Find the distribution function F(x). Mathematical expectation of a continuous random variable. Solution Example A random variable has a distribution density of the form

mathematical expectation discrete random variable called:

In the case of an infinite set of values, there is a series on the right side of (4.4), and we will consider only those values of X for which this series converges absolutely.

M(X) is the average expected value of the random variable. It has the following properties:

1) M(C)=C, where C=const

2) M(CX)=CM(X) (4.5)

3) M (X+Y)=M(X)+M(Y), for any X and Y.

4) M (XY)=M (X)M(Y) if X and Y are independent.

To estimate the degree of dispersion of the values of a random variable around its mean value M(X)= A concepts are introduced dispersionD(X) and the mean square (standard) deviation . dispersion called expected value squared difference (X- ), those. :

D(X)=M(X- ) 2 = p i ,

Where =M(X); defined as Square root from dispersion, i.e. .

To calculate the variance, use the formula:

(4.6)

Properties of variance and standard deviation:

1) D(C)=0, where C=const

2) D(CX)=C 2 D(X), (CX)= çCç(X) (4.7)

3) D(X+Y) =D(X)+D(Y),

if X and Y are independent.

The dimension of the quantities and coincides with the dimension of the random variable X itself, and the dimension of D(X) is equal to the square of the dimension of the random variable X.

4.3. Mathematical operations on random variables.

Let the random variable X take values with probabilities and the random variable Y take values with probabilities values of a random variable X. Therefore, its distribution law has the form table 4.2:

Table 4.2

			...
			...

Square random variable X, i.e. , is a new random variable that, with the same probabilities as the random variable X, takes values equal to the squares of its values.

Sum random variables X and Y is a new random variable that takes all values of the form with probabilities expressing the probability that the random variable X takes the value and Y - the value , that is

(4.8)

If the random variables X and Y are independent, then:

The difference and product of random variables X and Y are defined similarly.

Difference random variables X and Y is a new random variable that takes all values of the form , and work- all values of the form with probabilities determined by the formula (4.8), and if the random variables X and Y are independent, then by the formula (4.9).

4.4. Bernoulli and Poisson distributions.

Consider a sequence of n identical retests that satisfy the following conditions:

1. Each trial has two outcomes, called success and failure.

These two outcomes are mutually incompatible and opposite events.

2. The probability of success, denoted p, remains constant from trial to trial. The probability of failure is denoted by q.

3. All n trials are independent. This means that the probability of an event occurring in any of the n repeated trials does not depend on the results of other trials.

The probability that in n independent repeated trials, in each of which the probability of occurrence of an event is equal to , the event will occur exactly m times (in any sequence), is equal to

(4.10)

Expression (4.10) is called the Bernoulli formula.

The probabilities that the event will occur:

a) less than m times,

b) more than m times,

c) at least m times,

d) no more than m times - are found, respectively, according to the formulas:

Binomial is the law of distribution of a discrete random variable X - the number of occurrences of an event in n independent trials, in each of which the probability of the event occurring is equal to p; the probabilities of possible values X = 0,1,2,..., m,...,n are calculated using the Bernoulli formula (Table 4.3).

Table 4.3

Number of successes X=m				...	m	...	n
Probability P				...		...

Since the right side of formula (4.10) represents the general term of the binomial expansion, this distribution law is called binomial. For a random variable X, distributed according to the binomial law, we have.

Definition 13.1. The random variable X is called discrete, if it takes a finite or countable number of values.

Definition 13.2. The law of distribution of a random variable X is the set of pairs of numbers ( , ), where are the possible values of the random variable, and are the probabilities with which the random variable takes these values, i.e. =P( X= ), and =1.

The simplest form of specifying a discrete random variable is a table that lists the possible values of a random variable and their corresponding probabilities. Such a table is called near distribution discrete random variable.

X			…		…
R			…		…

The distribution series can be represented graphically. In this case, the abscissa is plotted along the ordinate, and the probability is plotted along the ordinate. Points with coordinates ( , ) are connected by segments and get a broken line called distribution polygon, which is one of the forms of specifying the law of distribution of a discrete random variable.

Example 13.3. Construct a distribution polygon of a random variable X with a distribution series

X
R	0,1	0,3	0,2	0,4

Definition 13.4. We say that a discrete random variable X has binomial distribution with parameters ( n,p) if it can take non-negative integer values k {1,2,…,n) with probabilities Р( X=x)= .

The distribution series has the form:

X			…	k	…	n
R			…		…

Sum of probabilities = =1.

Definition 13.5. It is said that the discrete form of the random variable X It has Poisson distribution with the parameter (>0), if it takes integer values k(0,1,2,…) with probabilities Р( X=k)= .

The distribution series has the form

X			…	k	…
R			…		…

Since the expansion in the Maclaurin series has the following form, then the sum of the probabilities = = =1.

Denote by X number of trials to be completed before the first occurrence of the event A in independent trials, if the probability of occurrence of A in each of them is equal to p (0<p <1), а вероятность непоявления . Возможными значениями X are natural numbers.

Definition 13.6. They say that the random variable X It has geometric distribution with parameter p (0<p <1), если она принимает натуральные значения k N with probabilities Р(Х=k)= , where . Distribution range:

X				…	n	…
R				…		…

The sum of probabilities = = =1.

Example 13.7. The coin is flipped 2 times. Compile a distribution series of a random variable X of the number of occurrences of the "coat of arms".

P 2 (0)= = ; P 2 (1)===0.5; P 2 (2) = = .

X
R

The distribution series will take the form:

Example 13.8. The gun is fired until the first hit on the target. The probability of hitting with one shot is 0.6. will hit on the 3rd shot.

Because the p=0,6, q=0,4, k=3, then P( A)= =0,4 2 *0,6=0,096.

14 Numerical characteristics of discrete random variables

The distribution law completely characterizes the random variable, but it is often unknown, so you have to limit yourself to less information. Sometimes it is even more profitable to use numbers (parameters) that describe the random variable in total. They're called numerical characteristics random variable. These include: mathematical expectation, variance, etc.

Definition 14.1. mathematical expectation A discrete random variable is called the sum of the products of all its possible values and their probabilities. Denote the mathematical expectation of a random variable X through M X=M( X)=E X.

If the random variable X takes a finite number of values, then M X= .

If the random variable X takes a countable number of values, then M X= ,

and the mathematical expectation exists if the series converges absolutely.

Remark 14.2. Mathematical expectation is a certain number approximately equal to a certain value of a random variable.

Example 14.3. Find the mathematical expectation of a random variable X, knowing its distribution series

X
R	0,1	0,6	0,3

M X=3*0,1+5*0,6+2*0,3=3,9.

Example 14.4. Find the mathematical expectation of the number of occurrences of an event A in one trial, if the probability of an event A is equal to p.

Random value X- number of occurrence of the event A in one test. It can take values =1 ( A happened) with a probability p and =0 with probability , i.e. distribution series

Hence MS=C*1=C.

Remark 14.6. The product of a constant value C by a discrete random variable X Defined as a discrete random variable C X, the possible values of which are equal to the products of the constant С and the possible values X, the probabilities of these values С X are equal to the probabilities of the corresponding possible values X.

Property 14.7. The constant factor can be taken out of the expectation sign:

M(S X)=C∙M X.

If the random variable X has a distribution number

X			…		…
R			…		…

Random variable distribution series

CX			…		…
R			…		…

M(S X)= = = С∙М( X).

Definition 14.8. Random variables , ,…, are called independent, if for , i=1,2,…,n

Р( , ,…, )= Р( ) Р( )… Р( ) (1)

If as = , i=1,2,…,n, then we obtain from (1)

R(< , < ,…, < }= Р{ < }Р{ < }… Р{ < }, откуда получается другая формула:

( , ,…, ) = () ()... () (2)

for the joint distribution function of random variables , ,…, , which can also be taken as a definition of the independence of a random variable.

Property 14.9. Mathematical expectation of the product of 2 independent random variables is equal to the product of their mathematical expectations:

M( XY)=M X∙M At.

Property 14.10. The mathematical expectation of the sum of 2 random variables is equal to the sum of their mathematical expectations:

M( X+Y)=M X+M At.

Remark 14.11. Properties 14.9 and 14.10 can be generalized to the case of several random variables.

Example 14.12. Find the mathematical expectation of the sum of the number of points that can fall out when throwing 2 dice.

Let X the number of points rolled on the first die, At the number of points rolled on the second die. They have the same distribution series:

X
R

Then M X=M At= (1+2+3+4+5+6)= = . M( X+Y)=2* =7.

Theorem 14.13. Mathematical expectation of the number of occurrences of the event A V n independent trials is equal to the product of the number of trials and the probability of an event occurring in each trial: M X=np.

Let X– number of occurrences of the event A V n independent tests. –number of occurrences of the event A V i- that test, i=1,2,…,n. Then = + +…+ . According to the properties of the mathematical expectation M X= . From example 14.4M X i=p,i=1,2,…,n, hence M X= =np.

Definition 14.14.dispersion random variable is called the number D X=M( X-M X) 2 .

Definition 14.15.Standard deviation random variable X called number =.

Remark 14.16. Dispersion is a measure of the spread of values of a random variable around its mathematical expectation. It is always non-negative. To calculate the variance, it is more convenient to use another formula:

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

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

Example 14.17. Find the variance of a random variable X, given by a number of distributions

X
P	0,1	0,6	0,3

M X=2*0.1+3*0.6+5*0.3=3.5; M( X 2)= 4*0,1+9*0,6+25*0,3=13,3;

D X=13.3-(3,5) 2 =1,05.

Dispersion Properties

Property 14.18. The dispersion of a constant value is 0:

DC = M(C-MC) 2 = M(C-C) 2 =0.

Property 14.19. The constant factor can be taken out of the dispersion sign by squaring it

D(C X) =C 2 D X.

D(CX)=M(C-CM X) 2 \u003d M (C (X- M X) 2) = C 2 M( X-M X) 2 = C 2 D X.

Property 14.20. The variance of the sum of 2 independent random variables is equal to the sum of the variances of these variables

D( X+Y)=D X+D Y.

D( X + Y)=M(( X+Y) 2) – (M( X+Y)) 2 = M( x2+ 2XY+Y2) - (M X+ M Y) 2 = =M( X) 2 +2M X M Y+M( Y 2)-(M( X) 2 +2M X M Y+M( Y) 2)= M( X 2)-(M X) 2 +M( Y 2)-(M Y) 2 = D X+D Y.

Corollary 14.21. The variance of the sum of several independent random variables is equal to the sum of their variances.

Theorem 14.22. Variance of the number of occurrences of an event A V n independent tests, in each of which the probability p) 2 =). Hence D +2 ,

Exercise 1. The distribution density of a continuous random variable X has the form:
Find:
a) parameter A ;
b) distribution function F(x) ;
c) the probability of hitting a random variable X in the interval ;
d) mathematical expectation MX and variance DX .
Plot the functions f(x) and F(x) .

Task 2. Find the variance of the random variable X given by the integral function.

Task 3. Find the mathematical expectation of a random variable X given a distribution function.

Task 4. The probability density of some random variable is given as follows: f(x) = A/x 4 (x = 1; +∞)
Find coefficient A , distribution function F(x) , mathematical expectation and variance, as well as the probability that a random variable takes a value in the interval . Plot f(x) and F(x) graphs.

Task. The distribution function of some continuous random variable is given as follows:

Determine the parameters a and b , find the expression for the probability density f(x) , the mathematical expectation and variance, as well as the probability that the random variable takes a value in the interval . Plot f(x) and F(x) graphs.

Let's find the distribution density function as a derivative of the distribution function.
F′=f(x)=a
Knowing that we will find the parameter a:

or 3a=1, whence a = 1/3
We find the parameter b from the following properties:
F(4) = a*4 + b = 1
1/3*4 + b = 1 whence b = -1/3
Therefore, the distribution function is: F(x) = (x-1)/3

Expected value.

Dispersion.

1 / 9 4 3 - (1 / 9 1 3) - (5 / 2) 2 = 3 / 4
Find the probability that a random variable takes a value in the interval
P(2< x< 3) = F(3) – F(2) = (1/3*3 - 1/3) - (1/3*2 - 1/3) = 1/3

Example #1. The probability distribution density f(x) of a continuous random variable X is given. Required:

Determine coefficient A .
find the distribution function F(x) .
schematically plot F(x) and f(x) .
find the mathematical expectation and variance of X .
find the probability that X takes a value from the interval (2;3).

f(x) = A*sqrt(x), 1 ≤ x ≤ 4.
Solution:

The random variable X is given by the distribution density f(x):

Find the parameter A from the condition:

or
14/3*A-1=0
Where,
A = 3 / 14

The distribution function can be found by the formula.

RANDOM VALUES

Example 2.1. Random value X given by the distribution function

Find the probability that as a result of the test X will take values between (2.5; 3.6).

Solution: X in the interval (2.5; 3.6) can be determined in two ways:

Example 2.2. At what values of the parameters A And IN function F(x) = A + Be - x can be a distribution function for non-negative values of a random variable X.

Solution: Since all possible values of the random variable X belong to the interval , then in order for the function to be a distribution function for X, the property should hold:

Answer: .

Example 2.3. The random variable X is given by the distribution function

Find the probability that, as a result of four independent trials, the value X exactly 3 times will take a value belonging to the interval (0.25; 0.75).

Solution: Probability of hitting a value X in the interval (0.25; 0.75) we find by the formula:

Example 2.4. The probability of the ball hitting the basket in one throw is 0.3. Draw up the law of distribution of the number of hits in three throws.

Solution: Random value X- the number of hits in the basket with three throws - can take on the values: 0, 1, 2, 3. The probabilities that X

Example 2.5. Two shooters make one shot at the target. The probability of hitting it by the first shooter is 0.5, the second - 0.4. Write down the law of distribution of the number of hits on the target.

Solution: Find the law of distribution of a discrete random variable X- the number of hits on the target. Let the event be a hit on the target by the first shooter, and - hit by the second shooter, and - respectively, their misses.

Let us compose the law of probability distribution of SV X:

Example 2.6. 3 elements are tested, working independently of each other. Durations of time (in hours) of failure-free operation of elements have distribution density functions: for the first: F 1 (t) =1-e- 0,1 t, for the second: F 2 (t) = 1-e- 0,2 t, for the third one: F 3 (t) =1-e- 0,3 t. Find the probability that in the time interval from 0 to 5 hours: only one element will fail; only two elements will fail; all three elements fail.

Solution: Let's use the definition of the generating function of probabilities:

The probability that in independent trials, in the first of which the probability of occurrence of an event A equals , in the second, etc., the event A appears exactly once, is equal to the coefficient at in the expansion of the generating function in powers of . Let's find the probabilities of failure and non-failure, respectively, of the first, second and third element in the time interval from 0 to 5 hours:

Let's create a generating function:

The coefficient at is equal to the probability that the event A will appear exactly three times, that is, the probability of failure of all three elements; the coefficient at is equal to the probability that exactly two elements will fail; coefficient at is equal to the probability that only one element will fail.

Example 2.7. Given a probability density f(x) random variable X:

Find the distribution function F(x).

Solution: We use the formula:

Thus, the distribution function has the form:

Example 2.8. The device consists of three independently operating elements. The probability of failure of each element in one experiment is 0.1. Compile the law of distribution of the number of failed elements in one experiment.

Solution: Random value X- the number of elements that failed in one experiment - can take the values: 0, 1, 2, 3. Probabilities that X takes these values, we find by the Bernoulli formula:

Thus, we obtain the following law of the probability distribution of a random variable X:

Example 2.9. There are 4 standard parts in a lot of 6 parts. 3 items were randomly selected. Draw up the law of distribution of the number of standard parts among the selected ones.

Solution: Random value X- the number of standard parts among the selected ones - can take the values: 1, 2, 3 and has a hypergeometric distribution. The probabilities that X

Where -- the number of parts in the batch;

-- the number of standard parts in the lot;

– number of selected parts;

-- the number of standard parts among those selected.

Example 2.10. The random variable has a distribution density

where and are not known, but , a and . Find and .

Solution: In this case, the random variable X has a triangular distribution (Simpson distribution) on the interval [ a, b]. Numerical characteristics X:

Hence, . Solving this system, we get two pairs of values: . Since, according to the condition of the problem, we finally have: .

Answer: .

Example 2.11. On average, for 10% of contracts, the insurance company pays the sums insured in connection with the occurrence of an insured event. Calculate the mathematical expectation and variance of the number of such contracts among four randomly selected ones.

Solution: The mathematical expectation and variance can be found using the formulas:

Possible values of SV (number of contracts (out of four) with the occurrence of an insured event): 0, 1, 2, 3, 4.

We use the Bernoulli formula to calculate the probabilities of a different number of contracts (out of four) for which the sums insured were paid:

The distribution series of CV (the number of contracts with the occurrence of an insured event) has the form:


	0,6561	0,2916	0,0486	0,0036	0,0001

Answer: , .

Example 2.12. Of the five roses, two are white. Write a distribution law for a random variable expressing the number of white roses among two taken at the same time.

Solution: In a sample of two roses, there may either be no white rose, or there may be one or two white roses. Therefore, the random variable X can take values: 0, 1, 2. The probabilities that X takes these values, we find by the formula:

Where -- number of roses;

-- number of white roses;

– the number of simultaneously taken roses;

-- the number of white roses among those taken.

Then the law of distribution of a random variable will be as follows:

Example 2.13. Among the 15 assembled units, 6 need additional lubrication. Draw up the law of distribution of the number of units in need of additional lubrication, among five randomly selected from the total number.

Solution: Random value X- the number of units that need additional lubrication among the five selected - can take the values: 0, 1, 2, 3, 4, 5 and has a hypergeometric distribution. The probabilities that X takes these values, we find by the formula:

Where -- the number of assembled units;

-- number of units requiring additional lubrication;

– the number of selected aggregates;

-- the number of units that need additional lubrication among the selected ones.

Then the law of distribution of a random variable will be as follows:

Example 2.14. Of the 10 watches received for repair, 7 need a general cleaning of the mechanism. Watches are not sorted by type of repair. The master, wanting to find a watch that needs cleaning, examines them one by one and, having found such a watch, stops further viewing. Find the mathematical expectation and variance of the number of hours watched.

Solution: Random value X- the number of units that need additional lubrication among the five selected - can take the following values: 1, 2, 3, 4. The probabilities that X takes these values, we find by the formula:

Then the law of distribution of a random variable will be as follows:

Now let's calculate the numerical characteristics of the quantity :

Answer: , .

Example 2.15. The subscriber has forgotten the last digit of the phone number he needs, but remembers that it is odd. Find the mathematical expectation and variance of the number of dials he made before hitting the desired number, if he dials the last digit at random and does not dial the dialed digit in the future.

Solution: Random variable can take values: . Since the subscriber does not dial the dialed digit in the future, the probabilities of these values are equal.

Let's compose a distribution series of a random variable:


	0,2

Let's calculate the mathematical expectation and variance of the number of dialing attempts:

Answer: , .

Example 2.16. The probability of failure during the reliability tests for each device of the series is equal to p. Determine the mathematical expectation of the number of devices that failed, if tested N appliances.

Solution: Discrete random variable X is the number of failed devices in N independent tests, in each of which the probability of failure is equal to p, distributed according to the binomial law. The mathematical expectation of the binomial distribution is equal to the product of the number of trials and the probability of an event occurring in one trial:

Example 2.17. Discrete random variable X takes 3 possible values: with probability ; with probability and with probability . Find and knowing that M( X) = 8.

Solution: We use the definitions of mathematical expectation and the law of distribution of a discrete random variable:

We find: .

Example 2.18. The technical control department checks products for standardity. The probability that the item is standard is 0.9. Each batch contains 5 items. Find the mathematical expectation of a random variable X- the number of batches, each of which contains exactly 4 standard products, if 50 batches are subject to verification.

Solution: In this case, all experiments conducted are independent, and the probabilities that each batch contains exactly 4 standard products are the same, therefore, the mathematical expectation can be determined by the formula:

where is the number of parties;

The probability that a batch contains exactly 4 standard items.

We find the probability using the Bernoulli formula:

Answer: .

Example 2.19. Find the variance of a random variable X– number of occurrences of the event A in two independent trials, if the probabilities of the occurrence of an event in these trials are the same and it is known that M(X) = 0,9.

Solution: The problem can be solved in two ways.

1) Possible CB values X: 0, 1, 2. Using the Bernoulli formula, we determine the probabilities of these events:

, , .

Then the distribution law X looks like:

From the definition of mathematical expectation, we determine the probability:

Let's find the variance of SW X:

2) You can use the formula:

Answer: .

Example 2.20. Mathematical expectation and standard deviation of a normally distributed random variable X are 20 and 5, respectively. Find the probability that as a result of the test X will take the value contained in the interval (15; 25).

Solution: Probability of hitting a normal random variable X on the section from to is expressed in terms of the Laplace function:

Example 2.21. Given a function:

At what value of the parameter C this function is the distribution density of some continuous random variable X? Find the mathematical expectation and variance of a random variable X.

Solution: In order for a function to be the distribution density of some random variable , it must be non-negative, and it must satisfy the property:

Hence:

Calculate the mathematical expectation using the formula:

Calculate the variance using the formula:

T is p. It is necessary to find the mathematical expectation and variance of this random variable.

Solution: The distribution law of a discrete random variable X - the number of occurrences of an event in independent trials, in each of which the probability of the occurrence of an event is , is called binomial. The mathematical expectation of the binomial distribution is equal to the product of the number of trials and the probability of the occurrence of the event A in one trial:

Example 2.25. Three independent shots are fired at the target. The probability of hitting each shot is 0.25. Determine the standard deviation of the number of hits with three shots.

Solution: Since three independent trials are performed, and the probability of occurrence of the event A (hit) in each trial is the same, we will assume that the discrete random variable X - the number of hits on the target - is distributed according to the binomial law.

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

Example 2.26. The average number of clients visiting the insurance company in 10 minutes is three. Find the probability that at least one customer arrives in the next 5 minutes.

Average number of customers arriving in 5 minutes: . .

Example 2.29. The waiting time for an application in the processor queue obeys an exponential distribution law with an average value of 20 seconds. Find the probability that the next (arbitrary) request will wait for the processor for more than 35 seconds.

Solution: In this example, the expectation , and the failure rate is .

Then the desired probability is:

Example 2.30. A group of 15 students holds a meeting in a hall with 20 rows of 10 seats each. Each student takes a seat in the hall at random. What is the probability that no more than three people will be in the seventh place in the row?

Solution:

Example 2.31.

Then according to the classical definition of probability:

Where -- the number of parts in the batch;

-- the number of non-standard parts in the lot;

– number of selected parts;

-- the number of non-standard parts among the selected ones.

Then the distribution law of the random variable will be as follows.

Examples of solving problems on the topic "Random variables".

Task 1 . There are 100 tickets issued in the lottery. One win of 50 USD was played. and ten wins of $10 each. Find the law of distribution of the value X - the cost of a possible gain.

Solution. Possible values of X: x 1 = 0; x 2 = 10 and x 3 = 50. Since there are 89 “empty” tickets, then p 1 = 0.89, the probability of winning is 10 c.u. (10 tickets) – p 2 = 0.10 and for a win of 50 c.u. –p 3 = 0.01. Thus:


	0,89	0,10	0,01

Easy to control: .

Task 2. The probability that the buyer has familiarized himself with the advertisement of the product in advance is 0.6 (p = 0.6). Selective quality control of advertising is carried out by polling buyers before the first one who has studied the advertisement in advance. Make a series of distribution of the number of interviewed buyers.

Solution. According to the condition of the problem p = 0.6. From: q=1 -p = 0.4. Substituting these values, we get: and construct a distribution series:


pi		0,24

Task 3. A computer consists of three independently operating elements: a system unit, a monitor, and a keyboard. With a single sharp increase in voltage, the probability of failure of each element is 0.1. Based on the Bernoulli distribution, draw up the distribution law for the number of failed elements during a power surge in the network.

Solution. Consider Bernoulli distribution(or binomial): the probability that in n tests, event A will appear exactly k once: , or:


	q n					p n

IN let's get back to the task.

Possible values of X (number of failures):

x 0 =0 - none of the elements failed;

x 1 =1 - failure of one element;

x 2 =2 - failure of two elements;

x 3 =3 - failure of all elements.

Since, by condition, p = 0.1, then q = 1 – p = 0.9. Using the Bernoulli formula, we get

, ,

, .

Control: .

Therefore, the desired distribution law:


	0,729	0,243	0,027	0,001

Task 4. Produced 5000 rounds. The probability that one cartridge is defective . What is the probability that there will be exactly 3 defective cartridges in the entire batch?

Solution. Applicable Poisson distribution: this distribution is used to determine the probability that, given a very large

number of trials (mass trials), in each of which the probability of event A is very small, event A will occur k times: , Where .

Here n \u003d 5000, p \u003d 0.0002, k \u003d 3. We find , then the desired probability: .

Task 5. When firing before the first hit with the probability of hitting p = 0.6 for a shot, you need to find the probability that the hit will occur on the third shot.

Solution. Let us apply the geometric distribution: let independent trials be performed, in each of which the event A has a probability of occurrence p (and non-occurrence q = 1 - p). Trials end as soon as event A occurs.

Under such conditions, the probability that event A will occur on the kth test is determined by the formula: . Here p = 0.6; q \u003d 1 - 0.6 \u003d 0.4; k \u003d 3. Therefore, .

Task 6. Let the law of distribution of a random variable X be given:

Find the mathematical expectation.

Solution. .

Note that the probabilistic meaning of the mathematical expectation is the average value of a random variable.

Task 7. Find the variance of a random variable X with the following distribution law:

Solution. Here .

The law of distribution of the square of X 2 :

X 2

Required variance: .

Dispersion characterizes the degree of deviation (scattering) of a random variable from its mathematical expectation.

Task 8. Let the random variable be given by the distribution:

			10m

Find its numerical characteristics.

Solution: m, m 2 ,

M 2 , m.

About a random variable X, one can say either - its mathematical expectation is 6.4 m with a variance of 13.04 m 2 , or - its mathematical expectation is 6.4 m with a deviation of m. The second formulation is obviously clearer.

Task 9. Random value X given by the distribution function:
.

Find the probability that, as a result of the test, the value X will take on a value contained in the interval .

Solution. The probability that X will take a value from a given interval is equal to the increment of the integral function in this interval, i.e. . In our case and , therefore