The random variable x is given by the distribution law. Distribution law of a random variable

Distribution law of random variable X:

Example No. 2. The probability of one shooter hitting the target with one shot for the first shooter is 0.8, for the second shooter – 0.85. The shooters fired one shot at the target. Considering hitting the target as independent events for individual shooters, find the probability of event A – exactly one hit on the target.
Solution.
Consider event A - one hit on the target. Possible options for this event to occur are as follows:

1. The first shooter hit, the second shooter missed: P(A/H1)=p 1 *(1-p 2)=0.8*(1-0.85)=0.12
2. The first shooter missed, the second shooter hit the target: P(A/H2)=(1-p 1)*p 2 =(1-0.8)*0.85=0.17
3. The first and second arrows hit the target independently of each other: P(A/H1H2)=p 1 *p 2 =0.8*0.85=0.68

Discrete random Variables are random variables that take only values ​​that are distant from each other and that can be listed in advance.
Law of distribution
The distribution law of a random variable is a relationship that establishes a connection between the possible values ​​of a random variable and their corresponding probabilities.
The distribution series of a discrete random variable is the list of its possible values ​​and the corresponding probabilities.
The distribution function of a discrete random variable is the function:
,
determining for each value of the argument x the probability that the random variable X will take a value less than this x.

Expectation of a discrete random variable
,
where is the value of a discrete random variable; - the probability of a random variable accepting X values.
If a random variable takes a countable set of possible values, then:
.
Mathematical expectation of the number of occurrences of an event in n independent trials:
,

Dispersion and standard deviation of a discrete random variable
Dispersion of a discrete random variable:
or .
Variance of the number of occurrences of an event in n independent trials
,
where p is the probability of the event occurring.
Standard deviation of a discrete random variable:
.

Example 1
Draw up a law of probability distribution for a discrete random variable (DRV) X – the number of k occurrences of at least one “six” in n = 8 throws of a pair of dice. Construct a distribution polygon. Find the numerical characteristics of the distribution (distribution mode, mathematical expectation M(X), dispersion D(X), standard deviation s(X)). Solution: Let us introduce the notation: event A – “when throwing a pair of dice, a six appeared at least once.” To find the probability P(A) = p of event A, it is more convenient to first find the probability P(Ā) = q of the opposite event Ā - “when throwing a pair of dice, a six never appeared.”
Since the probability of a “six” not appearing when throwing one die is 5/6, then according to the probability multiplication theorem
P(Ā) = q = = .
Respectively,
P(A) = p = 1 – P(Ā) = .
The tests in the problem follow the Bernoulli scheme, so d.s.v. magnitude X- number k the occurrence of at least one six when throwing two dice obeys the binomial law of probability distribution:

where = is the number of combinations of n By k.

The calculations carried out for this problem can be conveniently presented in the form of a table:
Probability distribution d.s.v. X º k (n = 8; p = ; q = )

Polygon (polygon) of probability distribution of a discrete random variable X shown in the figure:

Rice. Probability distribution polygon d.s.v. X=k.
The vertical line shows the mathematical expectation of the distribution M(X).

Let us find the numerical characteristics of the probability distribution of d.s.v. X. The distribution mode is 2 (here P 8(2) = 0.2932 maximum). The mathematical expectation by definition is equal to:
M(X) = = 2,4444,
Where xk = k– value taken by d.s.v. X. Variance D(X) we find the distribution using the formula:
D(X) = = 4,8097.
Standard deviation (RMS):
s( X) = = 2,1931.

Example2
Discrete random variable X given by the distribution law

Solution. If , then (third property).
If, then. Really, X can take the value 1 with probability 0.3.
If, then. Indeed, if it satisfies the inequality
, then equals the probability of an event that can occur when X will take the value 1 (the probability of this event is 0.3) or the value 4 (the probability of this event is 0.1). Since these two events are incompatible, then according to the addition theorem, the probability of an event is equal to the sum of the probabilities 0.3 + 0.1 = 0.4. If, then. Indeed, the event is certain, therefore its probability is equal to one. So, the distribution function can be written analytically as follows:

Graph of this function:
Let us find the probabilities corresponding to these values. By condition, the probabilities of failure of the devices are equal: then the probabilities that the devices will work during the warranty period are equal:




The distribution law has the form:

As is known, random variable is called a variable quantity that can take on certain values ​​depending on the case. Random variables are denoted by capital letters of the Latin alphabet (X, Y, Z), and their values ​​are denoted by corresponding lowercase letters (x, y, z). Random variables are divided into discontinuous (discrete) and continuous.

Discrete random variable is a random variable that takes only a finite or infinite (countable) set of values ​​with certain non-zero probabilities.

Distribution law of a discrete random variable is a function that connects the values ​​of a random variable with their corresponding probabilities. The distribution law can be specified in one of the following ways.

1 . The distribution law can be given by the table:

where λ>0, k = 0, 1, 2, … .

V) by using distribution functions F(x) , which determines for each value x the probability that the random variable X will take a value less than x, i.e. F(x) = P(X< x).

Properties of the function F(x)

3 . The distribution law can be specified graphically – distribution polygon (polygon) (see problem 3).

Note that to solve some problems it is not necessary to know the distribution law. In some cases, it is enough to know one or several numbers that reflect the most important features of the distribution law. This can be a number that has the meaning of the “average value” of a random variable, or a number showing the average size of the deviation of a random variable from its mean value. Numbers of this kind are called numerical characteristics of a random variable.

Basic numerical characteristics of a discrete random variable :

• Mathematical expectation (average value) of a discrete random variable M(X)=Σ x i p i.
  For binomial distribution M(X)=np, for Poisson distribution M(X)=λ
• Dispersion discrete random variable D(X)=M2 or D(X) = M(X 2)− 2. The difference X–M(X) is called the deviation of a random variable from its mathematical expectation.
  For binomial distribution D(X)=npq, for Poisson distribution D(X)=λ
• Standard deviation (standard deviation) σ(X)=√D(X).

Examples of solving problems on the topic “The law of distribution of a discrete random variable”

Task 1.

1000 lottery tickets were issued: 5 of them will win 500 rubles, 10 will win 100 rubles, 20 will win 50 rubles, 50 will win 10 rubles. Determine the law of probability distribution of the random variable X - winnings per ticket.

Solution. According to the conditions of the problem, the following values ​​of the random variable X are possible: 0, 10, 50, 100 and 500.

The number of tickets without winning is 1000 – (5+10+20+50) = 915, then P(X=0) = 915/1000 = 0.915.

Similarly, we find all other probabilities: P(X=0) = 50/1000=0.05, P(X=50) = 20/1000=0.02, P(X=100) = 10/1000=0.01 , P(X=500) = 5/1000=0.005. Let us present the resulting law in the form of a table:

Let's find the mathematical expectation of the value X: M(X) = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = (1+ 2+3+4+5+6)/6 = 21/6 = 3.5

Task 3.

The device consists of three independently operating elements. The probability of failure of each element in one experiment is 0.1. Draw up a distribution law for the number of failed elements in one experiment, construct a distribution polygon. Find the distribution function F(x) and plot it. Find the mathematical expectation, variance and standard deviation of a discrete random variable.

Solution. 1. The discrete random variable X = (the number of failed elements in one experiment) has the following possible values: x 1 = 0 (none of the device elements failed), x 2 = 1 (one element failed), x 3 = 2 (two elements failed ) and x 4 =3 (three elements failed).

Failures of elements are independent of each other, the probabilities of failure of each element are equal, therefore it is applicable Bernoulli's formula . Considering that, according to the condition, n=3, p=0.1, q=1-p=0.9, we determine the probabilities of the values:
P 3 (0) = C 3 0 p 0 q 3-0 = q 3 = 0.9 3 = 0.729;
P 3 (1) = C 3 1 p 1 q 3-1 = 3*0.1*0.9 2 = 0.243;
P 3 (2) = C 3 2 p 2 q 3-2 = 3*0.1 2 *0.9 = 0.027;
P 3 (3) = C 3 3 p 3 q 3-3 = p 3 =0.1 3 = 0.001;
Check: ∑p i = 0.729+0.243+0.027+0.001=1.

Thus, the desired binomial distribution law of X has the form:

We plot the possible values ​​of x i along the abscissa axis, and the corresponding probabilities p i along the ordinate axis. Let's construct points M 1 (0; 0.729), M 2 (1; 0.243), M 3 (2; 0.027), M 4 (3; 0.001). By connecting these points with straight line segments, we obtain the desired distribution polygon.

3. Let's find the distribution function F(x) = Р(Х

Graph of function F(x)

4. For binomial distribution X:
- mathematical expectation M(X) = np = 3*0.1 = 0.3;
- variance D(X) = npq = 3*0.1*0.9 = 0.27;
- standard deviation σ(X) = √D(X) = √0.27 ≈ 0.52.

Random variable A variable is called a variable that, as a result of each test, takes on one previously unknown value, depending on random reasons. Random variables are denoted by capital Latin letters: $X,\ Y,\ Z,\ \dots $ According to their type, random variables can be discrete And continuous.

Discrete random variable- this is a random variable whose values ​​can be no more than countable, that is, either finite or countable. By countability we mean that the values ​​of a random variable can be numbered.

Example 1 . Here are examples of discrete random variables:

a) the number of hits on the target with $n$ shots, here the possible values ​​are $0,\ 1,\ \dots ,\ n$.

b) the number of emblems dropped when tossing a coin, here the possible values ​​are $0,\ 1,\ \dots ,\ n$.

c) the number of ships arriving on board (a countable set of values).

d) the number of calls arriving at the PBX (countable set of values).

1. Law of probability distribution of a discrete random variable.

A discrete random variable $X$ can take values ​​$x_1,\dots ,\ x_n$ with probabilities $p\left(x_1\right),\ \dots ,\ p\left(x_n\right)$. The correspondence between these values ​​and their probabilities is called law of distribution of a discrete random variable. As a rule, this correspondence is specified using a table, the first line of which indicates the values ​​$x_1,\dots ,\ x_n$, and the second line contains the probabilities $p_1,\dots ,\ p_n$ corresponding to these values.

$\begin(array)(|c|c|)
\hline
X_i & x_1 & x_2 & \dots & x_n \\
\hline
p_i & p_1 & p_2 & \dots & p_n \\
\hline
\end(array)$

Example 2 . Let the random variable $X$ be the number of points rolled when tossing a die. Such a random variable $X$ can take the following values: $1,\ 2,\ 3,\ 4,\ 5,\ 6$. The probabilities of all these values ​​are equal to $1/6$. Then the law of probability distribution of the random variable $X$:

$\begin(array)(|c|c|)
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline

\hline
\end(array)$

Comment. Since in the distribution law of a discrete random variable $X$ the events $1,\ 2,\ \dots ,\ 6$ form a complete group of events, then the sum of the probabilities must be equal to one, that is, $\sum(p_i)=1$.

2. Mathematical expectation of a discrete random variable.

Expectation of a random variable sets its “central” meaning. For a discrete random variable, the mathematical expectation is calculated as the sum of the products of the values ​​$x_1,\dots ,\ x_n$ and the probabilities $p_1,\dots ,\ p_n$ corresponding to these values, that is: $M\left(X\right)=\sum ^n_(i=1)(p_ix_i)$. In English-language literature, another notation $E\left(X\right)$ is used.

Properties of mathematical expectation$M\left(X\right)$:

1. $M\left(X\right)$ lies between the smallest and largest values ​​of the random variable $X$.
2. The mathematical expectation of a constant is equal to the constant itself, i.e. $M\left(C\right)=C$.
3. The constant factor can be taken out of the sign of the mathematical expectation: $M\left(CX\right)=CM\left(X\right)$.
4. The mathematical expectation of the sum of random variables is equal to the sum of their mathematical expectations: $M\left(X+Y\right)=M\left(X\right)+M\left(Y\right)$.
5. The mathematical expectation of the product of independent random variables is equal to the product of their mathematical expectations: $M\left(XY\right)=M\left(X\right)M\left(Y\right)$.

Example 3 . Let's find the mathematical expectation of the random variable $X$ from example $2$.

$$M\left(X\right)=\sum^n_(i=1)(p_ix_i)=1\cdot ((1)\over (6))+2\cdot ((1)\over (6) )+3\cdot ((1)\over (6))+4\cdot ((1)\over (6))+5\cdot ((1)\over (6))+6\cdot ((1 )\over (6))=3.5.$$

We can notice that $M\left(X\right)$ lies between the smallest ($1$) and largest ($6$) values ​​of the random variable $X$.

Example 4 . It is known that the mathematical expectation of the random variable $X$ is equal to $M\left(X\right)=2$. Find the mathematical expectation of the random variable $3X+5$.

Using the above properties, we get $M\left(3X+5\right)=M\left(3X\right)+M\left(5\right)=3M\left(X\right)+5=3\cdot 2 +5=$11.

Example 5 . It is known that the mathematical expectation of the random variable $X$ is equal to $M\left(X\right)=4$. Find the mathematical expectation of the random variable $2X-9$.

Using the above properties, we get $M\left(2X-9\right)=M\left(2X\right)-M\left(9\right)=2M\left(X\right)-9=2\cdot 4 -9=-1$.

3. Dispersion of a discrete random variable.

Possible values ​​of random variables with equal mathematical expectations can disperse differently around their average values. For example, in two student groups the average score for the exam in probability theory turned out to be 4, but in one group everyone turned out to be good students, and in the other group there were only C students and excellent students. Therefore, there is a need for a numerical characteristic of a random variable that would show the spread of the values ​​of the random variable around its mathematical expectation. This characteristic is dispersion.

Variance of a discrete random variable$X$ is equal to:

$$D\left(X\right)=\sum^n_(i=1)(p_i(\left(x_i-M\left(X\right)\right))^2).\ $$

In English literature the notation $V\left(X\right),\ Var\left(X\right)$ is used. Very often the variance $D\left(X\right)$ is calculated using the formula $D\left(X\right)=\sum^n_(i=1)(p_ix^2_i)-(\left(M\left(X \right)\right))^2$.

Dispersion properties$D\left(X\right)$:

1. The variance is always greater than or equal to zero, i.e. $D\left(X\right)\ge 0$.
2. The variance of the constant is zero, i.e. $D\left(C\right)=0$.
3. The constant factor can be taken out of the sign of the dispersion provided that it is squared, i.e. $D\left(CX\right)=C^2D\left(X\right)$.
4. The variance of the sum of independent random variables is equal to the sum of their variances, i.e. $D\left(X+Y\right)=D\left(X\right)+D\left(Y\right)$.
5. The variance of the difference between independent random variables is equal to the sum of their variances, i.e. $D\left(X-Y\right)=D\left(X\right)+D\left(Y\right)$.

Example 6 . Let's calculate the variance of the random variable $X$ from example $2$.

$$D\left(X\right)=\sum^n_(i=1)(p_i(\left(x_i-M\left(X\right)\right))^2)=((1)\over (6))\cdot (\left(1-3.5\right))^2+((1)\over (6))\cdot (\left(2-3.5\right))^2+ \dots +((1)\over (6))\cdot (\left(6-3.5\right))^2=((35)\over (12))\approx 2.92.$$

Example 7 . It is known that the variance of the random variable $X$ is equal to $D\left(X\right)=2$. Find the variance of the random variable $4X+1$.

Using the above properties, we find $D\left(4X+1\right)=D\left(4X\right)+D\left(1\right)=4^2D\left(X\right)+0=16D\ left(X\right)=16\cdot 2=32$.

Example 8 . It is known that the variance of the random variable $X$ is equal to $D\left(X\right)=3$. Find the variance of the random variable $3-2X$.

Using the above properties, we find $D\left(3-2X\right)=D\left(3\right)+D\left(2X\right)=0+2^2D\left(X\right)=4D\ left(X\right)=4\cdot 3=12$.

4. Distribution function of a discrete random variable.

The method of representing a discrete random variable in the form of a distribution series is not the only one, and most importantly, it is not universal, since a continuous random variable cannot be specified using a distribution series. There is another way to represent a random variable - the distribution function.

Distribution function random variable $X$ is called a function $F\left(x\right)$, which determines the probability that the random variable $X$ will take a value less than some fixed value $x$, that is, $F\left(x\right )=P\left(X< x\right)$

Properties of the distribution function:

1. $0\le F\left(x\right)\le 1$.
2. The probability that the random variable $X$ will take values ​​from the interval $\left(\alpha ;\ \beta \right)$ is equal to the difference between the values ​​of the distribution function at the ends of this interval: $P\left(\alpha< X < \beta \right)=F\left(\beta \right)-F\left(\alpha \right)$
3. $F\left(x\right)$ - non-decreasing.
4. $(\mathop(lim)_(x\to -\infty ) F\left(x\right)=0\ ),\ (\mathop(lim)_(x\to +\infty ) F\left(x \right)=1\ )$.

Example 9 . Let us find the distribution function $F\left(x\right)$ for the distribution law of the discrete random variable $X$ from example $2$.

$\begin(array)(|c|c|)
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
\hline
\end(array)$

If $x\le 1$, then, obviously, $F\left(x\right)=0$ (including for $x=1$ $F\left(1\right)=P\left(X< 1\right)=0$).

If $1< x\le 2$, то $F\left(x\right)=P\left(X=1\right)=1/6$.

If $2< x\le 3$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)=1/6+1/6=1/3$.

If $3< x\le 4$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)=1/6+1/6+1/6=1/2$.

If $4< x\le 5$, то $F\left(X\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)=1/6+1/6+1/6+1/6=2/3$.

If $5< x\le 6$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)=1/6+1/6+1/6+1/6+1/6=5/6$.

If $x > 6$, then $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right) +P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)=1/6+1/6+1/6+1/6+ 1/6+1/6=1$.

So $F(x)=\left\(\begin(matrix)
0,\ at\ x\le 1,\\
1/6,at\ 1< x\le 2,\\
1/3,\ at\ 2< x\le 3,\\
1/2,at\ 3< x\le 4,\\
2/3,\ at\ 4< x\le 5,\\
5/6,\ at\ 4< x\le 5,\\
1,\ for\ x > 6.
\end(matrix)\right.$

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:

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)

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:

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:

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:

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:

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:

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.