While dialing a phone number, the subscriber forgot 4 digits. Calculating probabilities without constructing a space of elementary events

There are five balls of different sizes in the urn. What is the probability of drawing all the balls in ascending order if it is known that there are no identical balls?

Solution. Total number possible elementary outcomes experience is equal to the number of permutations of five elements, and the number of outcomes favorable to the event is equal to one.

Required probability:

.

Problem 17.

While dialing a phone number, the subscriber forgot the last two digits, and remembering that they were different, he dialed them for luck. What is the probability that he dialed the right number?

Solution. The total number of possible outcomes of the experiment is equal to the number of placements from 10 to 2, i.e. . The number of outcomes favorable to the event is equal to one.

Required probability:

.

Problem 18.

There are 15 notebooks in the desk drawer, 8 of them are squared. We took three notebooks at random. Find the probability that all three notebooks taken will be of the highest quality.

Solution. Since the order does not play a role here, the total number of all possible outcomes will be equal to the number of combinations of 15 by 3, i.e., and the number of favorable events will also be equal to the number of combinations of 8 by 3.

Required probability:

.

Problem 19.

There are 15 students in the group, 8 of whom are excellent students. 6 students were called at random (according to the list). Find the probability that 4 of the students called will be excellent students.

Solution. The number of possible outcomes of the experiment here is equal to the number of combinations of 15 by 6, .

We consider a combination to be favorable if there are 4 excellent students and 2 are not. 4 excellent students can be selected from 8 excellent students in different ways, while the remaining 6-4 = 2 students (not excellent students) are selected from 15-8 = 7 students in different ways.

If to each four excellent students we add one of the pairs

students who are not excellent students, we will get “favorable” groups of 6 people. Their number is equal to m =.

Required probability:

Problem 20.

The first difficulty that Pascal overcame in his correspondence with the Chevalier de Marais was that of an accurate count of cases. It was about a game in which three dice are thrown, and one of the players bets that the total on the thrown sides will be more than 10, and the other - that it will be equal to or less than 10. It is easy to see that the chances of both players are equal. But the difficulty was this. Patient accounting is very large number games showed the Chevalier de Marais that those who bet more than 10 more often win with 11 than with 12 points. However, Mere argued, 11 points can be obtained in six different ways (6-4-1; 6-3-2; 5-5-1; 5-4-2; 5-3-3; 4-4-3), and 12 points can also be obtained in six ways (6-5-1; 6-4-2; 6-3-3; 5-5-2; 5-4-3; 4-4-4). Pascal's answer is very simple: the combination 6-4-1 is not prime, but sixfold, since if the dice are numbered, or if each of the three dice is colored differently so that they can be distinguished, the value 6 can be obtained on each of the three dice , and the value 4 is on each of the remaining two, which already makes six combinations. In contrast, a combination such as 5-5-1 can only be achieved in three different ways, and a combination like 4-4-4 can only be achieved in a single way.

Therefore, if you want to know real number in various ways get 11 or get 12 points, then for each of these cases you need to add up the sum of those six numbers that correspond to the combinations,

whereas for the case of 12 points we have

From this we conclude that on average we get 11 points 27 times, while we get 12 points 25 times, and this result agrees perfectly with the observations of the Chevalier de Mere.

Example 4. While dialing a phone number, the subscriber forgot one digit and dialed it at random. Find the probability that the correct number is dialed.

Solution. Let us denote by A event – ​​the required number has been dialed. The subscriber could dial any of 10 digits. Therefore, the total number of possible elementary outcomes is 10. These outcomes are equally possible (the number is typed at random) and form full group(at least one digit will definitely be dialed), that is. There is only one number needed. Therefore for the event A A .

Example 5. While dialing a phone number, the subscriber forgot the last two digits and, remembering only that they were different, dialed them at random. Find the probability that the numbers you need.

Solution. Let us denote by IN event – ​​two required numbers have been dialed. There are only so many pairs you can collect different numbers, how many placements can be made of 10 digits by 2, that is . Therefore, the total number of equally possible elementary outcomes is . There is only one combination of two numbers needed. Therefore for the event A only one outcome is favorable. The required probability is equal to the ratio of the number of outcomes favorable to the event A to the number of all elementary outcomes: .

Example 6. In a batch of 10 parts there are 7 standard ones. Find the probability that among six parts taken at random, there are exactly 4 standard ones.

Solution. Let the event A– among the 6 parts taken, exactly 4 are standard. The total number of possible elementary test outcomes is equal to the number of ways in which 6 parts can be extracted from 10, that is, the number of combinations of 10 elements of 6 (). Let's count the number of outcomes favorable to the event A: 4 standard parts can be taken from 7 standard parts in different ways. In this case, the remaining 6-4=2 parts must be non-standard. They can be taken from 10-7=3 non-standard parts in ways. Therefore, the number of favorable outcomes is . The required probability is equal to the ratio of the number of outcomes favorable to the event A, to the number of all elementary outcomes.

While dialing a phone number, the subscriber forgot the last two digits and, remembering only that these digits were different, dialed them at random. Find the probability that the required numbers are dialed.

Task No. 2

Given a differential function of continuous random variable X:

Find integral function F(x)

Task No. 3

There are 3 white and 3 black balls in the urn. One ball at a time is taken out of the urn twice without replacing them. Find the probability of occurrence white ball on the second trial (event B) if the black ball was drawn on the first trial (event A).

Task No. 4

There are 3 boxes containing 10 parts each. The first box contains 8, the second 7 and the third 9 standard parts. One part is taken out at random from each box. Find the probability that all three of these removed parts will turn out to be standard.

Task No. 5
The probability of hitting the target when firing from three guns is as follows: = 0,8; = 0,7; = 0.9. Find the probability of at least one hit (event A) with one salvo from all guns.

Task No. 6

There are two sets of parts. The probability that the part of the first set is standard is 0.8, and the second is 0.9. Find the probability that a part taken at random (from a set taken at random) is standard.

Task No. 7

To participate in student qualifying sports competitions, 4 students were allocated from the first group of the course, 6 from the second, and 5 from the third group. The probabilities that a student of the first, second and third groups will be included in the institute’s team are respectively equal to 0.9; 0.7 and 0.8. A randomly selected student ended up in the national team as a result of the competition. Which group did this student most likely belong to?

Task No. 8

The probability that electricity consumption during one day will not exceed the established norm is 0.75. Find the probability that in the next 6 days, electricity consumption for 4 days will not exceed the norm.

Task No. 9

Find the probability that event A will occur exactly 80 times in 400 trials if the probability of this event occurring in each trial is 0.2.

Task No. 10

The probability of a shooter hitting a target with one shot is 0.75. Find the probability that with 100 shots the target will be hit: a) no less than 70 and no more than 80 times; b) no more than 70 times.

Task No. 11

A merchandiser examines 24 samples of goods. The probability that each of the samples will be considered fit for sale is 0.6. Find the most likely number of samples that a merchandiser recognizes as fit for sale.

Probability of an event occurring in each of 400 independent tests equal to 0.8. Find one positive number E, so that with probability 0.9876 absolute value the deviation of the relative frequency of occurrence of an event from its probability of 0.8 did not exceed E.

Task No. 13

The coin is tossed 5 times. Find the probability that the “coat of arms” will appear:

a) less than two times;

b) at least twice.

Task No. 14

The first urn contains 10 balls, 8 of which are white; The second urn contains 20 balls, 4 of which are white. One ball is drawn at random from each urn, and then one ball is drawn at random from these two balls. Find the probability that the white ball is drawn.

Task No. 15

How many independent trials must be performed with a probability of occurrence of an event in each trial equal to 0.4 so that the most probable number of occurrence of an event in these trials is equal to 25?

Task No. 16

The discrete random variable X is specified by the distribution law.

Find: variance D(X), mean standard deviation(X) and construct a distribution polygon.

Task No. 17

The textbook was published in a circulation of 100,000 copies. The probability that the textbook is bound incorrectly is 0.0001. Find the probability that the circulation contains exactly 5 defective books.

Task No. 18

A list of possible values ​​of a discrete random variable X is given: and also known mathematical expectations this quantity and its squares:

M(X)=2.3 and M(X )=5,9.

Find the probabilities corresponding possible values X.

Task No. 19

The random variable X is specified by the integral function

Find the probability that, as a result of the test, the value of X will take a value contained in the interval (-1;1)

Task No. 20
A discrete random variable is specified by a distribution law

Find the integral function and plot its graph.

Task No. 21

Continuous random variable X is given differential function
in the interval (0; π/3); outside this interval f(x)=0. Find the probability that X will take a value belonging to the interval (
)

Task No. 22

The discrete random variable X is specified by the distribution law:

Find central points first, second, third and fourth orders

Task No. 23

Write a binomial law for the distribution of a discrete random variable X - the number of occurrences of an even number of points on two dice.

Task No. 24

Find the variance and standard deviation of a discrete random variable X specified by the distribution law:

The probability of occurrence of event a in each trial is ½. Using Chebyshev's inequality, estimate the probability that the number X of occurrences of event A will be in the range from 40 to 60 if 100 independent trials are performed.

Task No. 27

Build a histogram relative frequencies By given distribution samples

The sample is specified as a frequency distribution

Task No. 29

For built interval series calculate confidence interval at γ=0.99 and t=2.861

Task No. 30

The sample is specified as a frequency distribution