The sum of independent events. Theorems of addition and multiplication of probabilities: main tasks

There will also be tasks for independent decision to which you can see the answers.

General statement of the problem: the probabilities of some events are known, but the probabilities of other events that are associated with these events need to be calculated. In these problems, there is a need for such operations on probabilities as addition and multiplication of probabilities.

For example, two shots were fired while hunting. Event A- hitting a duck from the first shot, event B- hit from the second shot. Then the sum of events A and B- hit from the first or second shot or from two shots.

Tasks of a different type. Several events are given, for example, a coin is tossed three times. It is required to find the probability that either all three times the coat of arms will fall out, or that the coat of arms will fall out at least once. This is a multiplication problem.

Addition of probabilities of incompatible events

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

Sum of events A and B designate A + B or AB. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B- an event that occurs if and only if an event occurs during the observation A or event B, or at the same time A and B.

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

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

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

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

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

and events V:

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

The theorem of addition of probabilities for several incompatible events. If the events make up the complete set of events, then the sum of their probabilities is equal to 1:

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

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

The probabilities of opposite events are usually denoted in small letters. p and q. In particular,

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

Example 2 The target in the dash is divided into 3 zones. The probability that a certain shooter will shoot at a target in the first zone is 0.15, in the second zone - 0.23, in the third zone - 0.17. Find the probability that the shooter hits the target and the probability that the shooter misses the target.

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

Find the probability that the shooter misses the target:

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

Addition of probabilities of mutually joint events

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

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

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

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

Similarly:

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

When using formula (8), it should be taken into account that the events A and V may be:

  • mutually independent;
  • mutually dependent.

Probability formula for mutual independent events:

Probability formula for mutually dependent events:

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

Example 3 In auto racing, when driving in the first car, the probability of winning, when driving in the second car. Find:

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

1) The probability that the first car will win does not depend on the result of the second car, so the events A(first car wins) and V(second car wins) - independent events. Find the probability that both cars win:

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

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

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

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

Probability multiplication

Multiplication of probabilities is used when the probability of a logical product of events is to be calculated.

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

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

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

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

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

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

Example 7 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability that the letters will form the word "end".

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

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

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

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

Definition. Product or intersection events A and B call an event consisting in the simultaneous occurrence of events both A and B. The designation of the work: AB or A B.

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 A and B are called incompatible if their product is an impossible event, i.e. AB = V.

Events A - the loss of the coat of arms and B - the loss of a number during a single toss of a coin cannot occur simultaneously, their product is an impossible event, events A and B are incompatible.

The concepts of sum and product of events have a clear geometric interpretation.

Rice. 6.4. Geometric interpretation of the product (a) and sum (b) of two joint events

Let event A be a set of points in area A; event B is a set of points in the area B. The shaded area corresponds to the event AB in Fig. 6.4, a; event in Fig. 6.4, b.

For incompatible events A and B we have: AB = V (Fig. 6.5, a). Event A + B corresponds to the shaded area in Fig. 6.5, b.

Rice. 6.5. Geometric interpretation of the product (a) and sum (b) of two incompatible events

Events and call opposite, if they are incompatible and in total constitute a reliable event, i.e.

For example, let's make one shot at the target: event - the shooter hit the target, did not hit; a coin is tossed: the event is heads, − numbers are rolled; schoolchildren write a test: an event - not a single mistake in control work, - there are errors in the control work; the student came to take the test: event A - passed the test, - did not pass the test.

There are boys and girls in the class, excellent students, good students and three students who study English and German. Let the event M be a boy, O be an excellent student, A be a student English language. 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. A dice is thrown - 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 event A be the appearance of an odd number, and event B be the appearance of a number that is a multiple of three. Find the outcomes that make up each of the events: U, A, A + B, 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 makes up the space of elementary events. It is clear that the event , event

An event is the occurrence of either an odd number or a multiple of three. When listing the outcomes, it is taken into account that each outcome in the set can be contained only once.



An event is the occurrence of both an odd number and a multiple of three.

Example. Checked homework three students. Let the event be the completion of the task by the th student, What is the meaning of the events: and ?

Solution. An event is the completion of a 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 - the task was not completed by any student: neither the first, nor the second, nor the third. An event is the completion of a task by three students: the first, the second, and the third.

When considering the joint occurrence of several events, there are 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.

Definition. Event A is called independent from event B, if the probability of event A does not change depending on whether event B occurred or not. Otherwise, event A is called dependent on event B. Two events A and B 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.

Theorem. (Probability multiplications) The probability of the product of two independent events is equal to the product of the probabilities of these events:

P(A B)=P(A) P(B)

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. The student takes three exams. Probability successful delivery 0.9 for the first exam, 0.65 for the second, and 0.35 for the third. Find the probability that he fails at least one exam.

Solution: Denote A - the event student did not pass at least one exam. Then P(A) = 1- P(ùA), where ùA is the opposite event the student has passed all exams. Since the passing of each exam does not depend on other exams, then Р(А)=1-Р(ùА)= 1-0.9*0.65*0.35=0.7953.

Definition. The probability of event A, calculated assuming that event B occurs, is called conditional probability event A, subject to the occurrence of B and is denoted by P 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 happened:

P (A B) \u003d P (A) P A (B) \u003d P (B) P B (A). (*)

Example. 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 A is that the first time you got a bad ticket, event B - the second time you get a good ticket. Then A·B - the student will pass the exam (under the specified circumstances). Events A and B are dependent, because the probability of choosing a successful ticket on the second attempt depends on the outcome of the first choice. Therefore, we use formula (6):

P (A B) \u003d P (A) RA (B) \u003d (4/34) * (30/33) \u003d 20/187

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?!

Theorem. (Extended addition theorem) 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 (product):

P(A+B)=P(A)+P(B)-P(A B).

Example. Two students solve a problem. The probability that the first student will solve the problem (event A) is 0.9; the probability that the second student will solve the problem (event B) is 0.8. What is the probability that the problem will be solved?

When searching for the probabilities of events, we used classic definition probabilities.

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

distance learning

Gorki, 2012

Addition and multiplication of probabilities. Repeated

independent tests

  1. Addition of probabilities

The sum of two joint events A and V called an event WITH, consisting in the occurrence of at least one of the events A or V. Similarly, the sum of several joint events is an event consisting in the occurrence of at least one of these events.

The sum of two disjoint events A and V called an event WITH, consisting in the occurrence or event A, or events V. Similarly, the sum of several incompatible events is an event consisting in the occurrence of any one of these events.

The theorem of addition of 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 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 . A box contains 2 white, 3 red and 5 blue balls. The balls are shuffled and one is drawn at random. What is the probability that the ball is colored?

Solution . Let's denote the events:

A=(color ball removed);

B=(white ball drawn);

C=(red ball drawn);

D=(blue ball removed).

Then A= C+ D. Since the events C, D are incompatible, then we use the theorem of addition of probabilities of incompatible events: .

Example 2 . An urn contains 4 white balls and 6 black balls. 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\u003d (balls of the same color are taken out);

B\u003d (white balls are taken out);

C= (black balls are taken out).

Because A= B+ C and events V and WITH are incompatible, then by the theorem of addition of probabilities of incompatible events
. Event Probability V is equal to
, where
4,

. Substitute k and n into the formula and get
Similarly, we find the probability of an 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\u003d (among the drawn cards there are at least three aces);

B\u003d (among the drawn cards there are three aces);

C= (among the drawn cards there are four aces).

Because A= B+ C, and the events V and WITH inconsistent, then
. Let's find the probabilities of events V and WITH:


,
. Therefore, the probability that among the drawn cards there are at least three aces is equal to

0.0022.

  1. Probability multiplication

work two events A and V called an event WITH, consisting in the joint occurrence of these events:
. This definition extends to any finite number of events.

The two events are called independent if the probability of occurrence of one of them does not depend on whether the other event occurred or not. Events , , … , called collectively independent , if the probability of occurrence of each of them does not depend on whether other events occurred or did not occur.

Example 4 . Two arrows shoot at a target. Let's denote the events:

A=(first shooter hit the target);

B= (the second shooter hit the target).

Obviously, the probability of hitting the target by the first shooter does not depend on whether the second shooter hit or missed, and vice versa. Therefore, the events A and V independent.

The theorem of multiplication of 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 events that are independent in the aggregate: .

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 the same time. Determine the probability that there will be two hits on the target.

Solution . Let's denote the events:

A

B

C=(both arrows will hit the target).

Because
, and the events A and V independent, then
, i.e. .

Events A and V called dependent if the probability of occurrence of one of them depends on whether the other event occurred or not. Probability of an event A provided that the event V it's already here, it's called conditional probability and denoted
or
.

Example 6 . An urn contains 4 white and 7 black balls. Balls are drawn from the urn. Let's denote the events:

A=(white ball removed) ;

B=(black ball removed).

Before you start drawing balls from the urn
. One ball is drawn from the urn and it turns out to be black. Then the probability of the event A after the event V will be different, equal . This means that the probability of an event A event dependent V, i.e. these events will be dependent.

The theorem of multiplication of probabilities of dependent events is valid: the probability of the product of two dependent events is equal to the product of the probability of one of them by the conditional probability of the other, calculated on the assumption that the first event has already occurred, i.e. or .

Example 7 . An urn contains 4 white balls and 8 red balls. Two balls are drawn at random from it. Find the probability that both balls are black.

Solution . Let's denote the events:

A=(black ball drawn first);

B=(a black ball is drawn second).

Events A and V dependent because
, a
. Then
.

Example 8 . Three arrows 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 two hits will occur if each shooter fires one shot.

Solution . Let's denote the events:

A=(there will be two hits on the target);

B=(first shooter hits 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 , then using the addition theorem for the probabilities of incompatible events and the theorem for multiplying the probabilities of independent events, we get:

Let the events
form a complete group of events of some trial, and the events A can only occur with one of these events. If the probabilities and conditional probabilities of an event are known A, then the probability of event A is calculated by the formula:

Or
. This formula is called total probability formula , and the events
hypotheses .

Example 9 . The assembly line receives 700 parts from the first machine and 300 parts from the second. The first machine gives 0.5% rejects, and the second - 0.7%. Find the probability that the item taken is defective.

Solution . Let's denote the events:

A=(the item taken will be defective);

= (the part is made on the first machine);

= (the part is made on the second machine).

The probability that the part was made on the first machine is
. For the second machine
. By the condition, the probability of obtaining 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 part taken will be defective is calculated by the total probability formula

If an event is known to have occurred as a result of a test A, then the probability that this event occurred with the hypothesis
, is equal to
, where
- total probability of the 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 . Parts of the same type for cars are produced at two factories and go to the store. The first plant produces 80% of the total number of parts, and the second - 20%. The production of the first plant contains 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 is made in the second factory.

Solution . Let's denote the events:

A=(purchased a standard part);

= (the part is made at the first factory);

= (the part is made at the second factory).

According to the example
,
,
and
. Calculate the total probability of an event A: 0.91. The probability that the part is manufactured at the second plant is calculated 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. 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 - to replace individual components. Three tractors are randomly selected. Find the probability that no more than two selected tractors need an engine replacement.

    The 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 assemble bearings. The probability that the bearing assembled by the first worker is of the highest quality is 0.7, the second - 0.8, and 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 are of the highest quality.

    The probability of winning on a lottery ticket of the first issue is 0.2, the second - 0.3 and the third - 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 of interest to him 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 of interest to the accountant is contained in no more than two directories.

    Three machines make parts. The first automaton produces a part of the highest quality with a probability of 0.9, the second with a probability of 0.7, and the third with a probability of 0.6. One item 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 manufacturing a non-standard part for the first machine is 0.03, for the second - 0.02. The processed parts are stacked in one place. Among them, 67% are from the first machine, and the rest from the second. A randomly taken part 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 defective. In the second box there are 10 capacitors, of which 3 are faulty. Capacitors were transferred to one box. Find the probability that a capacitor taken at random from the box is good.

    On three machines, the same type of parts are made, which are fed to a common conveyor. Among all the details, 20% from the first machine, 30% from the second and 505 from the third. The probability of manufacturing a standard part on the first machine is 0.8, on the second - 0.6 and on the third - 0.7. The part taken was standard. Find the probability that this part is made on the third machine.

    The picker receives 40% of the parts from the factory for assembly A, and the rest - from the factory V. The probability that the part from the factory A- the highest quality, equal to 0.8, and from the factory V– 0.9. The picker randomly took one part and it was not of the highest quality. Find the probability that this part is from the factory V.

    10 students from the first group and 8 students from the second were selected to participate in student sports competitions. The probability that a student from the first group will get into the national team of the academy is 0.8, and from the second - 0.7. A randomly selected student was selected for the national team. Find the probability that he is from the first group.

Addition theorem

Consider incompatible random events.

It is known that incompatible random events $A$ and $B$ in the same trial have probabilities $P\left(A\right)$ and $P\left(B\right)$ respectively. Let us find the probability of the sum $A+B$ of these events, i.e. the probability of occurrence of at least one of them.

Suppose that in this test the number of all equally possible elementary events is $n$. Of these, events $A$ and $B$ are favored by $m_(A)$ and $m_(B)$ elementary events, respectively. Since the events $A$ and $B$ are incompatible, the event $A+B$ is favored by $m_(A) +m_(B)$ elementary events. We have $P\left(A+B\right)=\frac(m_(A) +m_(B) )(n) =\frac(m_(A) )(n) +\frac(m_(B) ) (n) =P\left(A\right)+P\left(B\right)$.

Theorem 1

The probability of the sum of two incompatible events is equal to the sum of their probabilities.

Note 1

Consequence 1. The probability of the sum of any number of incompatible events is equal to the sum of the probabilities of these events.

Consequence 2. The sum of the probabilities of a complete group of incompatible events (the sum of the probabilities of all elementary events) is equal to one.

Consequence 3. The sum of the probabilities of opposite events is equal to one, since they form a complete group of incompatible events.

Example 1

The probability that it will never rain in the city for some time is $p=0.7$. Find the probability $q$ that during the same time it will rain in the city at least once.

The events "for some time it never rained in the city" and "for some time it rained in the city at least once" are opposite. Therefore $p+q=1$, whence $q=1-p=1-0.7=0.3$.

Consider joint random events.

It is known that joint random events $A$ and $B$ in the same trial have probabilities of occurrence $P\left(A\right)$ and $P\left(B\right)$ respectively. Let us find the probability of the sum $A+B$ of these events, i.e. the probability of occurrence of at least one of them.

Suppose that in this test the number of all equally possible elementary events is $n$. Of these, events $A$ and $B$ are favored by $m_(A)$ and $m_(B)$ elementary events, respectively. Since the events $A$ and $B$ are joint, then out of the total number of $m_(A) +m_(B) $ elementary events, a certain number $m_(AB) $ favors both the event $A$ and the event $B$, that is, their joint occurrence (the product of events $A\cdot B$). This quantity $m_(AB)$ entered both $m_(A)$ and $m_(B)$. So event $A+B$ is favored by $m_(A) +m_(B) -m_(AB) $ elementary events. We have: $P\left(A+B\right)=\frac(m_(A) +m_(B) -m_(AB) )(n) =\frac(m_(A) )(n) +\frac (m_(B) )(n) -\frac(m_(AB) )(n) =P\left(A\right)+P\left(B\right)-P\left(A\cdot B\right )$.

Theorem 2

The probability of the sum of two joint events is equal to the sum of the probabilities of these events minus the probability of their product.

Comment. If the events $A$ and $B$ are incompatible, then their product $A\cdot B$ is an impossible event whose probability is $P\left(A\cdot B\right)=0$. Therefore, the formula for adding the probabilities of incompatible events is a special case of the formula for adding the probabilities of joint events.

Example 2

Find the probability that when two dice are thrown at the same time, the number 5 will come up at least once.

When throwing two dice at the same time, the number of all equally possible elementary events is equal to $n=36$, since six digits of the second die can fall on each digit of the first dice. Of these, the event $A$ - the number 5 rolled on the first die - occurs 6 times, the event $B$ - the number 5 rolled on the second die - also occurs 6 times. Of all twelve times, the number 5 appears once on both dice. So $P\left(A+B\right)=\frac(6)(36) +\frac(6)(36) -\frac(1)(36) =\frac(11)(36) $.

Probability multiplication theorem

Consider independent events.

Events $A$ and $B$ that occur in two successive trials are called independent if the probability of occurrence of event $B$ does not depend on whether event $A$ took place or did not take place.

For example, suppose there are 2 white and 2 black balls in an urn. The test is to extract the ball. The event $A$ is "a white ball is drawn in the first trial". Probability $P\left(A\right)=\frac(1)(2) $. After the first test, the ball was put back and a second test was carried out. Event $B$ -- ``white ball drawn in second trial''. Probability $P\left(B\right)=\frac(1)(2) $. The probability $P\left(B\right)$ does not depend on whether the event $A$ took place or not, hence the events $A$ and $B$ are independent.

It is known that independent random events $A$ and $B$ of two consecutive trials have probabilities $P\left(A\right)$ and $P\left(B\right)$ respectively. Let us find the probability of the product $A\cdot B$ of these events, that is, the probability of their joint occurrence.

Suppose that in the first trial the number of all equally possible elementary events is $n_(1) $. Of these, $A$ is favored by $m_(1)$ elementary events. Let us also assume that in the second test the number of all equally possible elementary events is $n_(2) $. Of these, event $B$ is favored by $m_(2)$ elementary events. Now consider a new elementary event, which consists in the successive occurrence of events from the first and second trials. The total number of such equally probable elementary events is equal to $n_(1) \cdot n_(2) $. Since the events $A$ and $B$ are independent, from this number the joint occurrence of the event $A$ and the event $B$ (the products of the events $A\cdot B$) is favored by $m_(1) \cdot m_(2) $ events . We have: $P\left(A\cdot B\right)=\frac(m_(1) \cdot m_(2) )(n_(1) \cdot n_(2) ) =\frac(m_(1) ) (n_(1) ) \cdot \frac(m_(2) )(n_(2) ) =P\left(A\right)\cdot P\left(B\right)$.

Theorem 3

The probability of the product of two independent events is equal to the product of the probabilities of these events.

Consider dependent events.

In two consecutive trials, events $A$ and $B$ occur. An event $B$ is said to be dependent on the event $A$ if the probability of occurrence of the event $B$ depends on whether the event $A$ took place or not. Then the probability of the event $B$, which was calculated under the condition that the event $A$ took place, is called the conditional probability of the event $B$ under the condition $A$ and is denoted by $P\left(B/A\right)$.

For example, suppose there are 2 white and 2 black balls in an urn. The test is the extraction of the ball. The event $A$ is "a white ball is drawn in the first trial". Probability $P\left(A\right)=\frac(1)(2) $. After the first test, the ball is not put back and the second test is performed. Event $B$ -- ``white ball drawn in second trial''. If a white ball was drawn in the first trial, then the probability is $P\left(B/A\right)=\frac(1)(3) $. If a black ball was drawn in the first trial, then the probability is $P\left(B/\overline(A)\right)=\frac(2)(3) $. Thus, the probability of the event $B$ depends on whether the event $A$ took place or not, therefore, the event $B$ depends on the event $A$.

Assume that events $A$ and $B$ occur in two consecutive trials. It is known that the event $A$ has the probability of occurrence $P\left(A\right)$. It is also known that the event $B$ is dependent on the event $A$ and its conditional probability under condition $A$ is equal to $P\left(B/A\right)$.

Theorem 4

The probability of the product of the event $A$ and the event $B$ dependent on it, that is, the probability of their joint occurrence, can be found by the formula $P\left(A\cdot B\right)=P\left(A\right)\cdot P\left(B/A\right)$.

The symmetric formula $P\left(A\cdot B\right)=P\left(B\right)\cdot P\left(A/B\right)$ is also valid, where the event $A$ is assumed to be dependent on the event $ B$.

For the conditions of the last example, we find the probability that the white ball will be drawn in both trials. Such an event is a product of the events $A$ and $B$. Its probability is $P\left(A\cdot B\right)=P\left(A\right)\cdot P\left(B/A\right)=\frac(1)(2) \cdot \frac( 1)(3) =\frac(1)(6) $.

  • Theorem. The probability of the sum of incompatible events and is equal to the sum of the probabilities of these events:

  • Consequence 1. Using the method of mathematical induction, formula (3.10) can be generalized to any number of pairwise incompatible events:

  • Consequence 2. Since the opposite events are incompatible, and their sum is a reliable event, then, using (3.10), we have:

  • Often, when solving problems, formula (3.12) is used in the form:

    (3.13)

    Example 3.29. In the experiment with throwing a dice, find the probabilities of getting more than 3 and less than 6 on the upper bound of the number of points.

    Let us denote the events associated with the loss of one point on the upper face of the dice through U 1 , two points across U 2 ,…, six points through U 6 .

    Let the event U- loss on the upper face of the die number of points more than 3 and less than 6. This event will occur if at least one of the events occurs U 4 or U 5 , therefore, it can be represented as the sum of these events: . Because events U 4 and U 5 are inconsistent, then to find the probability of their sum we use formula (3.11). Considering that the probabilities of events U 1 , U 2 ,…,U 6 are equal, we get:

  • Comment. Previously, problems of this type were solved by counting the number of favorable outcomes. Indeed, the event U is favored by two outcomes, and only six elementary outcomes, therefore, using the classical approach to the concept of probability, we obtain:

    However, the classical approach to the concept of probability, in contrast to the theorem on the probability of the sum of incompatible events, is applicable only for equally possible outcomes.

    Example 3.30. The shooter's chance of hitting the target is 0.7. What is the probability that the shooter misses the target?

    Let the event be the shooter hitting the target, then the event that the shooter does not hit the target is the opposite event to the event, because as a result of each test, one and only one of these events always occurs. Using formula (3.13), we obtain:

  • 3.2.10. Probability of producing events

  • Definition. The event is called dependent from the event if the probability of an event depends on whether the event occurred or not.

    Definition. The probability of an event, given that the event has occurred, is called conditional probability events and is marked

    Theorem. The probability of a product of events is equal to the product of the probability of one of them by the conditional probability of the other, calculated under the condition that the first took place:

  • Condition of independence of an event from an event can be written in the form It follows from this statement that for independent events the following relation holds:

  • i.e., the probability of the product of independent events and is equal to the product of their probabilities.

    Comment. The probability of the product of several events is equal to the product of the probabilities of these events, and the probability of each next event in order is calculated under the condition that all previous ones took place:

  • If the events are independent, then we have:

  • Example 3.31. There are 5 white and 3 black balls in a box. Two balls are drawn at random from it without replacement. Find the probability that both balls are white.

    Let the event be the appearance of a white ball on the first draw, the appearance of a white ball on the second draw. Given that, (the probability of a second white ball appearing, provided that the first ball drawn was white and was not returned to the box). Since the events and are dependent, we can find the probability of their product using the formula (3.15):

  • Example 3.32. The probability of hitting the target by the first shooter is 0.8; the second - 0.7. Each shooter fired at a target. What is the probability that at least one shooter hits the target? What is the probability that one shooter will hit the target?

    Let the event be a hit on the target by the first shooter, - by the second. All possible options can be represented as tables 3.5, where "+" means that the event happened, and "-" - did not happen.

    Table 3.5

  • Let an event be a hit by at least one shooter on the target. Then the event is the sum of independent events and, therefore, it is impossible to apply the theorem on the probability of the sum of incompatible events in this situation.

    Consider an event opposite to the event that will occur when no shooter hits the target, i.e., is the product of independent events. Using formulas (3.13) and (3.15), we obtain:

  • Let the event be a hit by one shooter on the target. This event can be represented as follows:

    The events and are independent, the events and are also independent. Events that are products of events are incompatible. Using formulas (3.10) and (3.15) we obtain:

  • Properties of operations of addition and multiplication of events:

  • 3.2.11. Total Probability Formula. Bayes formula

  • Let an event occur only together with one of pairwise incompatible events (hypotheses),,...,, forming a complete group, i.e.

    The probability of an event is found by the formula full probability:

  • If the event has already happened, then the probabilities of the hypotheses can be overestimated by the formula Bayes:

    (3.17)

    Example 3.33. There are two identical urns with balls. The first urn contains 5 white and 10 black balls, and the second urn contains 3 white and 7 black balls. One urn is chosen at random and one ball is drawn from it.

      Find the probability that this ball is white.

      A white ball is drawn from an urn at random. Find the probability that the ball was drawn from the first urn.