Problems on the classical determination of probability.
Examples of solutions

In the third lesson we will look at various problems involving the direct application of the classical definition of probability. To effectively study the materials in this article, I recommend that you familiarize yourself with the basic concepts probability theory And basics of combinatorics. The task of classically determining probability with a probability tending to one will be present in your independent/control work on terver, so let’s get ready for serious work. You may ask, what's so serious about this? ...just one primitive formula. I warn you against frivolity - thematic tasks are quite diverse, and many of them can easily confuse you. In this regard, in addition to working through the main lesson, try to study additional tasks on the topic that are in the piggy bank ready-made solutions for higher mathematics. Solution techniques are solution techniques, but “friends” still “need to be known by sight,” because even a rich imagination is limited and there are also enough standard tasks. Well, I’ll try to sort out as many of them as possible in good quality.

Let's remember the classics of the genre:

The probability of an event occurring in a certain test is equal to the ratio , where:

– total number of all equally possible, elementary outcomes of this test, which form full group of events;

- quantity elementary outcomes favorable to the event.

And immediately an immediate pit stop. Do you understand the underlined terms? This means clear, not intuitive understanding. If not, then it’s still better to return to the 1st article on probability theory and only after that move on.

Please do not skip the first examples - in them I will repeat one fundamentally important point, and also tell you how to correctly format a solution and in what ways this can be done:

Problem 1

An urn contains 15 white, 5 red and 10 black balls. 1 ball is drawn at random, find the probability that it will be: a) white, b) red, c) black.

Solution: The most important prerequisite for using the classical definition of probability is ability to count the total number of outcomes.

There are a total of 15 + 5 + 10 = 30 balls in the urn, and obviously the following facts are true:

– retrieving any ball is equally possible (equal opportunity outcomes), while the outcomes elementary and form full group of events (i.e., as a result of the test, one of the 30 balls will definitely be removed).

Thus, the total number of outcomes:

Consider the event: – a white ball will be drawn from the urn. This event is favored elementary outcomes, therefore, according to the classical definition:
– the probability that a white ball will be drawn from the urn.

Oddly enough, even in such a simple task one can make a serious inaccuracy, which I already focused on in the first article on probability theory. Where is the pitfall here? It is incorrect to argue here that “since half the balls are white, then the probability of drawing a white ball» . The classic definition of probability refers to ELEMENTARY outcomes, and the fraction must be written down!

With other points, similarly, consider the following events:

– a red ball will be drawn from the urn;
– a black ball will be drawn from the urn.

An event is favored by 5 elementary outcomes, and an event is favored by 10 elementary outcomes. So the corresponding probabilities are:

A typical check of many server tasks is carried out using theorems on the sum of probabilities of events forming a complete group. In our case, the events form a complete group, which means the sum of the corresponding probabilities must necessarily be equal to one: .

Let's check if this is true: that's what I wanted to make sure of.

Answer:

In principle, the answer can be written down in more detail, but personally, I’m used to putting only numbers there - for the reason that when you start “stamping out” problems in hundreds and thousands, you try to reduce the writing of the solution as much as possible. By the way, about brevity: in practice, the “high-speed” design option is common solutions:

Total: 15 + 5 + 10 = 30 balls in the urn. According to the classical definition:
– the probability that a white ball will be drawn from the urn;
– the probability that a red ball will be drawn from the urn;
– the probability that a black ball will be drawn from the urn.

Answer:

However, if there are several points in the condition, then it is often more convenient to formulate the solution in the first way, which takes a little more time, but at the same time “lays everything out on the shelves” and makes it easier to navigate the problem.

Let's warm up:

Problem 2

The store received 30 refrigerators, five of which have a manufacturing defect. One refrigerator is selected at random. What is the probability that it will be without a defect?

Select the appropriate design option and check the sample at the bottom of the page.

In the simplest examples, the number of common and the number of favorable outcomes lie on the surface, but in most cases you have to dig up the potatoes yourself. A canonical series of problems about a forgetful subscriber:

Problem 3

When dialing a phone number, the subscriber forgot the last two digits, but remembers that one of them is zero and the other is odd. Find the probability that he will dial the correct number.

Note : zero is an even number (divisible by 2 without a remainder)

Solution: First we find the total number of outcomes. By condition, the subscriber remembers that one of the digits is zero, and the other digit is odd. Here it is more rational not to be tricky with combinatorics and use method of direct listing of outcomes . That is, when making a solution, we simply write down all the combinations:
01, 03, 05, 07, 09
10, 30, 50, 70, 90

And we count them - in total: 10 outcomes.

There is only one favorable outcome: the correct number.

According to the classical definition:
– probability that the subscriber will dial the correct number

Answer: 0,1

Decimal fractions look quite appropriate in probability theory, but you can also adhere to the traditional Vyshmatov style, operating only with ordinary fractions.

Advanced task for independent solution:

Problem 4

The subscriber has forgotten the PIN code for his SIM card, but remembers that it contains three “fives”, and one of the numbers is either a “seven” or an “eight”. What is the probability of successful authorization on the first try?

Here you can also develop the idea of ​​​​the likelihood that the subscriber will face punishment in the form of a puk code, but, unfortunately, the reasoning will go beyond the scope of this lesson

The solution and answer are below.

Sometimes listing combinations turns out to be a very painstaking task. In particular, this is the case in the next, no less popular group of problems, where 2 dice are rolled (less often - larger quantities):

Problem 5

Find the probability that when throwing two dice the total number will be:

a) five points;
b) no more than four points;
c) from 3 to 9 points inclusive.

Solution: find the total number of outcomes:

Ways the side of the 1st die can fall out And in different ways the side of the 2nd cube can fall out; By rule for multiplying combinations, Total: possible combinations. In other words, each the face of the 1st cube can be ordered a couple with each the edge of the 2nd cube. Let us agree to write such a pair in the form , where is the number rolled on the 1st die, is the number rolled on the 2nd die. For example:

– the first dice scored 3 points, the second dice scored 5 points, total points: 3 + 5 = 8;
– the first dice scored 6 points, the second dice scored 1 point, total points: 6 + 1 = 7;
– 2 points rolled on both dice, sum: 2 + 2 = 4.

Obviously, the smallest amount is given by a pair, and the largest by two “sixes”.

a) Consider the event: – when throwing two dice, 5 points will appear. Let's write down and count the number of outcomes that favor this event:

Total: 4 favorable outcomes. According to the classical definition:
– the desired probability.

b) Consider the event: – no more than 4 points will be rolled. That is, either 2, or 3, or 4 points. Again we list and count the favorable combinations, on the left I will write down the total number of points, and after the colon - the suitable pairs:

Total: 6 favorable combinations. Thus:
– the probability that no more than 4 points will be rolled.

c) Consider the event: – 3 to 9 points will roll, inclusive. Here you can take the straight road, but... for some reason you don’t want to. Yes, some pairs have already been listed in the previous paragraphs, but there is still a lot of work to be done.

What's the best way to proceed? In such cases, a roundabout path turns out to be rational. Let's consider opposite event: – 2 or 10 or 11 or 12 points will be rolled.

What's the point? The opposite event is favored by a significantly smaller number of couples:

Total: 7 favorable outcomes.

According to the classical definition:
– the probability that you will roll less than three or more than 9 points.

In addition to direct listing and counting of outcomes, various combinatorial formulas. And again an epic problem about the elevator:

Problem 7

3 people entered the elevator of a 20-story building on the first floor. And let's go. Find the probability that:

a) they will exit on different floors
b) two will exit on the same floor;
c) everyone will get off on the same floor.

Our exciting lesson has come to an end, and finally, I once again strongly recommend that if not solve, then at least figure out additional problems on the classical determination of probability. As I already noted, “hand padding” matters too!

Further along the course - Geometric definition of probability And Probability addition and multiplication theorems and... luck in the main thing!

Solutions and Answers:

Task 2: Solution: 30 – 5 = 25 refrigerators have no defect.

– the probability that a randomly selected refrigerator does not have a defect.
Answer :

Task 4: Solution: find the total number of outcomes:
ways you can select the place where the dubious number is located and on every Of these 4 places, 2 digits (seven or eight) can be located. According to the rule of multiplication of combinations, the total number of outcomes: .
Alternatively, the solution can simply list all the outcomes (fortunately there are few of them):
7555, 8555, 5755, 5855, 5575, 5585, 5557, 5558
There is only one favorable outcome (correct pin code).
Thus, according to the classical definition:
– probability that the subscriber logs in on the 1st attempt
Answer :

Task 6: Solution: find the total number of outcomes:
numbers on 2 dice can appear in different ways.

a) Consider the event: – when throwing two dice, the product of the points will be equal to seven. There are no favorable outcomes for a given event, according to the classical definition of probability:
, i.e. this event is impossible.

b) Consider the event: – when throwing two dice, the product of the points will be at least 20. The following outcomes are favorable for this event:

Total: 8
According to the classical definition:
– the desired probability.

c) Consider the opposite events:
– the product of points will be even;
– the product of points will be odd.
Let's list all the outcomes favorable to the event:

Total: 9 favorable outcomes.
According to the classical definition of probability:
Opposite events form a complete group, therefore:
– the desired probability.

Answer :

Problem 8: Solution: let's calculate the total number of outcomes: 10 coins can fall in different ways.
Another way: ways the 1st coin can fall And ways the 2nd coin can fall And … And ways the 10th coin can fall. According to the rule of multiplying combinations, 10 coins can fall ways.
a) Consider the event: – heads will appear on all coins. This event is favored by a single outcome, according to the classical definition of probability: .
b) Consider the event: – 9 coins will land heads, and one coin will land tails.
There are coins that can land on heads. According to the classical definition of probability: .
c) Consider the event: – heads will appear on half of the coins.
Exists unique combinations of five coins that can land heads. According to the classical definition of probability:
Answer :

Combinatorics studies ways to count the number of elements in finite sets. Combinatorics formulas are used to directly calculate probabilities.
Sets of elements consisting of the same different elements and differing from each other only in their order are called permutations these elements. The number of possible permutations from n elements are denoted by , and this number is equal to n! (read "en-factorial"):
\(P_n=n\) (1.3.1)
Where
. (1.3.2)

Remark 1. For the empty set, the convention is accepted: the empty set can be ordered in only one way; by definition believe.

Placements are called sets made up of n various elements according to m elements that differ either in the composition of the elements or in their order. The number of all possible placements is determined by the formula
. (1.3.3)

Combinations from n various elements according to m are called sets containing m elements from among n given, and which differ in at least one element. Number of combinations of n elements by m stand for: or . This number is expressed by the formula

. (1.3.4)

Remark 2. By definition, assume .

For the number of combinations the equalities are valid:
, , (1.3.5)
. (1.3.6)

The last equality is sometimes formulated as the following theorem about finite sets:
The number of all subsets of a set consisting of them n elements, equals .
Note that the numbers of permutations, placements and combinations are related by the equality

Remark 3. It was assumed above that all n elements are different. If some elements are repeated, then in this case sets with repetitions are calculated using other formulas.

For example, if among n elements are elements of one type, elements of another type, etc., then the number of permutations with repetitions is determined by the formula
(1.3.7)
Where .

Number of placements by m elements with repetitions from n elements is equal
, that is
with repeat (1.3.8)
Number of combinations with repetitions from n elements by m elements is equal to the number of combinations without repetitions from n + m- 1 elements each m elements, that is
from repeat (1.3.9)

When solving combinatorics problems, the following rules are used.

Sum rule. If some object A can be selected from a set of objects in m ways, and another object B can be selected in n ways, then either A or B can be selected in m + n ways.

Product rule. If object A can be selected from a variety of objects m methods and after each such choice, object B can be selected n ways, then a pair of objects (A, B) in the specified order can be selected in ways.

The classical scheme for calculating probabilities is suitable for solving a number of purely practical problems. Let's consider, for example, a certain set of elements of volume N. These can be products, each of which is suitable or defective, or seeds, each of which can be viable or not. Situations of this kind are described by an urn scheme: there are N balls in the urn, of which M are blue and (N - M) are red.

From an urn containing N balls, in which there are M blue balls, n balls are drawn. You need to determine the probability that m blue balls will be found in a sample of size n. Let us denote by A the event “there are m blue balls in a sample of size n”, then
(1.3.10)

Example 1. In how many different ways can three persons be selected for three different positions out of ten candidates?

Solution. Let's use formula (1.3.3). For n = 10, m = 3 we get
.

Example 2. In how many different ways can 5 people fit on a bench?

Solution. According to formula (1.3.1) with n=5 we find
P 5 =5!=1·2·3·4·5=120.

Example 3. In how many ways can three persons be selected for three identical positions out of ten candidates?

Solution. In accordance with formula (1.3.4) we find

Example 4. How many different six-digit numbers can be written using the digits 1; 1; 1; 2; 2; 2?

Solution. Here you need to find the number of permutations with repetitions, which is determined by formula (1.3.7). With k = 2, n 1 = 3, n 2 = 3, n = 6, using this formula we obtain

Example 5. How many different permutations of letters can be made in the words: castle, rotor, axe, bell?

Solution. In the word castle, all the letters are different, there are five in total. In accordance with formula (1.3.1) we obtain P 5 = 5! = 1·2·3·4·5 = 120. In a word rotor, consisting of five letters, letters p And o are repeated twice. To calculate various permutations, we use formula (1.3.7). For n = 5, n 1 = 2, n 2 = 2, using this formula we find

The letter in the word ax O is repeated twice, so

In the seven letter word bell, the letter To appears twice, letter O- three times, letter l- twice. In accordance with formula (13.7) with n = 7, n 1 = 2, n 2 = 3, n з = 2 we obtain

Example 6. The letters I, K, M, N, S are written on five identical cards. The cards are shuffled and randomly placed in a row. What is the probability that the word MINSK will appear?

Solution. From five different elements you can create P5 permutations:
. This means that there will be a total of 120 possible outcomes, but only one favorable for a given event. Hence,

Example 7. From the letters of the word rotor, composed using a split alphabet, 3 letters are randomly selected sequentially and placed in a row. What is the probability that the word will come out torus?

Solution. To distinguish identical letters from each other, we provide them with numbers: p 1 , p 2 , 0 1 , 0 2. The total number of elementary outcomes is equal to: . Word rotor will work in cases ( then 1 r 1, then 1 r 2, then 2 r 1, then 2 r 2). The required probability is equal to

When calculating the number of favorable cases, we used the product rule: the letter m you can select one way, letter O- two, a letter R- two ways.

Example 8. The letters of the word are written on six cards of the same shape and size. talent- one letter on each card. The cards are thoroughly mixed. they are taken out at random and placed on the table one after the other. What is the probability of getting the floor again? talent?

Solution. Let's number the cards with letters:

The word talent (513246) will not change if the letters A rearrange, but according to the arrangement of the cards you will get a different combination: talent (523146). If in each of these two combinations we do the same with the letter t, we will get 2 more different combinations of cards with the word talent. This means that the appearance of the word talent 4 elementary outcomes are favorable. The total number of possible elementary outcomes is equal to the number of permutations of 6 elements: n = 6! = 720. Therefore, the required probability

.

Remark: This probability can also be found using formula (1.3.7), which for n = 6, n 1 = 1, n 2 = 1, n 3 = 2, n 4 = 2 takes view:

. Thus, P = 1/180.

Example 9. Letters are written on five identical cards: on two cards l, on the other three And. These cards are placed at random in
row. What is the probability that this will produce the word lilies?

Solution. Let's find the number of permutations of these five letters with repetitions.
Using formula (1.3.7) for n = 5, n 1 = 2, n 2 = 3 we obtain

This is the total number of equally possible outcomes of the experiment; this event A - “the appearance of the word lily” is favored by one. In accordance with formula (1.2.1) we obtain

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

Solution. The total number of possible Ix 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 elements each ().

We determine the number of outcomes favorable to event A - “among 6 taken parts, 4 are standard.” Four standard parts out of seven standard ones can be taken in different ways, while the remaining 6 - 4 = 2 parts must be non-standard; There are ways to take 2 non-standard parts out of 10 - 7 = 3 non-standard parts. Therefore, the number of favorable outcomes is equal to .

The required probability is equal to the ratio of the number of outcomes favorable to the event to the number of all elementary outcomes:

Remark. The last formula is a special case of formula (1.3.10): N= 10, M= 7, n = 6, m = 4.

Example 11. Among 25 students in a group of 10 girls, 5 tickets are drawn. Find the probability that there will be 2 girls among the ticket holders.

Solution. The number of all equally possible cases of distributing 5 tickets among 25 students is equal to the number of combinations of 25 elements of 5, that is. The number of groups of three boys out of 15 who can receive tickets is . Each such triplet can be combined with any pair of ten girls, and the number of such pairs is equal to . Consequently, the number of groups of 5 students formed from a group of 25 students, each of which will include three boys and two girls, is equal to the product. This product is equal to the number of favorable cases of distributing five tickets among the students of the group so that three tickets go to boys and two tickets to girls. In accordance with formula (1.2.1) we find the required probability

Remark. The last formula is a special case of formula (1.3.10): N= 25, M= 15, n = 5, m = 3.

Example 12. A box contains 15 red, 9 blue and 6 green balls. 6 balls are drawn at random. What is the probability that 1 green, 2 blue and 3 red balls are drawn (event A)?

Solution. There are only 30 balls in the box. For this experiment, the number of all equally possible elementary outcomes will be . Let's count the number of elementary outcomes favorable to event A. Three red balls out of 15 can be chosen in ways, two blue balls out of 9 can be chosen in ways, one green out of 6 can be chosen in ways
The number of favorable outcomes is equal to the product

The required probability is determined by formula (1.3.10):

Example 14. The dice are tossed 10 times. What is the probability that sides 1, 2, 3, 4, 5, 6 will appear 2, 3, 1, 1, 1, 2 times, respectively (event A)?

Solution. We calculate the number of outcomes favorable for event A using formula (1.3.7):
The number of all elementary outcomes in this experiment is n = 6 10, therefore

Tasks
1. The letters B, E, R, S, T are written on 5 identical cards. These cards are randomly placed in a row. What is the probability that the word BREST will appear?
2. There are 4 blue and 5 red balls in a box. 2 balls are drawn at random from the box. Find the probability that these balls are different colors.
3. There are 4 women and 3 men in the team. 4 tickets to the theater are being raffled off among brigade members. What is the probability that among the ticket holders there will be 2 women and 2 men?
4. There are 10 balls in a box, of which 2 are white, 3 are red and 5 are blue. 3 balls are drawn at random. Find the probability that all 3 balls are different colors.
5. The letters l, m, o, o, t are written on five identical cards. What is the probability that, taking out the cards one at a time, we will get the word hammer in the order they appeared?
6. From a batch containing 10 products, of which 3 are defective, 3 products are selected at random. Find the probability that one product in the resulting sample is defective.
7. Out of ten tickets, two are winning. What is the probability that among five tickets chosen at random, one is a winner?

Answers
1. 1/120. 2. 5/9. 3. 18/35. 4 . 0,25. 5 . 1/60. 6 . 21/40. 7 . 5/9.

Questions
1. What are called permutations?
2. What form is used to calculate the number of permutations of n different elements?
3. What are placements called?
4. What formula is used to calculate the number of placements of n different elements by m elements?
5. What are combinations called?
6. What formula do you use to calculate the number of combinations of n elements of m elements?
7. What equality relates the numbers of permutations, placements and combinations?
8. What formula is used to calculate the number of permutations of n elements if some elements are repeated?
9. What formula determines the number of placements of m elements with repetitions of n elements?
10. What formula determines the number of combinations with repetitions of n elements of m elements?

Tough teacher, urgently need to solve problems on probability theory in 1 day, topic "Probability Theory (Mathematics)"

1. The telephone number consists of six digits. Find the probability that all numbers are different. 2. There are 10 products in the batch, four of which are non-standard. Four items are taken at random. Find the probability that among the taken products there are more standard ones than non-standard ones. 3. Ten people randomly sit on a ten-seat bench. Find the probability that 2 specific persons will be nearby. 4. A point is selected at random inside a square with vertices. Find the probability of the following event: 5. Two shooters independently fired one shot at the target. It is known that the probability of hitting the target for one of the shooters is 0.6; and for the other - 0.7. Find the probability that at least one of the shooters will miss the target. 6. Before passing the first round of the competition, each applicant is given three tasks: a text for artistic reading, a theme for presentation in pantomime, a poem for vocal performance to his own melody. When passing the competition, it is proposed to perform two numbers out of three. The selection of numbers is random. The competitor estimates that he will pass the first round in literary reading with a probability of 0.9; when performing pantomime – 0.3; when performing a vocal task – 0.5. What is the probability of passing the first round for a contestant with such preparation? 7. The first urn contains 10 balls, 8 of which are white; The second urn contains 15 balls, 4 of which are white. Two balls were drawn at random from the first urn, and then a ball from the second urn was transferred into it. After this, a ball was drawn from the first urn. Find the probability that this ball is white. 8. Out of 18 shooters, 5 hit the target with a probability of 0.6; 7 – with probability 0.7; 4 – with probability 0.8; 2 – with probability 0.5. The randomly selected shooter missed the target. Which group does this shooter most likely belong to? 9. The probability of hitting the target with one shot is 0.7. Find the probability that with 20 independent shots the target will be hit no more than 14 times. 10. There are 5 coins in your pocket, approximately the same to the touch: three - 2 rubles each and two - 10 rubles each. Without looking, they pull out 2 coins. A random variable is the total number of rubles extracted. For a random variable: a) construct a distribution series, b) find the mathematical expectation and variance, c) find the probability of the event (at least 4, but not more than 12 rubles were extracted). 11. A technician called to your home can appear at any time from 10 a.m. to 6 p.m. The client, having waited until 14 o'clock, left for 1 hour. Considering the arrival time of the master to be a random variable distributed uniformly, find the probability density, the distribution function. Determine the probability that the master (his arrival is obligatory) will not find the client at home? Construct probability density graphs and distribution functions.

1. The telephone number consists of six digits. Find the probability that all numbers are different. 2. There are 10 products in the batch, four of which are non-standard. Four items are taken at random. Find the probability that among the taken products there are more standard ones than non-standard ones. 3. Ten people randomly sit on a ten-seat bench. Find the probability that 2 specific persons will be nearby. More details

§ 7. Application of combinatorics to calculating probability

If from the total volume n sampling is made k elements with return, then the probability of obtaining each specific sample is considered equal to .

If the sample is made without returning, then this probability is equal to .

Let the occurrence of event A consist in the appearance of a sample with some additional restrictions and the number of such samples is equal to m. Then in the case of sampling with return we have:

in case of sampling without return:

Example 1. A three-digit number is chosen at random without a zero in its decimal notation. What is the probability that the selected number has exactly two identical digits?

Solution. Let's imagine that the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written on 9 identical cards and these cards are placed in an urn. Choosing a three-digit number at random is equivalent to sequentially drawing, returning 3 cards from the urn and writing down the numbers in the order they appear. Consequently, the number of all elementary outcomes of the experiment is 93 = 729. The number of favorable cases for the event A of interest to us is calculated as follows: 2 different numbers x and y can be chosen in different ways; if x and y are selected, then https://pandia.ru/text/78/365/images/image007_10.gif" width="115 height=41" height="41"> can be made from them.

Example 2. From the letters of the word “rotor”, composed using a split alphabet, 3 letters are sequentially extracted at random and put in a row. What is the probability that the word "tor" will come out?

Solution. To distinguish identical letters from each other, we provide them with numbers: p1, p2, o1, o2. Then the total number of elementary outcomes is equal to: . The word “torus” will appear in 1 × 2 ×2 = 4 cases (to1р1, then1р2, then2р1, then2р2)..gif" width="24" height="25 src="> and we assume that they all have equal probabilities .

Example 3. In a batch of N parts there are n defective ones. What is the probability that among k parts selected at random there will be s defective ones?

Solution. The number of all elementary outcomes is equal to . To calculate the number of favorable cases, we reason as follows: from n defective parts one can select s parts in s ways, and from N - n non-defective parts one can select k – s non-defective parts in ways; According to the product rule, the number of favorable cases is equal to ×. The required probability is:

.

Example 4. There are 4 women and 3 men in the team. 4 tickets to the theater are being raffled off among brigade members. What is the probability that among the ticket holders there will be 2 women and 2 men?

Solution. Let us apply a statistical selection scheme. Out of 7 team members, 4 people can be selected = 35 ways, therefore, the number of all elementary outcomes of the test is 35..gif" width="28" height="34">= 3 ways. Then the number of favorable cases will be equal to 6 × 3 = 18..gif" width="21" height="41"> . How many white balls are there in the urn?

150. There are n white and m black balls in an urn. K balls (k>m) are drawn at random. What is the probability that there are only white balls left in the urn?

151. From an urn containing N balls, one ball is removed N times, each time the removed ball is returned. What is the probability that all the balls are drawn once?

152. A full deck of cards (52 sheets) is divided at random into 2 equal parts (26 cards each). Find the probabilities of the following events:

A – there will be 2 aces in each part;

B – in one of the parts there will not be a single ace;

C – one of the parts will have exactly one ace.

153. An urn contains a white, b black and c red balls. All the balls are taken out of this urn one by one without returning and their colors are recorded. Find the probability that white appears before black in this list.

154. There are 2 urns: the first contains a white and b black balls; the second with white and d black. A ball is drawn from each urn. Find the probability that both balls will be white (event A) and the probability that the balls will be different colors (event B).

155. 2n teams are divided into 2 subgroups of n teams. Find the probability that the 2 strongest teams will end up: a) in different subgroups (event A); b) into one subgroup (event B).

156. 3 cards are drawn at random from a deck of 36 cards. Determine the probability that the sum of points in these cards is 21 if the jack is 2 points, the queen is 3, the king is 4, the ace is 11, and the remaining cards are 6, 7, 8, 9, 10 points, respectively.

157. The owner of one Sportloto lottery card (6 out of 49) crosses out 6 numbers. What is the probability that they will guess:

a) all 6 numbers in the next circulation;

b) 5 or 6 numbers;

c) at least 3 numbers?

158. A bus with 15 passengers has to make 20 stops. Assuming that all possible ways of distributing passengers among stops are equally possible, find the probability that no 2 passengers get off at the same stop.

159. From the numbers 1, 2, …, N, r different numbers (r £ N) are chosen at random. Find the probability that r consecutive numbers will be selected.

160. Several cards are drawn from a full deck of cards (52 sheets). How many cards must be drawn in order to assert with a probability greater than 0.5 that among them there will be cards of the same suit?

161. There are n balls that are randomly scattered over m holes. Find the probability that exactly k1 balls will fall into the first hole, k2 balls into the second, etc., and km of balls into the m-th hole, if k1+k2+…+km=n.

162. In the conditions of the previous problem, find the probability that in one of the holes (it doesn’t matter which one) there will be k1 balls, and in the other - k2 balls, etc., in the m-th - km balls (numbers k1, k2, ... ,km are assumed to be different).

163. From the set (1, 2,…, N), the numbers x1 and x2 are selected sequentially without returning. Find p(x2 > x1).

1 manuscripts are divided into 30 folders (one manuscript occupies 3 folders). Find the probability that 6 randomly discarded folders do not contain a single manuscript in its entirety.

165. What is the probability that in a company of r people at least two will have the same birthday? (For simplicity, it is assumed that February 29 is not a birthday).

166. Using a table of lg n values! and the condition of the previous problem, calculate the probabilities for r = 22, 23, 60.

167.You set out to find a person whose birthday coincides with yours. How many strangers would you have to interview so that the probability of meeting such a person would be no less than 0.5?

168. For the State Loan, 6 main draws and one additional draw are drawn annually, taking place after the main fifth one. Out of 100,000 episodes, 170 episodes win in each main run, and 230 episodes in each additional run. Find the probability of winning one bond in the first 10 years: a) in the main circulation; b) in an additional edition; c) in any edition.

1. A full deck of cards (52 sheets) is divided at random into 2 equal parts (26 cards each). Find the probabilities of the following events: A – there will be 2 aces in each part; B – in one of the parts there will not be a single ace; C – one of the parts will have exactly one ace.

2. 5 military personnel are selected at random from a group of 4 officers and 12 soldiers. What is the probability that there will be no more than two officers in the group?

3. Find the probability that a participant in the “Sportloto 6 out of 45” lottery, who bought one ticket, will correctly guess: a) 2 numbers, b) 6 numbers.

4. Three people are randomly placed in 8 train cars. What is the probability that all of them: a) will enter the same carriage, b) will enter carriage No. 3, c) will be placed in different carriages?

5. Among a batch of 50 products, there are 5 defective ones. In order to control this batch, 5 products are selected. If among them there is more than one defective, then the entire batch of products is rejected. What is the probability that a batch of products will be rejected?

6. Of the 20 laboratory employees, 5 people must go on a business trip. What is the probability that among the seconded employees there will not be 3 laboratory managers (the head, his deputy and the chief engineer)?

7. 12 students are randomly seated in the first 12 seats of one row of the stalls. What is the probability that students M and N will sit next to each other?

8. The post office sells 6 types of postcards. The buyer purchased 4 postcards. Find the probability that these postcards are: a) of the same type; b) of various types.

9. From a group consisting of 7 men and 4 women, 5 people must be selected. What is the probability that among these selected people there will be at least three women.

10. There are 10 light bulbs in the box, 3 of which are burnt out. Find the probability that out of 5 light bulbs taken at random from a box, 2 light bulbs will be on.

11. There are 15 students in the group. Of these, 12 are girls, the rest are boys. It is known that two students must be called to the board. What is the probability that among them there will be: a) one girl and one boy; b) two girls?

12. There are 10 carriages of different products at the station. The cars are marked with numbers from 1 to 10. Find the probability that among the 5 cars selected for control opening there will be cars with numbers 2 and 5?

13. 20 boxes of components for one type of computer were brought to the warehouse, but among them there were 4 boxes of components for another type of computer. We took 6 boxes at random. What is the probability that among 6 boxes there will be: a) one box of incomplete parts; b) at least one box of incomplete parts?

14. Of 20 joint stock companies, 4 are bankrupt. The citizen purchased one share each of six joint-stock companies. What is the probability that among the purchased shares 2 will be bankrupt shares?

15. There are 5 blue, 4 red and 3 green pencils in a box. 3 pencils are taken out at random. What is the probability that: a) they are all the same color, b) they are all different colors, c) among them there are 2 red and 1 green pencil.

16. There are 8 new and 10 used cars at the rental point. We rented three cars at random. What is the probability that all the cars rented are: a) all new; b) 1 new and 2 used?

17. The letters A, A, I, M, L, N are written on separate cards. Find the probability that, choosing cards at random one after another: a) the word “MINA” will be obtained; b) “RASPBERRY”; c) "BURBT".

18. In an envelope among 100 photographs there is one wanted one. 10 cards are drawn at random from the envelope. Find the probability that the right one will be among them?

19. There are 10 televisions in the store, 4 of which are defective. The batch is randomly divided into 2 equal parts, which are sent to two consumers. What is the probability that defective products will go equally to two consumers?

20. In a group of 20 students, 9 are low achievers. Two people are chosen at random from the group. What is the probability that among them: a) there is only one low-performing student; b) at least one underperforming student?

21. There are 7 radio tubes, among which 3 are faulty, seemingly no different from the working ones. Two lamps are selected at random. What is the probability that: a) both lamps will be in good working order; b) one is working; c) is at least one working?

22. There are 20 buses in the fleet of two brands: 12 and 8, respectively. The probability of each brand of bus going on an excursion is the same. What is the probability that after 18 buses go on an excursion, the following buses remain in the fleet: a) of the first brand; b) one brand; c) different brands?

23. A bus with 15 passengers has to make 20 stops. Assuming that all possible ways of distributing passengers among stops are equally possible, find the probability that no 2 passengers get off at the same stop.

24. There are 12 students in the group, including 3 excellent students. 9 students were selected at random from the list. Find the probability that among the selected students: a) 3 excellent students; b) at least 3 excellent students.

25. There are 5 identical products in a box, and 3 of them are painted. 2 items were removed at random. Find the probability that among two extracted products there will be: a) one painted product; b) two painted products; c) at least one painted product.

26. There are 15 textbooks arranged in random order on a library shelf, 5 of which are bound. The librarian takes 3 textbooks at random. Find the probability that the binding will contain: a) at least one of the taken textbooks; b) 2 textbooks will not be bound.

27. 5 people sit randomly on a five-seater bench. What is the probability that 3 specific people will be nearby?.

28. The mechanism includes two identical parts. The mechanism will not work if both parts supplied are undersized. The assembler has 10 parts, 3 of which are less than the standard. Determine the probability that the mechanism will work normally if the assembler picks two parts at random.

29. A flower shop sells 8 asparagus and 5 geraniums. What is the probability that among 5 plants sold: a) 2 asparagus; b) all geraniums?

30. 8 chess players, including 3 grandmasters, are divided into 2 teams of 4 people by drawing lots. What is the probability that: a) two grandmasters will be in one team, and another one will be in another; will all 3 grandmasters be on the same team?