Determination of event probability and statistical distribution.

In the practical application of probability theory, one often encounters problems in which the same experiment or similar experiments are repeated repeatedly. As a result of each experiment, some event may or may not appear, and we are not interested in the result of each individual experiment, but in the total number of occurrences of the event as a result of a series of experiments. For example, if a group of shots is fired at the same target, we are usually not interested in the result of each shot, but in the total number of hits. IN similar tasks requires the ability to determine the probability of any given number of occurrences of an event as a result of a series of experiments. Such tasks will be considered in this chapter. They can be solved quite simply in the case when the experiments are independent.

Several experiments are called independent if the probability of one or another outcome of each experiment does not depend on what outcomes the other experiments had. For example, several successive coin tosses constitute independent experiments. Several successive removals of a card from the deck constitute independent experiments, provided that the removed card is returned to the deck each time and the cards are shuffled; otherwise, these are dependent experiences. Several shots constitute independent experiments only if the aiming is done anew before each shot; in the case when aiming is carried out once before the entire shooting or is carried out continuously during the shooting process (firing in a burst, bombing in a series), the shots represent dependent experiments. Independent experiments can be carried out in the same or different conditions. In the first case, the probability of an event changes from experience to experience. A particular theorem applies to the first case, and to the second - general theorem about repeating experiments. We will start with a particular theorem, as it is more elementary. First of all, let's look at a specific example.

Example. Three independent shots are fired at the target, the probability of hitting it with each shot is equal to . Find the probability that with these three shots we will get exactly two hits.

Solution. Let us denote the event that exactly two projectiles hit the target. This event can happen in three ways:

1) hit on the first shot, hit on the second, miss on the third;

2) hit on the first shot, miss on the second, hit on the third;

3) miss on the first shot, hit on the second, hit on the third.

Therefore, an event can be represented as the sum of products of events:

where - hits on the first, second, third shots, respectively, - miss on the first, second, third shots.

Considering that the three listed versions of the event are incompatible, and the events included in the products are independent, using the theorems of addition and multiplication we obtain:

or, denoting ,

Likewise, listing all possible options, in which the event of interest to us may appear given number times, we can solve the following general problem.

Independent experiments are carried out, in each of which some event may or may not appear; the probability of an event occurring in each experiment is equal to , and the probability of non-occurrence is . We need to find the probability that an event will appear exactly once in these experiments.

Let us consider the event that the event will appear exactly once in the experiments. This event can come true in various ways. Let us decompose an event into the sum of products of events consisting of the appearance or non-appearance of an event in a separate experience. We will denote the occurrence of an event in the i-th experiment; - non-occurrence of an event in the i-th experiment.

Obviously, each variant of the occurrence of an event (each member of the sum) must consist of m occurrences of the event and non-occurrences, i.e. from events and events with different indices. Thus,

Moreover, in each work the event must appear once, but must appear once.

The number of all combinations of this kind is equal, i.e. the number of ways in which one can choose from experiments in which the event occurred. The probability of each such combination, according to the multiplication theorem for independent events, is equal to . Since combinations are incompatible with each other, then, according to the addition theorem, the probability of an event is equal to

Determining event probability and statistical distribution

Task 1

Light bulbs mixed in the box same size and shapes: 150 W - 8 pieces and 100 W - 13. Three lamps were taken out of the box at random. Find the probability that among them:

a) only one 150 W lamp; b) two lamps of 150 W;

c) at least two lamps of 150 W each; d) at least one 150 W lamp;

f) all lamps are of the same power.

a) event F1 - out of three randomly taken lamps, only one will be 150 W:

b) event F2 - out of three randomly taken lamps, two lamps will be 150 W each:

c) event F3 - out of three randomly taken lamps, at least 2 will be 150 W each:

d) event F4 - out of three randomly taken parts there will be at least one 150 W lamp:

e) event F5 - out of three randomly taken lamps, all three will be of the same power

Task 2

Three independent shots are fired at the aircraft. The probability of a hit on the first shot is 0.4, on the second - 0.5, on the third - 0.6. Three hits are enough to disable an aircraft. With two hits it fails with a probability of 0.7, with one hit - with a probability of 0.4.

1. Find the probability that the plane will be disabled as a result of three shots.

2. As a result of three shots, the plane was not disabled. How many hits were most likely on the plane?

1) Consider the hypotheses:

H1 - out of three shots there will be no hits

H2 - out of three shots there will be exactly one hit

H3 - out of three shots there will be two hits

H4 - out of three shots there will be three hits

and event

F - the plane will be disabled.

Because the plane was not disabled, i.e. event F has occurred, then the probabilities of the hypotheses are determined using the Bayes formula

0,121+0,380,6+0,380,3+0,120=0,462

Thus, most likely the plane was hit once.

Task 3

According to statistics in the city of N, on average, 18% of new businesses that open cease their activities within a year.

1. What is the probability that out of 6 randomly selected new enterprises in city N, by the end of the year of activity there will remain:

a) exactly 4; b) 4; c) less than 4; d) at least one enterprise?

2. Calculate the probability that out of one hundred newly opened enterprises in city N will cease to operate by the end of the year:

a) 15; b) at least 15; c) no more than 21; d) at least 13, but not more than 23 enterprises.

n=6q=0.18p=1-q=1-0.18=0.82

n value<10, поэтому для расчетов воспользуемся формулой Бернулли:

a) exactly 4 enterprises will remain:

b) more than 4 enterprises will remain:

P(more than 4)=P6(5;6)=P6(5)+P6(6)

P(more than 4)=0.4004+0.304=0.7044

c) less than 4 enterprises will remain:

P(less than 4)=1-P(at least 4)=1-P6(4;6)=1-(0.2197+0.4004+0.304)=0.0759

d) at least one enterprise will remain

P(at least 1)=1-P(none)=1-P6(0)=1-0.186=0.999966

n=100p=0.18q=0.82

The value n=100 is quite large, so for calculations we will use the local and integral Laplace formulas:

a) exactly 15 enterprises will cease their activities:

where, and (x) is the local Laplace function

From the table we find that

(-0,78)=(0,78)=0,2943,

b) at least 15 enterprises will cease their activities, i.e. from 15 to 100:

Pn(k1;k2)Ф(x2)-Ф(x1),

where and, and Ф(x) is the Laplace integral function

From the table of function values ​​Ф(x) we find that Ф(-0.78)=-Ф(0.78)=-0.2823, and Ф(21.34)=0.5, P100(15;100) 0.5+0.2823=0.7823

c) no more than 21 enterprises will cease their activities: i.e. from 0 to 21:

From the table of function values ​​Ф(x) we find that Ф(-4.69)=-Ф(4.69)=-0.499999, and Ф(0.78)=0.2823, P100(0;21) 0.2823+0.499999=0.782299

d) at least 13, but not more than 23 enterprises will cease their activities:

From the table of function values ​​Ф(x) we find that Ф(1,3)=0.4032,

P100(13;23)0.4032+0.4032=0.8064

Task 4

Two accountants independently fill out identical statements. The first accountant makes errors on average in 8%, the second - in 12% of all documents. The number of statements completed by the first accountant is 1, the second - 2. A random variable (r.v.) is considered - the number of statements filled out by two accountants without errors.

1. Compile a series of distributions of r.v. and present it graphically.

3. Calculate mathematical expectation(mean) M, variance

D and mean square (standard) deviation ().

4. Determine the probabilities: a) P; b) P; c) P

1) Let us determine the possible values ​​of the random variable X and their probabilities:

X=0: 0.920.882=0.712448

X=1: 0.080.882+0.92(0.120.88+0.880.12)=0.256256

X=2: 0.920.122+0.08(0.120.88+0.880.12)=0.030144

X=3: 0.080.122=0.001152

Examination:

0,712488+0,256256+0,030144+0,001152=1

Let's write down the distribution series

Let us depict the distribution series graphically in the form of a polygon

2) Let's create a distribution function:

Let's plot the distribution function

3) The mathematical expectation and variance are found by the formula:

D(X)=0.3872-0.322=0.2848

4) Find the required probabilities:

P(X

Р(XMX+1)=1-Р(X<1,32)=1-F(1,32)=1-0,968704=0,031296

P(-0.2137

Task 5

Between two settlements located at a distance of L = 9 km from each other, a bus runs with stops on demand anywhere. The distance (in km) traveled by a certain passenger who boards the bus at the beginning of the route is random with the distribution density

1. Set the unknown constant C and plot the function p(x).

2. Find the distribution function of r.v. and build its graph.

3. Calculate the mathematical expectation (average value) M, variance D and standard deviation ().

4. How many times is the number of disembarks from the beginning of the route to the middle place of the passenger’s trip greater than the number of disembarks from this place to the end of the bus route?

1) To find the constant C, we use the property of the distribution density:

Let's plot the distribution density

2) Find the distribution function

a) if x<0, то F(x)=0, т.к. значений, меньших 0, случайная величина не принимает.

b) if 0x<9, то

c) if x>3, then

due to the distribution density property

Finally we get:

Let's plot F(x):

3) the mathematical expectation is calculated using the formula

The variance is calculated using the formula:

DX=24.3-4.52=4.05

Average standard deviation equals:

P(X

P(XMX)=1-P(X

Those. the number of disembarks from the beginning of the route to the middle place of the passenger's trip and the number of disembarks from this place to the end of the bus route are equal.

Task 6

When carrying cargo by helicopters, cables are used that are made of synthetic materials based on new chemical technologies. As a result of 25 tensile tests of the cable, the following data were obtained (in tons):

2.948 , 3.875, 5.526, 5.422, 4.409, 4.314, 5.150, 2.451, 5.226, 4.105, 3.280, 5.732, 3.249, 3.408, 7.204, 5.174, 6.222, 5.276, 5.853, 4.420, 6.525, 2.127, 5.264, 4.647, 5.591

Necessary:

1. Determine the characteristic being studied and its type (discrete or continuous).

2. Depending on the type of attribute, construct a polygon or a histogram of relative frequencies.

3. Based on a visual analysis of the polygon (histogram), formulate a hypothesis about the distribution law of the characteristic being studied.

4. Calculate the sample characteristics of the characteristic: mean, dispersion and standard deviation.

5. Using Pearson’s chi-square goodness-of-fit test, check the compliance of the sample data with the distribution law put forward in paragraph 3 at a significance level of 0.01.

6. For the general mean and variance, construct confidence intervals corresponding to a confidence probability of 0.99.

7. With a reliability of 0.99, test the hypothesis of equality:

a) general average value 5C;

b) general dispersion value C 2, where C = 1.09.

Sample values ​​according to job option

1. The type of attribute is continuous, because a random variable can take any value from a certain interval.

2. Let's build a histogram of relative frequencies. Let's determine the number of intervals:

where n is the number of values ​​and k is the number of intervals.

IN in this case there are 25 values, so the number of intervals is:

k=1+1.44ln 25 5.6.

Let's take the number of intervals to be 5.

Let's determine the size of one interval:

Let us determine the relative frequencies for each interval. It is convenient to carry out calculations in the table

Let's build a histogram

3. Based on visual analysis, we can put forward a hypothesis about the distribution of the characteristic according to the normal law.

4. Let’s determine the sample characteristics of the trait being studied.

a) sample average:

b) sample variance:

c) sample standard deviation

5. Let’s check the hypothesis that the sample data corresponds to a normal distribution

Let’s determine the ends of the intervals using the formula, for which we’ll create a table

Let's find the theoretical probabilities pi and theoretical frequencies. We will write the calculation results in the table

Let's calculate the observed value of the Pearson criterion. To do this, let's create a table:

Based on the significance level =0.01 and the number of degrees of freedom k=n-3=5-3=2, we find from the table of critical points: =9.2

Because , then there is no reason to reject the hypothesis about the normal distribution of the critical mass for rupture.

6. Construct a confidence interval for the general mean and general variance

The maximum sampling error for the average is calculated using the formula:

where t is the confidence coefficient, which depends on the probability with which the statement is made.

The confidence coefficient is found from the relation 2Ф(t)=p, where Ф(х) is the Laplace integral function.

According to the condition p=0.99,

The boundaries within which the general average falls are given by the inequalities:

5.1225 - 0.7034 a 5.1225 + 0.7034

Let's find the interval estimate of the variance:

According to the table of critical points of the distribution, we find that =42.98, a =10.86, then the confidence interval for the variance will be:

a) let’s check the hypothesis that the general average is equal to 5.45.

We put forward hypotheses:

Because the variance of the population is unknown, then we calculate the expression

Using the table of values ​​of Student's critical points, we find the critical value

tcr(;n-1)=tcr(0.01;24)=2.8

Because 1.201<2,8, то нет оснований отклонить гипотезу о равенстве генеральной средней значению 5,45.

b) Let's check the hypothesis that the general variance is equal to 1.1881.

We put forward hypotheses:

Calculate the expression

Using the table of values ​​of critical points of the Chi-square distribution, we find the critical value (;n-1)=(0.01;24)=43

Because 37.5<43, то нет оснований отклонить гипотезу о равенстве генеральной дисперсии значению 1,1881.

References

probability statistical variance mathematical

1. Gmurman V.E. A guide to solving problems in probability theory and mathematical statistics: A textbook for university students. - M.: Higher School, 2002.

2. Semenov A.T. Probability theory and mathematical statistics: Educational and methodological complex. - Novosibirsk: NGAEiU, 2003.

2. Three independent experiments are carried out, in each of which the event appears with probability 0.2. Compile a series of distributions of the number of occurrences of an event in three experiments. Find M(X) and D(X) of this random variable.

3. Independent random variables X and Y are specified by distribution tables:

Find:
a) M(X), M(Y), D(X), D(Y);
b) tables of distribution of random variables Z1 = 2X + Y and Z2=XY;
c) M (Z1), M (Z2), D (Z1), D(Z2) directly from distribution tables and based on the properties of mathematical expectation and dispersion.

4. An automatic machine stamps parts. The probability that the part will be defective is 0.01. Create a series of distribution of defective parts out of 200 manufactured. Find M(X) of this random variable.

5. The number of equipment elements that failed during a time T is a random variable X, distributed exponentially (λ = 0.2). Indicate the density and distribution function, construct their graphs, find the average number of elements that can fail during time T. What is the probability that the number of failed elements is between 3 and 10?

6. The load G on the rod obeys the normal distribution law with parameters a = 250 kg; σ = 50 kg. What is the probability that the load will not exceed 380 kg? What is the probability of loads from 100 to 200 kg?

The total probability formula allows you to find the probability of an event A, which can only occur with each of n mutually exclusive events that form a complete system, if their probabilities are known, and conditional probabilities events A relative to each of the system events are equal.

Events are also called hypotheses; they are mutually exclusive. Therefore, in the literature you can also find their designation not by the letter B, and the letter H(hypothesis).

To solve problems with such conditions, it is necessary to consider 3, 4, 5 or in the general case n possibility of an event occurring A- with every event.

Using the theorems of addition and multiplication of probabilities, we obtain the sum of the products of the probability of each of the events of the system by conditional probability events A regarding each of the system events. That is, the probability of an event A can be calculated using the formula

or in general

,

which is called total probability formula .

Total probability formula: examples of problem solving

Example 1. There are three identical-looking urns: the first has 2 white balls and 3 black, the second has 4 white and one black, the third has three white balls. Someone approaches one of the urns at random and takes out one ball from it. Taking advantage total probability formula, find the probability that this ball will be white.

Solution. Event A- the appearance of a white ball. We put forward three hypotheses:

The first urn is selected;

The second urn is selected;

The third urn is selected.

Conditional probabilities of an event A regarding each of the hypotheses:

, , .

We apply the total probability formula, resulting in the required probability:

.

Example 2. At the first plant, out of every 100 light bulbs, an average of 90 standard light bulbs are produced, at the second - 95, at the third - 85, and the products of these factories constitute, respectively, 50%, 30% and 20% of all light bulbs supplied to stores in a certain area. Find the probability of purchasing a standard light bulb.

Solution. Let us denote the probability of purchasing a standard light bulb by A, and the events that the purchased light bulb was manufactured at the first, second and third factories, respectively, through . By condition, the probabilities of these events are known: , , and conditional probabilities of the event A regarding each of them: , , . These are the probabilities of purchasing a standard light bulb, provided that it was manufactured at the first, second, and third factories, respectively.

Event A will occur if an event occurs K- the light bulb is manufactured at the first plant and is standard, or an event L- the light bulb is manufactured in a second plant and is standard, or an event M- the light bulb was manufactured at the third plant and is standard. Other possibilities for the event to occur A No. Therefore, the event A is the sum of events K, L And M, which are incompatible. Using the probability addition theorem, we imagine the probability of an event A in the form

and by the probability multiplication theorem we get

that is, special case of the total probability formula.

Substituting the probability values ​​into the left side of the formula, we obtain the probability of the event A :

Example 3. The plane is landing at the airfield. If the weather permits, the pilot lands the plane, using, in addition to instruments, also visual observation. In this case, the probability of a safe landing is equal to . If the airfield is covered with low clouds, then the pilot lands the plane, guided only by instruments. In this case, the probability of a safe landing is equal to; . Devices that provide blind landing are reliable (probability of failure-free operation) P. In the presence of low clouds and failed blind landing instruments, the probability of a successful landing is equal to; . Statistics show that in k% of landings the airfield is covered with low clouds. Find total probability of an event A- safe landing of the plane.

Solution. Hypotheses:

There is no low clouds;

There are low clouds.

Probabilities of these hypotheses (events):

;

Conditional probability.

We will again find the conditional probability using the formula of total probability with hypotheses

Blind landing devices are operational;

The blind landing instruments failed.

Probabilities of these hypotheses:

According to the total probability formula

Example 4. The device can operate in two modes: normal and abnormal. Normal mode is observed in 80% of all cases of operation of the device, and abnormal mode - in 20% of cases. Probability of device failure within a certain time t equal to 0.1; in abnormal 0.7. Find full probability failure of the device over time t.

Solution. We again denote the probability of device failure through A. So, regarding the operation of the device in each mode (event), the probabilities are known according to the condition: for normal mode this is 80% (), for abnormal mode - 20% (). Probability of event A(that is, device failure) depending on the first event (normal mode) is equal to 0.1 (); depending on the second event (abnormal mode) - 0.7 ( ). We substitute these values ​​into the total probability formula (that is, the sum of the products of the probability of each of the events of the system by the conditional probability of the event A regarding each of the events of the system) and before us is the required result.