The dispersed random variable x is specified by the distribution law. Theoretical material for modules "probability theory and mathematical statistics"

Educational institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

Guidelines

to study the topic “Random Variables” by students of the Faculty of Accounting for Correspondence Education (NISPO)

Gorki, 2013

Random variables

Discrete and continuous random variables

One of the main concepts in probability theory is the concept random variable . Random variable is a quantity that, as a result of testing, takes only one of its many possible values, and it is not known in advance which one.

There are random variables discrete and continuous . Discrete random variable (DRV) is a random variable that can take on a finite number of values isolated from each other, i.e. if the possible values of this quantity can be recalculated. Continuous random variable (CNV) is a random variable, all possible values of which completely fill a certain interval of the number line.

Random variables are denoted by capital letters of the Latin alphabet X, Y, Z, etc. Possible values of random variables are indicated by the corresponding small letters.

Record
means "the probability that a random variable X will take a value of 5, equal to 0.28.”

Example 1 . The dice are thrown once. In this case, numbers from 1 to 6 may appear, indicating the number of points. Let us denote the random variable X=(number of points rolled). This random variable as a result of the test can take only one of six values: 1, 2, 3, 4, 5 or 6. Therefore, the random variable X there is DSV.

Example 2 . When a stone is thrown, it travels a certain distance. Let us denote the random variable X=(stone flight distance). This random variable can take any, but only one, value from a certain interval. Therefore, the random variable X there is NSV.

Distribution law of a discrete random variable

A discrete random variable is characterized by the values it can take and the probabilities with which these values are taken. The correspondence between possible values of a discrete random variable and their corresponding probabilities is called law of distribution of a discrete random variable .

If all possible values are known
random variable X and probabilities
appearance of these values, then it is believed that the law of distribution of DSV X is known and can be written in table form:

The DSV distribution law can be depicted graphically if points are depicted in a rectangular coordinate system
,
, …,
and connect them with straight line segments. The resulting figure is called a distribution polygon.

Example 3 . Grain intended for cleaning contains 10% weeds. 4 grains were selected at random. Let us denote the random variable X=(number of weeds among the four selected). Construct the DSV distribution law X and distribution polygon.

Solution . According to the example conditions. Then:

Let's write down the distribution law of DSV X in the form of a table and construct a distribution polygon:

Expectation of a discrete random variable

The most important properties of a discrete random variable are described by its characteristics. One of these characteristics is expected value random variable.

Let the DSV distribution law be known X:

Mathematical expectation DSV X is the sum of the products of each value of this quantity and the corresponding probability:
.

The mathematical expectation of a random variable is approximately equal to the arithmetic mean of all its values. Therefore, in practical problems, the average value of this random variable is often taken as the mathematical expectation.

Example 8 . The shooter scores 4, 8, 9 and 10 points with probabilities of 0.1, 0.45, 0.3 and 0.15. Find the mathematical expectation of the number of points with one shot.

Solution . Let us denote the random variable X=(number of points scored). Then . Thus, the expected average number of points scored with one shot is 8.2, and with 10 shots - 82.

Main properties mathematical expectation are:

, Where
,
.

, Where X And Y are independent random variables.

Difference
called deviation random variable X from its mathematical expectation. This difference is a random variable and its mathematical expectation is zero, i.e.
.

Variance of a discrete random variable

To characterize a random variable, in addition to the mathematical expectation, we also use dispersion , which makes it possible to estimate the dispersion (spread) of the values of a random variable around its mathematical expectation. When comparing two homogeneous random variables with equal mathematical expectations, the “best” value is considered to be the one that has less spread, i.e. less dispersion.

Variance random variable X is called the mathematical expectation of the squared deviation of a random variable from its mathematical expectation: .

In practical problems, an equivalent formula is used to calculate the variance.

The main properties of the dispersion are:

X is given by the law of probability distribution: Then its standard deviation is equal to ... 0.80

Solution:
The standard deviation of the random variable X is defined as , where the variance of a discrete random variable can be calculated using the formula . Then , and

Solution:
A(a ball drawn at random is black) we apply the total probability formula: Here is the probability that a white ball was transferred from the first urn to the second urn; – the probability that a black ball was transferred from the first urn to the second urn; – the conditional probability that the drawn ball is black if a white ball was moved from the first urn to the second; – the conditional probability that the drawn ball is black if a black ball was moved from the first urn to the second.

The discrete random variable X is given by the law of probability distribution: Then the probability equal...

Solution:
The variance of a discrete random variable can be calculated using the formula. Then

Or . Solving the last equation, we get two roots and

Topic: Determination of probability
In a batch of 12 parts, there are 5 defective parts. Three parts were selected at random. Then the probability that there are no suitable parts among the selected parts is equal to...

Solution:
To calculate event A (there are no suitable parts among the selected parts), we use the formula where n m– the number of elementary outcomes favorable to the occurrence of event A. In our case, the total number of possible elementary outcomes is equal to the number of ways in which three details can be extracted from the 12 available, that is.

And the total number of favorable outcomes is equal to the number of ways in which three defective parts can be extracted from five, that is.

The bank issues 44% of all loans to legal entities, and 56% to individuals. The probability that a legal entity will not repay the loan on time is 0.2; and for an individual this probability is 0.1. Then the probability that the next loan will be repaid on time is...

0,856

Solution:
To calculate the probability of an event A(the issued loan will be repaid on time) apply the total probability formula: . Here is the probability that the loan was issued to a legal entity; – the probability that the loan was issued to an individual; – the conditional probability that the loan will be repaid on time if it was issued to a legal entity; – the conditional probability that the loan will be repaid on time if it was issued to an individual. Then

Topic: Laws of probability distribution of discrete random variables
For a discrete random variable X

0,655

Topic: Determination of probability
The die is rolled twice. Then the probability that the sum of the rolled points is not less than nine is...

Solution:
To calculate the event (the sum of the points rolled will be at least nine), we use the formula , where is the total number of possible elementary outcomes of the test, and m– the number of elementary outcomes favorable to the occurrence of the event A. In our case it is possible elementary test outcomes, of which favorable are outcomes of the form , , , , , , , and , that is. Hence,

Topic: Laws of probability distribution of discrete random variables

the probability distribution function has the form:

Then the value of the parameter can be equal to...

			0,7

			0,85
			0,6

Solution:
A-priory . Therefore, and . These conditions are satisfied, for example, by the value

Topic: Numerical characteristics of random variables
A continuous random variable is specified by a probability distribution function:

Then its variance is...

Solution:
This random variable is distributed uniformly in the interval. Then its variance can be calculated using the formula . That is

Topic: Total probability. Bayes formulas
The first urn contains 6 black balls and 4 white balls. The second urn contains 2 white and 8 black balls. One ball was taken from a random urn, which turned out to be white. Then the probability that this ball was drawn from the first urn is...

Solution:
A(a ball drawn at random is white) according to the total probability formula: . Here is the probability that the ball is drawn from the first urn; is the probability that the ball is drawn from the second urn; – the conditional probability that the drawn ball is white if it is drawn from the first urn; is the conditional probability that the drawn ball is white if it is drawn from the second urn.
Then .
Now let's calculate the conditional probability that this ball was drawn from the first urn using Bayes' formula:

Topic: Numerical characteristics of random variables
Discrete random variable X is given by the law of probability distribution:

Then its variance is...

			7,56
			3,2
			3,36
			6,0

Solution:
The variance of a discrete random variable can be calculated using the formula

Topic: Laws of probability distribution of discrete random variables

Solution:
A-priory . Then
a) at , ,
b) at , ,
c) at , ,
d) at , ,
d) at , .
Hence,

Topic: Determination of probability
A point is thrown at random inside a circle of radius 4. Then the probability that the point will be outside the square inscribed in the circle is...

Topic: Determination of probability
In a batch of 12 parts, there are 5 defective parts. Three parts were selected at random. Then the probability that there are no defective parts among the selected parts is equal to...

Solution:
To calculate the event (there are no defective parts among the selected parts), we use the formula, where n is the total number of possible elementary test outcomes, and m– the number of elementary outcomes favorable to the occurrence of the event. In our case, the total number of possible elementary outcomes is equal to the number of ways in which three details can be extracted from the 12 available, that is. And the total number of favorable outcomes is equal to the number of ways in which three non-defective parts can be extracted from seven, that is. Hence,

Topic: Total probability. Bayes formulas

			0,57
			0,43
			0,55
			0,53

Solution:
To calculate the probability of an event A
Then

Topic: Laws of probability distribution of discrete random variables
A discrete random variable is specified by the probability distribution law:

Then the probability equal...

Solution:
Let's use the formula . Then

Topic: Total probability. Bayes formulas

			0,875
			0,125
			0,105
			0,375

Solution:
Let's first calculate the probability of the event A
.
.

Topic: Numerical characteristics of random variables

Then its mathematical expectation is...

Solution:
Let's use the formula . Then .

Topic: Determination of probability

Solution:

Topic: Numerical characteristics of random variables
A continuous random variable is specified by the probability distribution density . Then the mathematical expectation a and the standard deviation of this random variable are equal to ...

Solution:
The probability distribution density of a normally distributed random variable has the form , Where , . That's why .

Topic: Laws of probability distribution of discrete random variables
A discrete random variable is specified by the probability distribution law:

Then the values a And b may be equal...

Solution:
Since the sum of the probabilities of possible values is equal to 1, then . The answer satisfies this condition: .

Topic: Determination of probability
A smaller circle of radius 5 is placed in a circle of radius 8. Then the probability that a point thrown at random into the larger circle will also fall into the smaller circle is...

Solution:
To calculate the probability of the desired event, we use the formula , where is the area of the smaller circle, and is the area of the larger circle. Hence, .

Topic: Total probability. Bayes formulas
The first urn contains 3 black balls and 7 white balls. The second urn contains 4 white balls and 5 black balls. One ball was transferred from the first urn to the second urn. Then the probability that a ball drawn at random from the second urn will be white is...

			0,47
			0,55
			0,35
			0,50

Solution:
To calculate the probability of an event A(a ball drawn at random is white) apply the total probability formula: . Here is the probability that a white ball was transferred from the first urn to the second urn; – the probability that a black ball was transferred from the first urn to the second urn; – the conditional probability that the drawn ball is white if a white ball was moved from the first urn to the second; – the conditional probability that the drawn ball is white if a black ball is moved from the first urn to the second.
Then

Topic: Laws of probability distribution of discrete random variables
For a discrete random variable:

the probability distribution function has the form:

Then the value of the parameter can be equal to...

			0,7

			0,85
			0,6

TASK N 10 report an error
Topic: Total probability. Bayes formulas
The bank issues 70% of all loans to legal entities, and 30% to individuals. The probability that a legal entity will not repay the loan on time is 0.15; and for an individual this probability is 0.05. A message was received indicating that the loan was not repaid. Then the probability that the legal entity did not repay this loan is...

			0,875
			0,125
			0,105
			0,375

Solution:
Let's first calculate the probability of the event A(the issued loan will not be repaid on time) according to the total probability formula: . Here is the probability that the loan was issued to a legal entity; – the probability that the loan was issued to an individual; – the conditional probability that the loan will not be repaid on time if it was issued to a legal entity; – the conditional probability that the loan will not be repaid on time if it was issued to an individual. Then
.
Now let’s calculate the conditional probability that this loan was not repaid by a legal entity, using the Bayes formula:
.

TASK N 11 report an error
Topic: Determination of probability
In a batch of 12 parts, there are 5 defective parts. Three parts were selected at random. Then the probability that there are no suitable parts among the selected parts is equal to...

Solution:
To calculate the event (there are no suitable parts among the selected parts), we use the formula, where n is the total number of possible elementary test outcomes, and m– the number of elementary outcomes favorable to the occurrence of the event. In our case, the total number of possible elementary outcomes is equal to the number of ways in which three details can be extracted from the 12 available, that is. And the total number of favorable outcomes is equal to the number of ways in which three defective parts can be extracted from five, that is. Hence,

TASK N 12 report an error
Topic: Numerical characteristics of random variables
A continuous random variable is specified by the probability distribution density:

Then its variance is...

Solution:
The variance of a continuous random variable can be calculated using the formula

Then

Topic: Laws of probability distribution of discrete random variables
A discrete random variable is specified by the probability distribution law:

Then its probability distribution function has the form...

Solution:
A-priory . Then
a) at , ,
b) at , ,
c) at , ,
d) at , ,
d) at , .
Hence,

Topic: Total probability. Bayes formulas
There are three urns containing 5 white and 5 black balls, and seven urns containing 6 white and 4 black balls. One ball is drawn from a random urn. Then the probability that this ball is white is...

			0,57
			0,43
			0,55
			0,53

Solution:
To calculate the probability of an event A(a ball drawn at random is white) apply the total probability formula: . Here is the probability that a ball is drawn from the first series of urns; – the probability that the ball is drawn from the second series of urns; – the conditional probability that the drawn ball is white if it is drawn from the first series of urns; – the conditional probability that the drawn ball is white if it is drawn from the second series of urns.
Then .

Topic: Laws of probability distribution of discrete random variables
A discrete random variable is specified by the probability distribution law:

Then the probability equal...

Topic: Determination of probability
The die is rolled twice. Then the probability that the sum of the points drawn is ten is...

Definition 2.3. A random variable, denoted by X, is called discrete if it takes on a finite or countable set of values, i.e. set – a finite or countable set.

Let's consider examples of discrete random variables.

1. Two coins are tossed once. The number of emblems in this experiment is a random variable X. Its possible values are 0,1,2, i.e. – a finite set.

2. The number of ambulance calls within a given period of time is recorded. Random value X– number of calls. Its possible values are 0, 1, 2, 3, ..., i.e. =(0,1,2,3,...) is a countable set.

3. There are 25 students in the group. On a certain day, the number of students who came to class is recorded - a random variable X. Its possible values: 0, 1, 2, 3, ...,25 i.e. =(0, 1, 2, 3, ..., 25).

Although all 25 people in example 3 cannot miss classes, the random variable X can take this value. This means that the values of a random variable have different probabilities.

Let's consider a mathematical model of a discrete random variable.

Let a random experiment be carried out, which corresponds to a finite or countable space of elementary events. Let us consider the mapping of this space onto the set of real numbers, i.e., let us assign to each elementary event a certain real number , . The set of numbers can be finite or countable, i.e. or

A system of subsets, which includes any subset, including a one-point one, forms an -algebra of a numerical set ( – finite or countable).

Since any elementary event is associated with certain probabilities p i(in the case of finite everything), and , then each value of a random variable can be associated with a certain probability p i, such that .

Let X is an arbitrary real number. Let's denote R X (x) the probability that the random variable X took a value equal to X, i.e. P X (x)=P(X=x). Then the function R X (x) can take positive values only for those values X, which belong to a finite or countable set , and for all other values the probability of this value P X (x) = 0.

So, we have defined the set of values, -algebra as a system of any subsets and for each event ( X = x) compared the probability for any, i.e. constructed a probability space.

For example, the space of elementary events of an experiment consisting of tossing a symmetrical coin twice consists of four elementary events: , where

When the coin was tossed twice, two tails appeared; when the coin was tossed twice, two coats of arms fell;

On the first toss of the coin, a hash came up, and on the second, a coat of arms;

On the first toss of the coin, the coat of arms came up, and on the second, the hash mark.

Let the random variable X– number of grating dropouts. It is defined on and the set of its values . All possible subsets, including single-point ones, form an algebra, i.e. =(Ø, (1), (2), (0,1), (0,2), (1,2), (0,1,2)).

Probability of an event ( X=x i}, і = 1,2,3, we define as the probability of the occurrence of an event that is its prototype:

Thus, on elementary events ( X = x i) set a numerical function R X, So .

Definition 2.4. The law of distribution of a discrete random variable is a set of pairs of numbers (x i, р i), where x i are the possible values of the random variable, and р i are the probabilities with which it takes these values, and .

The simplest form of specifying the distribution law of a discrete random variable is a table that lists the possible values of the random variable and the corresponding probabilities:

			…		…
			…		…

Such a table is called a distribution series. To give the distribution series a more visual appearance, it is depicted graphically: on the axis Oh dots x i and draw perpendiculars of length from them p i. The resulting points are connected and a polygon is obtained, which is one of the forms of the distribution law (Fig. 2.1).

Thus, to specify a discrete random variable, you need to specify its values and the corresponding probabilities.

Example 2.2. The cash slot of the machine is triggered each time a coin is inserted with the probability R. Once it is triggered, the coins do not come down. Let X– the number of coins that must be inserted before the machine’s cash slot is triggered. Construct a series of distribution of a discrete random variable X.

Solution. Possible values of a random variable X: x 1 = 1, x 2 = 2,..., x k = k, ... Let's find the probabilities of these values: p 1– the probability that the money receiver will operate the first time it is lowered, and p 1 = p; p 2 – the probability that two attempts will be made. To do this, it is necessary that: 1) the money receiver does not work on the first attempt; 2) on the second try it worked. The probability of this event is (1–р)р. Likewise and so on, . Distribution range X will take the form

	1	2	3	…	To	…
	R	qp	q 2 p		q r -1 p

Note that the probabilities r k form a geometric progression with the denominator: 1–p=q, q<1, therefore this probability distribution is called geometric.

Let us further assume that a mathematical model has been constructed experiment described by a discrete random variable X, and consider calculating the probabilities of the occurrence of arbitrary events.

Let an arbitrary event contain a finite or countable set of values x i: A= {x 1, x 2,..., x i, ...) .Event A can be represented as a union of incompatible events of the form: . Then, using Kolmogorov’s axiom 3 , we get

since we determined the probabilities of the occurrence of events to be equal to the probabilities of the occurrence of events that are their prototypes. This means that the probability of any event , , can be calculated using the formula, since this event can be represented in the form of a union of events, where .

Then the distribution function F(x) = Р(–<Х<х) is found by the formula. It follows that the distribution function of a discrete random variable X is discontinuous and increases in jumps, i.e. it is a step function (Fig. 2.2):

If the set is finite, then the number of terms in the formula is finite, but if it is countable, then the number of terms is countable.

Example 2.3. The technical device consists of two elements that operate independently of each other. The probability of failure of the first element during time T is 0.2, and the probability of failure of the second element is 0.1. Random value X– the number of failed elements during time T. Find the distribution function of the random variable and plot its graph.

Solution. The space of elementary events of an experiment consisting of studying the reliability of two elements of a technical device is determined by four elementary events , , , : – both elements are operational; – the first element is working, the second is faulty; – the first element is faulty, the second is working; – both elements are faulty. Each of the elementary events can be expressed through elementary events of spaces And , where – the first element is operational; – the first element has failed; – the second element is working; – the second element has failed. Then, and since the elements of a technical device work independently of each other, then

8. What is the probability that the values of a discrete random variable belong to the interval?

Discrete called a random variable that can take on individual, isolated values with certain probabilities.

EXAMPLE 1. The number of times the coat of arms appears in three coin tosses. Possible values: 0, 1, 2, 3, their probabilities are equal respectively:

P(0) = ; Р(1) = ; Р(2) = ; Р(3) = .

EXAMPLE 2. The number of failed elements in a device consisting of five elements. Possible values: 0, 1, 2, 3, 4, 5; their probabilities depend on the reliability of each element.

Discrete random variable X can be given by a distribution series or a distribution function (the integral distribution law).

Near distribution is the set of all possible values Xi and their corresponding probabilities Ri = P(X = xi), it can be specified as a table:

x i				x n
p i				р n

At the same time, the probabilities Ri satisfy the condition

Ri= 1 because

where is the number of possible values n may be finite or infinite.

Graphical representation of the distribution series called the distribution polygon . To construct it, possible values of the random variable ( Xi) are plotted along the x-axis, and the probabilities Ri- along the ordinate axis; points Ai with coordinates ( Xi,рi) are connected by broken lines.

Distribution function random variable X called function F(X), whose value at the point X is equal to the probability that the random variable X will be less than this value X, that is

F(x) = P(X< х).

Function F(X) For discrete random variable calculated by the formula

F(X) = Ri , (1.10.1)

where the summation is carried out over all values i, for which Xi< х.

EXAMPLE 3. From a batch containing 100 products, of which there are 10 defective, five products are randomly selected to check their quality. Construct a series of distributions of a random number X defective products contained in the sample.

Solution. Since in the sample the number of defective products can be any integer ranging from 0 to 5 inclusive, then the possible values Xi random variable X are equal:

x 1 = 0, x 2 = 1, x 3 = 2, x 4 = 3, x 5 = 4, x 6 = 5.

Probability R(X = k) that the sample contains exactly k(k = 0, 1, 2, 3, 4, 5) defective products, equals

P (X = k) = .

As a result of calculations using this formula with an accuracy of 0.001, we obtain:

R 1 = P(X = 0) @ 0,583;R 2 = P(X = 1) @ 0,340;R 3 = P(X = 2) @ 0,070;

R 4 = P(X = 3) @ 0,007;R 5 = P(X= 4) @ 0;R 6 = P(X = 5) @ 0.

Using equality to check Rk=1, we make sure that the calculations and rounding were done correctly (see table).

x i
p i

EXAMPLE 4. Given a distribution series of a random variable X :

x i
p i

Find the probability distribution function F(X) of this random variable and construct it.

Solution. If X£10 then F(X)= P(X<X) = 0;

if 10<X£20 then F(X)= P(X<X) = 0,2 ;

if 20<X£30 then F(X)= P(X<X) = 0,2 + 0,3 = 0,5 ;

if 30<X£40 then F(X)= P(X<X) = 0,2 + 0,3 + 0,35 = 0,85 ;

if 40<X£50 then F(X)= P(X<X) = 0,2 + 0,3 + 0,35 + 0,1=0,95 ;

If X> 50, then F(X)= P(X<X) = 0,2 + 0,3 + 0,35 + 0,1 + 0,05 = 1.

In applications of probability theory, the quantitative characteristics of the experiment are of primary importance. A quantity that can be quantitatively determined and which, as a result of an experiment, can take on different values depending on the case is called random variable.

Examples of random variables:

1. The number of times an even number of points appears in ten throws of a die.

2. The number of hits on the target by a shooter who fires a series of shots.

3. The number of fragments of an exploding shell.

In each of the examples given, the random variable can only take on isolated values, that is, values that can be numbered using a natural series of numbers.

Such a random variable, the possible values of which are individual isolated numbers, which this variable takes with certain probabilities, is called discrete.

The number of possible values of a discrete random variable can be finite or infinite (countable).

Law of distribution A discrete random variable is a list of its possible values and their corresponding probabilities. The distribution law of a discrete random variable can be specified in the form of a table (probability distribution series), analytically and graphically (probability distribution polygon).

When carrying out an experiment, it becomes necessary to evaluate the value being studied “on average.” The role of the average value of a random variable is played by a numerical characteristic called mathematical expectation, which is determined by the formula

Where x 1 , x 2 ,.. , x n– random variable values X, A p 1 ,p 2 , ... , p n– the probabilities of these values (note that p 1 + p 2 +…+ p n = 1).

Example. Shooting is carried out at the target (Fig. 11).

A hit in I gives three points, in II – two points, in III – one point. The number of points scored in one shot by one shooter has a distribution law of the form

To compare the skill of shooters, it is enough to compare the average values of the points scored, i.e. mathematical expectations M(X) And M(Y):

M(X) = 1 0,4 + 2  0,2 + 3  0,4 = 2,0,

M(Y) = 1 0,2 + 2  0,5 + 3  0,3 = 2,1.

The second shooter gives on average a slightly higher number of points, i.e. it will give better results when fired repeatedly.

Let us note the properties of the mathematical expectation:

1. The mathematical expectation of a constant value is equal to the constant itself:

M(C) = C.

2. The mathematical expectation of the sum of random variables is equal to the sum of the mathematical expectations of the terms:

M =(X 1 + X 2 +…+ X n)= M(X 1)+ M(X 2)+…+ M(X n).

3. The mathematical expectation of the product of mutually independent random variables is equal to the product of the mathematical expectations of the factors

M(X 1 X 2 … X n) = M(X 1)M(X 2)… M(X n).

4. The mathematical negation of the binomial distribution is equal to the product of the number of trials and the probability of an event occurring in one trial (task 4.6).

M(X) = pr.

To assess how a random variable “on average” deviates from its mathematical expectation, i.e. In order to characterize the spread of values of a random variable in probability theory, the concept of dispersion is used.

Variance random variable X is called the mathematical expectation of the squared deviation:

D(X) = M[(X - M(X)) 2 ].

Dispersion is a numerical characteristic of the dispersion of a random variable. From the definition it is clear that the smaller the dispersion of a random variable, the more closely its possible values are located around the mathematical expectation, that is, the better the values of the random variable are characterized by its mathematical expectation.

From the definition it follows that the variance can be calculated using the formula

It is convenient to calculate the variance using another formula:

D(X) = M(X 2) - (M(X)) 2 .

The dispersion has the following properties:

1. The variance of the constant is zero:

D(C) = 0.

2. The constant factor can be taken out of the dispersion sign by squaring it:

D(CX) = C 2 D(X).

3. The variance of the sum of independent random variables is equal to the sum of the variance of the terms:

D(X 1 + X 2 + X 3 +…+ X n)= D(X 1)+ D(X 2)+…+ D(X n)

4. The variance of the binomial distribution is equal to the product of the number of trials and the probability of the occurrence and non-occurrence of an event in one trial:

D(X) = npq.

In probability theory, a numerical characteristic equal to the square root of the variance of a random variable is often used. This numerical characteristic is called the mean square deviation and is denoted by the symbol

It characterizes the approximate size of the deviation of a random variable from its average value and has the same dimension as the random variable.

4.1. The shooter fires three shots at the target. The probability of hitting the target with each shot is 0.3.

Construct a distribution series for the number of hits.

Solution. The number of hits is a discrete random variable X. Each value x n random variable X corresponds to a certain probability P n .

The distribution law of a discrete random variable in this case can be specified near distribution.

In this problem X takes values 0, 1, 2, 3. According to Bernoulli's formula

Let's find the probabilities of possible values of the random variable:

R 3 (0) = (0,7) 3 = 0,343,

R 3 (1) =0,3(0,7) 2 = 0,441,

R 3 (2) =(0,3) 2 0,7 = 0,189,

R 3 (3) = (0,3) 3 = 0,027.

By arranging the values of the random variable X in increasing order, we obtain the distribution series:

X n

Note that the amount

means the probability that the random variable X will take at least one value from among the possible ones, and this event is reliable, therefore

4.2 .There are four balls in the urn with numbers from 1 to 4. Two balls are taken out. Random value X– the sum of the ball numbers. Construct a distribution series of a random variable X.

Solution. Random variable values X are 3, 4, 5, 6, 7. Let's find the corresponding probabilities. Random variable value 3 X can be accepted in the only case when one of the selected balls has the number 1, and the other 2. The number of possible test outcomes is equal to the number of combinations of four (the number of possible pairs of balls) of two.

Using the classical probability formula we get

Likewise,

R(X= 4) =R(X= 6) =R(X= 7) = 1/6.

The sum 5 can appear in two cases: 1 + 4 and 2 + 3, so

X has the form:

Find the distribution function F(x) random variable X and plot it. Calculate for X its mathematical expectation and variance.

Solution. The distribution law of a random variable can be specified by the distribution function

F(x) = P(X x).

Distribution function F(x) is a non-decreasing, left-continuous function defined on the entire number line, while

F (- )= 0,F (+ )= 1.

For a discrete random variable, this function is expressed by the formula

Therefore in this case

Distribution function graph F(x) is a stepped line (Fig. 12)

	F(x)

Expected valueM(X) is the weighted arithmetic average of the values X 1 , X 2 ,……X n random variable X with scales ρ 1, ρ 2, …… , ρ n and is called the mean value of the random variable X. According to the formula

M(X)= x 1 ρ 1 + x 2 ρ 2 +……+ x n ρ n

M(X) = 3·0.14+5·0.2+7·0.49+11·0.17 = 6.72.

Dispersion characterizes the degree of dispersion of the values of a random variable from its average value and is denoted D(X):

D(X)=M[(HM(X)) 2 ]= M(X 2) –[M(X)] 2 .

For a discrete random variable, the variance has the form

or it can be calculated using the formula

Substituting the numerical data of the problem into the formula, we get:

M(X 2) = 3 2 ∙ 0,14+5 2 ∙ 0,2+7 2 ∙ 0,49+11 2 ∙ 0,17 = 50,84

D(X) = 50,84-6,72 2 = 5,6816.

4.4. Two dice are rolled twice at the same time. Write the binomial law of distribution of a discrete random variable X- the number of occurrences of an even total number of points on two dice.

Solution. Let us introduce a random event

A= (two dice with one throw resulted in a total of even number of points).

Using the classical definition of probability we find

R(A)= ,

Where n - the number of possible test outcomes is found according to the rule

multiplication:

n = 6∙6 =36,

m - number of people favoring the event A outcomes - equal

m= 3∙6=18.

Thus, the probability of success in one trial is

ρ = P(A)= 1/2.

The problem is solved using a Bernoulli test scheme. One challenge here would be to roll two dice once. Number of such tests n = 2. Random variable X takes values 0, 1, 2 with probabilities

R 2 (0) =,R 2 (1) =∙,R 2 (2) =

The required binomial distribution of a random variable X can be represented as a distribution series:

X n
ρ n

4.5 . In a batch of six parts there are four standard ones. Three parts were selected at random. Construct a probability distribution of a discrete random variable X– the number of standard parts among those selected and find its mathematical expectation.

Solution. Random variable values X are the numbers 0,1,2,3. It's clear that R(X=0)=0, since there are only two non-standard parts.

R(X=1) =
=1/5,

R(X= 2) =
= 3/5,

R(X=3) =
= 1/5.

Distribution law of a random variable X Let's present it in the form of a distribution series:

X n
ρ n

Expected value

M(X)=1 ∙ 1/5+2 ∙ 3/5+3 ∙ 1/5=2.

4.6 . Prove that the mathematical expectation of a discrete random variable X- number of occurrences of the event A V n independent trials, in each of which the probability of an event occurring is equal to ρ – equal to the product of the number of trials by the probability of the occurrence of an event in one trial, that is, to prove that the mathematical expectation of the binomial distribution

M(X) =n . ρ ,

and dispersion

D(X) =n.p. .

Solution. Random value X can take values 0, 1, 2..., n. Probability R(X= k) is found using Bernoulli’s formula:

R(X=k)= R n(k)= ρ To (1-ρ ) n- To

Distribution series of a random variable X has the form:

X n
ρ n	q n	ρq n- 1	ρq n- 2		ρ n

Where q= 1- ρ .

For the mathematical expectation we have the expression:

M(X)=ρq n - 1 +2 ρ 2 q n - 2 +…+.n ρ n

In the case of one test, that is, with n= 1 for random variable X 1 – number of occurrences of the event A- the distribution series has the form:

X n
ρ n

M(X 1)= 0∙q + 1 ∙ p = p

D(X 1) = p – p 2 = p(1- p) = pq.

If X k – number of occurrences of the event A in which test, then R(X To)= ρ And

X=X 1 +X 2 +….+X n .

From here we get

M(X)=M(X 1 )+M(X 2)+ … +M(X n)= nρ,

D(X)=D(X 1)+D(X 2)+ ... +D(X n)=npq.

4.7. The quality control department checks products for standardness. The probability that the product is standard is 0.9. Each batch contains 5 products. Find the mathematical expectation of a discrete random variable X- the number of batches, each of which will contain 4 standard products - if 50 batches are subject to inspection.

Solution. The probability that there will be 4 standard products in each randomly selected batch is constant; let's denote it by ρ .Then the mathematical expectation of the random variable X equals M(X)= 50∙ρ.

Let's find the probability ρ according to Bernoulli's formula:

ρ=P 5 (4)== 0,94∙0,1=0,32.

M(X)= 50∙0,32=16.

4.8 . Three dice are thrown. Find the mathematical expectation of the sum of the dropped points.

Solution. You can find the distribution of a random variable X- the sum of the dropped points and then its mathematical expectation. However, this path is too cumbersome. It is easier to use another technique, representing a random variable X, the mathematical expectation of which needs to be calculated, in the form of a sum of several simpler random variables, the mathematical expectation of which is easier to calculate. If the random variable X i is the number of points rolled on i– th bones ( i= 1, 2, 3), then the sum of points X will be expressed in the form

X = X 1 + X 2 + X 3 .

To calculate the mathematical expectation of the original random variable, all that remains is to use the property of mathematical expectation

M(X 1 + X 2 + X 3 )= M(X 1 )+ M(X 2)+ M(X 3 ).

It's obvious that

R(X i = K)= 1/6, TO= 1, 2, 3, 4, 5, 6, i= 1, 2, 3.

Therefore, the mathematical expectation of the random variable X i looks like

M(X i) = 1/6∙1 + 1/6∙2 +1/6∙3 + 1/6∙4 + 1/6∙5 + 1/6∙6 = 7/2,

M(X) = 3∙7/2 = 10,5.

4.9. Determine the mathematical expectation of the number of devices that failed during testing if:

a) the probability of failure for all devices is the same R, and the number of devices under test is equal to n;

b) probability of failure for i – of the device is equal to p i , i= 1, 2, … , n.

Solution. Let the random variable X is the number of failed devices, then

X = X 1 + X 2 + … + X n ,

X i =

It's clear that

R(X i = 1)= R i , R(X i = 0)= 1– R i ,i= 1, 2, … ,n.

M(X i)= 1∙R i + 0∙(1-R i)=P i ,

M(X)=M(X 1)+M(X 2)+ … +M(X n)=P 1 +P 2 + … + P n .

In case “a” the probability of device failure is the same, that is

R i =p,i= 1, 2, … ,n.

M(X)= n.p..

This answer could be obtained immediately if we notice that the random variable X has a binomial distribution with parameters ( n, p).

4.10. Two dice are thrown simultaneously twice. Write the binomial law of distribution of a discrete random variable X - the number of rolls of an even number of points on two dice.

Solution. Let

A=(rolling an even number on the first die),

B =(rolling an even number on the second dice).

Getting an even number on both dice in one throw is expressed by the product AB. Then

R (AB) = R(A)∙R(IN) =
.

The result of the second throw of two dice does not depend on the first, so Bernoulli's formula applies when

n = 2,p = 1/4, q = 1– p = 3/4.

Random value X can take values 0, 1, 2 , the probability of which can be found using Bernoulli’s formula:

R(X= 0)= P 2 (0) = q 2 = 9/16,

R(X= 1)= P 2 (1)= C ,R∙q = 6/16,

R(X= 2)= P 2 (2)= C , R 2 = 1/16.

Distribution series of a random variable X:

4.11. The device consists of a large number of independently operating elements with the same very small probability of failure of each element over time t. Find the average number of refusals over time t elements, if the probability that at least one element will fail during this time is 0.98.

Solution. Number of people who refused over time t elements – random variable X, which is distributed according to Poisson's law, since the number of elements is large, the elements work independently and the probability of failure of each element is small. Average number of occurrences of an event in n tests equals

M(X) = n.p..

Since the probability of failure TO elements from n expressed by the formula

R n (TO)
,

where  = n.p., then the probability that not a single element will fail during the time t we get at K = 0:

R n (0)= e -  .

Therefore, the probability of the opposite event is in time t at least one element fails – equal to 1 - e -  . According to the conditions of the problem, this probability is 0.98. From Eq.

1 - e -  = 0,98,

e -  = 1 – 0,98 = 0,02,

from here  = -ln 0,02  4.

So, in time t operation of the device, on average 4 elements will fail.

4.12 . The dice are rolled until a “two” comes up. Find the average number of throws.

Solution. Let's introduce a random variable X– the number of tests that must be performed until the event of interest to us occurs. The probability that X= 1 is equal to the probability that during one throw of the dice a “two” will appear, i.e.

R(X= 1) = 1/6.

Event X= 2 means that on the first test the “two” did not come up, but on the second it did. Probability of event X= 2 is found by the rule of multiplying the probabilities of independent events:

R(X= 2) = (5/6)∙(1/6)

Likewise,

R(X= 3) = (5/6) 2 ∙1/6, R(X= 4) = (5/6) 2 ∙1/6

etc. We obtain a series of probability distributions:


					(5/6) To ∙1/6

The average number of throws (trials) is the mathematical expectation

M(X) = 1∙1/6 + 2∙5/6∙1/6 + 3∙(5/6) 2 ∙1/6 + … + TO (5/6) TO -1 ∙1/6 + … =

1/6∙(1+2∙5/6 +3∙(5/6) 2 + … + TO (5/6) TO -1 + …)

Let's find the sum of the series:

TOg TO -1 = (g TO) g 
.

Hence,

M(X) = (1/6) (1/ (1 – 5/6) 2 = 6.

Thus, you need to make an average of 6 throws of the dice until a “two” comes up.

4.13. Independent tests are carried out with the same probability of occurrence of the event A in every test. Find the probability of an event occurring A, if the variance of the number of occurrences of an event in three independent trials is 0.63 .

Solution. The number of occurrences of an event in three trials is a random variable X, distributed according to the binomial law. The variance of the number of occurrences of an event in independent trials (with the same probability of occurrence of the event in each trial) is equal to the product of the number of trials by the probabilities of the occurrence and non-occurrence of the event (problem 4.6)

D(X) = npq.

By condition n = 3, D(X) = 0.63, so you can R find from equation

0,63 = 3∙R(1-R),

which has two solutions R 1 = 0.7 and R 2 = 0,3.