayed balls left in the box?
Example 7. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. Find the probability that the letters will form the word "end".
Example 8. From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards will be of different suits.
Example 9. The same task as in example 8, but each card after being removed is returned to the deck.
More complex problems, in which you need to use both addition and multiplication of probabilities, as well as calculate the product of several events, can be found on the page "Various problems involving addition and multiplication of probabilities".
The probability that at least one of the mutually independent events will occur can be calculated by subtracting from 1 the product of the probabilities of opposite events, that is, using the formula:
Example 10. Cargo is delivered by three modes of transport: river, rail and road transport. The probability that the cargo will be delivered by river transport is 0.82, by rail 0.87, by road transport 0.90. Find the probability that the cargo will be delivered by at least one of three types transport.
