The classic definition of probability will solve the exam. Probability theory on the exam in mathematics

The probability of an event $A$ is the ratio of the number of outcomes favorable for $A$ to the number of all equally possible outcomes

$P(A)=(m)/(n)$, where $n$ is the total number of possible outcomes and $m$ is the number of outcomes favoring $A$.

The probability of an event is a number from the segment $$

Taxi company available $50$ cars. $35$ of them are black, the rest are yellow. Find the probability that a car will arrive at a random call yellow color.

Find the number of yellow cars:

In total, there are $50$ cars, that is, one out of fifty will come to the call. There are $15$ of yellow cars, therefore, the probability of arrival of a yellow car is $(15)/(50)=(3)/(10)=0.3$

Answer:$0.3$

Opposite events

Two events are said to be opposite if this test they are incompatible and one of them is bound to happen. The probabilities of opposite events add up to 1. An event opposite to the event $A$ is written $((A))↖(-)$.

$P(A)+P((A))↖(-)=1$

Independent events

Two events $A$ and $B$ are called independent if the probability of occurrence of each of them does not depend on whether the other event occurred or not. Otherwise, the events are called dependent.

The probability of the product of two independent events $A$ and $B$ is equal to the product of these probabilities:

$P(A B)=P(A) P(B)$

Ivan Ivanovich bought two different lottery tickets. The probability that the first one wins lottery ticket, is equal to $0.15$. The probability that the second lottery ticket will win is $0.12. Ivan Ivanovich participates in both draws. Assuming that the draws are held independently of each other, find the probability that Ivan Ivanovich wins in both draws.

Probability $P(A)$ - wins the first ticket.

Probability $P(B)$ - wins the second ticket.

Events $A$ and $B$ are independent events. That is, to find the probability that both events will occur, you need to find the product of the probabilities

$P(A B)=P(A) P(B)$

$P=0.15 0.12=0.018$

Answer: $0.018

Incompatible events

Two events $A$ and $B$ are said to be incompatible if there are no outcomes favoring both event $A$ and event $B$. (Events that cannot happen at the same time)

The probability of the sum of two incompatible events $A$ and $B$ is equal to the sum of the probabilities of these events:

$P(A+B)=P(A)+P(B)$

In the algebra exam, the student gets one question out of all the exams. The probability that this is a question on the topic " Quadratic equations", is equal to $0.3$. The probability that this is a question on the topic " Irrational equations", is equal to $0.18$. There are no questions related to these two topics at the same time. Find the probability that the student will get a question on one of these two topics on the exam.

These events are called incompatible, since the student will get a question EITHER on the topic “Quadricular Equations”, OR on the topic “Irrational Equations”. Topics cannot be caught at the same time. The probability of the sum of two incompatible events $A$ and $B$ is equal to the sum of the probabilities of these events:

$P(A+B)=P(A)+P(B)$

$P \u003d 0.3 + 0.18 \u003d 0.48 $

Answer: $0.48

Joint Events

Two events are said to be joint if the occurrence of one of them does not exclude the occurrence of the other in the same trial. Otherwise, the events are called incompatible.

The probability of the sum of two joint events $A$ and $B$ is equal to the sum of the probabilities of these events minus the probability of their product:

$P(A+B)=P(A)+P(B)-P(A B)$

There are two identical coffee machines in the lobby of the cinema. The probability that the machine will run out of coffee by the end of the day is $0.6$. The probability that both machines run out of coffee is $0.32$. Find the probability that at least one of the vending machines will run out of coffee by the end of the day.

Let's denote the events, let:

$A$ = coffee will end in the first machine,

$B$ = coffee will end in the second machine.

$A B =$ coffee will run out in both vending machines,

$A + B =$ coffee will run out in at least one vending machine.

By convention, $P(A) = P(B) = 0.6; P(A B) = $0.32.

The events $A$ and $B$ are joint, the probability of the sum of two joint events is equal to the sum of the probabilities of these events, reduced by the probability of their product:

$P(A + B) = P(A) + P(B) − P(A B) = 0.6 + 0.6 − 0.32 = 0.88$

Presented to date in the open bank of USE problems in mathematics (mathege.ru), the solution of which is based on only one formula, which is a classical definition of probability.

The easiest way to understand the formula is with examples.
Example 1 There are 9 red balls and 3 blue ones in the basket. The balls differ only in color. At random (without looking) we get one of them. What is the probability that the ball chosen in this way will be blue?

Comment. In probability problems, something happens (in this case, our action of pulling the ball) that can have different result- outcome. It should be noted that the result can be viewed in different ways. "We pulled out a ball" is also a result. "We pulled out the blue ball" is the result. "We drew this particular ball out of all possible balls" - this least generalized view of the result is called the elementary outcome. It is the elementary outcomes that are meant in the formula for calculating the probability.

Decision. Now we calculate the probability of choosing a blue ball.
Event A: "the chosen ball turned out to be blue"
Total number of all possible outcomes: 9+3=12 (number of all balls we could draw)
Number of outcomes favorable for event A: 3 (the number of such outcomes in which event A occurred - that is, the number of blue balls)
P(A)=3/12=1/4=0.25
Answer: 0.25

Let us calculate for the same problem the probability of choosing a red ball.
The total number of possible outcomes will remain the same, 12. The number of favorable outcomes: 9. The desired probability: 9/12=3/4=0.75

The probability of any event always lies between 0 and 1.
Sometimes in everyday speech (but not in probability theory!) The probability of events is estimated as a percentage. The transition between mathematical and conversational assessment is done by multiplying (or dividing) by 100%.
So,
In this case, the probability is zero for events that cannot happen - improbable. For example, in our example, this would be the probability of drawing a green ball from the basket. (The number of favorable outcomes is 0, P(A)=0/12=0 if counted according to the formula)
Probability 1 has events that will absolutely definitely happen, without options. For example, the probability that "the chosen ball will be either red or blue" is for our problem. (Number of favorable outcomes: 12, P(A)=12/12=1)

We've looked at a classic example that illustrates the definition of probability. All similar USE tasks according to probability theory are solved by applying this formula.
Instead of red and blue balls, there can be apples and pears, boys and girls, learned and unlearned tickets, tickets containing and not containing a question on a certain topic (prototypes , ), defective and high-quality bags or garden pumps (prototypes , ) - the principle remains the same.

Slightly differ in the formulation of the problem of the theory USE probabilities, where you need to calculate the probability of an event occurring on a specific day. ( , ) As in the previous tasks, you need to determine what is an elementary outcome, and then apply the same formula.

Example 2 The conference lasts three days. On the first and second days, 15 speakers each, on the third day - 20. What is the probability that the report of Professor M. will fall on the third day, if the order of the reports is determined by lottery?

What is the elementary outcome here? - Assigning a professor's report to one of all possible serial numbers for a speech. 15+15+20=50 people participate in the draw. Thus, Professor M.'s report can receive one of 50 numbers. This means that there are only 50 elementary outcomes.
What are the favorable outcomes? - Those in which it turns out that the professor will speak on the third day. That is, the last 20 numbers.
According to the formula, the probability P(A)= 20/50=2/5=4/10=0.4
Answer: 0.4

The drawing of lots here is the establishment of a random correspondence between people and ordered places. In Example 2, matching was considered in terms of which of the places a particular person could take. You can approach the same situation from the other side: which of the people with what probability could get to a particular place (prototypes , , , ):

Example 3 5 Germans, 8 Frenchmen and 3 Estonians participate in the draw. What is the probability that the first (/second/seventh/last - it doesn't matter) will be a Frenchman.

The number of elementary outcomes is the number of all possible people who could, by lot, get into given place. 5+8+3=16 people.
Favorable outcomes - the French. 8 people.
Desired probability: 8/16=1/2=0.5
Answer: 0.5

The prototype is slightly different. There are tasks about coins () and dice () that are somewhat more creative. Solutions to these problems can be found on the prototype pages.

Here are some examples of coin tossing or dice tossing.

Example 4 When we toss a coin, what is the probability of getting tails?
Outcomes 2 - heads or tails. (it is believed that the coin never falls on the edge) Favorable outcome - tails, 1.
Probability 1/2=0.5
Answer: 0.5.

Example 5 What if we flip a coin twice? What is the probability that it will come up heads both times?
The main thing is to determine which elementary outcomes we will consider when tossing two coins. After tossing two coins, one of the following results can occur:
1) PP - both times it came up tails
2) PO - first time tails, second time heads
3) OP - the first time heads, the second time tails
4) OO - heads up both times
There are no other options. This means that there are 4 elementary outcomes. Only the first one is favorable, 1.
Probability: 1/4=0.25
Answer: 0.25

What is the probability that two tosses of a coin will land on tails?
The number of elementary outcomes is the same, 4. Favorable outcomes are the second and third, 2.
Probability of getting one tail: 2/4=0.5

In such problems, another formula may come in handy.
If with one toss of a coin options we have 2 results, then for two throws the results will be 2 2=2 2 =4 (as in example 5), for three throws 2 2 2=2 3 =8, for four: 2 2 2 2 =2 4 =16, … for N throws there are 2·2·...·2=2 N possible outcomes.

So, you can find the probability of getting 5 tails out of 5 coin tosses.
The total number of elementary outcomes: 2 5 =32.
Favorable outcomes: 1. (RRRRRR - all 5 times tails)
Probability: 1/32=0.03125

The same is true for the dice. With one throw, there are 6 possible results. So, for two throws: 6 6=36, for three 6 6 6=216, etc.

Example 6 We throw a dice. What is the probability of getting an even number?

Total outcomes: 6, according to the number of faces.
Favorable: 3 outcomes. (2, 4, 6)
Probability: 3/6=0.5

Example 7 Throw two dice. What is the probability that the total rolls 10? (round to hundredths)

There are 6 possible outcomes for one die. Hence, for two, according to the above rule, 6·6=36.
What outcomes will be favorable for a total of 10 to fall out?
10 must be decomposed into the sum of two numbers from 1 to 6. This can be done in two ways: 10=6+4 and 10=5+5. So, for cubes, options are possible:
(6 on the first and 4 on the second)
(4 on the first and 6 on the second)
(5 on the first and 5 on the second)
In total, 3 options. Desired probability: 3/36=1/12=0.08
Answer: 0.08

Other types of B6 problems will be discussed in one of the following "How to Solve" articles.

V-6-2014 (all 56 prototypes from the USE bank)

Be able to build and explore the simplest mathematical models(probability theory)

1. In a random experiment, two dice are thrown. Find the probability of getting 8 points in total. Round the result to the nearest hundredth. Decision: The number of outcomes in which 8 points will fall out as a result of a roll of dice is 5: 2+6, 3+5, 4+4, 5+3, 6+2. Each of the dice can fall out in six ways, so the total number of outcomes is 6 6 = 36. Therefore, the probability that 8 points will fall out in total is 5: 36=0.138…=0.14

2. In a random experiment, a symmetrical coin is thrown twice. Find the probability that heads come up exactly once. Solution: There are 4 possible outcomes of the experiment: heads-heads, heads-tails, tails-heads, tails-tails. Heads come up exactly once in two cases: heads-tails and tails-heads. Therefore, the probability that heads will fall out exactly 1 time is 2: 4 = 0.5.

3. 20 athletes participate in the gymnastics championship: 8 from Russia, 7 from the USA, the rest from China. The order in which the gymnasts perform is determined by lot. Find the probability that the athlete who competes first is from China. Solution: Participates in the championshipathletes from China. Then the probability that the athlete who performs first will be from China is 5: 20 = 0.25

4. On average, out of 1,000 garden pumps sold, 5 leak. Find the probability that one randomly selected pump does not leak. Solution: On average, out of 1,000 garden pumps sold, 1,000 - 5 = 995 do not leak. This means that the probability that one pump randomly selected for control does not leak is 995: 1000 = 0.995

5. The factory produces bags. On average, for every 100 quality bags, there are eight bags with hidden defects. Find the probability that the purchased bag will be of high quality. Round the result to the nearest hundredth. Solution: According to the condition, for every 100 + 8 = 108 bags, there are 100 quality bags. This means that the probability that the purchased bag will be of high quality is 100: 108 \u003d 0.925925 ... \u003d 0.93

6. 4 athletes from Finland, 7 athletes from Denmark, 9 athletes from Sweden and 5 athletes from Norway participate in the shot put competition. The order in which the athletes compete is determined by lot. Find the probability that the last player to compete is from Sweden.. Solution : In total, 4 + 7 + 9 + 5 = 25 athletes take part in the competition. So the probability that the athlete who competes last will be from Sweden is 9: 25 = 0.36

7. Scientific conference is held in 5 days. A total of 75 reports are planned - the first three days, 17 reports each, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by a draw. What is the probability that Professor M.'s report will be scheduled for the last day of the conference? Solution: During the first three days 51 reports will be read, 24 reports are planned for the last two days. Therefore, 12 reports are scheduled for the last day. This means that the probability that the report of Professor M. will be scheduled for the last day of the conference is 12: 75 = 0.16

8. The contest of performers is held in 5 days. A total of 80 performances were announced - one from each country. On the first day there are 8 performances, the rest are distributed equally between the remaining days. The order of performances is determined by a draw. What is the probability that the performance of the representative of Russia will take place on the third day of the competition? Solution: Scheduled for the third dayspeeches. This means that the probability that the performance of a representative from Russia will be scheduled for the third day of the competition is 18: 80 = 0.225

9. 3 scientists from Norway, 3 from Russia and 4 from Spain came to the seminar. The order of reports is determined by a draw. Find the probability that the eighth will be the report of a scientist from Russia. Solution: In total, 3 + 3 + 4 = 10 scientists take part in the seminar, which means that the probability that the eighth scientist will be from Russia is 3:10 = 0.3.

10. Before the start of the first round of the badminton championship, the participants are randomly divided into game pairs by drawing lots. In total, 26 badminton players participate in the championship, including 10 participants from Russia, including Ruslan Orlov. Find the probability that in the first round Ruslan Orlov will play with any badminton player from Russia? Solution: In the first round Ruslan Orlov can play with 26 − 1 = 25 badminton players, of which 10 − 1 = 9 from Russia. This means that the probability that in the first round Ruslan Orlov will play with any badminton player from Russia is 9: 25 = 0.36

11. There are only 55 tickets in the collection of biology tickets, 11 of them contain a question on botany. Find the probability that a student will get a question on botany in a randomly selected exam ticket. Solution: 11: 55 = 0.2

12. 25 athletes compete at the diving championship, among them 8 jumpers from Russia and 9 jumpers from Paraguay. The order of performances is determined by a draw. Find the probability that the sixth jumper will be from Paraguay.

13. Two factories produce the same glass for car headlights. The first factory produces 30% of these glasses, the second - 70%. The first factory produces 3% of defective glasses, and the second - 4%. Find the probability that a glass accidentally bought in a store will be defective.

Decision. Convert %% to fractions.

Event A - "Purchased glasses from the first factory." P(A)=0.3

Event B - "Glasses from the second factory are purchased." P(B)=0.7

Event X - "Windows are defective".

P(A and X) = 0.3*0.03=0.009

P(B and X) = 0.7*0.04=0.028 According to the total probability formula: P = 0.009+0.028 = 0.037

14. If grandmaster A. plays white, then he wins grandmaster B. with a probability of 0.52. If A. plays black, then A. beats B. with a probability of 0.3. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times. Decision: 0,52 * 0,3 = 0,156.

15. Vasya, Petya, Kolya and Lyosha cast lots - who should start the game. Find the probability that Petya will start the game.

Solution: Random experiment - casting lots.
In this experiment, the elementary event is the participant who wins the lot.
We list the possible elementary events:
(Vasya), (Petya), (Kolya), (Lesha).
There will be 4 of them, i.e. N=4. The lot implies that all elementary events are equally possible.
The event A= (Petya won the lot) is favored by only one elementary event (Petya). Therefore N(A)=1.
Then P(A)=0.25 Answer: 0.25.

16. 16 teams participate in the World Championship. By drawing lots, they must be divided into four groups of four teams each. Mixed in the box are cards with group numbers: 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4. Team captains draw one card at a time. What is the probability that the Russian team will be in the second group? Decision: There are 16 outcomes in total. with number 2 will be 4. So 4: 16=0.25

17. At the geometry exam, the student gets one question from the list of exam questions. The probability that this is an inscribed circle question is 0.2. The probability that this is a Parallelogram question is 0.15. There are no questions related to these two topics at the same time. Find the probability that the student will get a question on one of these two topics on the exam.

= (question on the topic "Inscribed circle"),
= (a question on the topic "Parallelogram").
Events and are incompatible, since by condition there are no questions in the list related to these two topics at the same time.
Event= (question on one of these two topics) is their union:.
We apply the formula for adding the probabilities of incompatible events:
.

18.B mall two identical vending machines sell coffee. The probability that the machine will run out of coffee by the end of the day is 0.3. The probability that both machines will run out of coffee is 0.12. Find the probability that by the end of the day there will be coffee left in both vending machines.

Let's define events
= (coffee will end in the first machine),
= (coffee will end in the second machine).
According to the task and .
Using the formula for adding probabilities, we find the probability of an event
and = (coffee will end in at least one of the machines):

.
Therefore, the probability of the opposite event (coffee will remain in both machines) is equal to
.

19. A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hit the targets the first three times and missed the last two. Round the result to the nearest hundredth.

In this problem, it is assumed that the result of each next shot does not depend on the previous ones. Therefore, the events “hit on the first shot”, “hit on the second shot”, etc. independent.
The probability of each hit is. So the probability of each miss is. We use the formula for multiplying the probabilities of independent events. We get that the sequence
= (hit, hit, hit, miss, miss) has a probability
=
= . Answer: .

20. There are two payment machines in the store. Each of them can be faulty with a probability of 0.05, regardless of the other automaton. Find the probability that at least one automaton is serviceable.

This problem also assumes the independence of the operation of automata.
Find the probability of the opposite event
= (both machines are faulty).
To do this, we use the formula for multiplying the probabilities of independent events:
.
So the probability of an event= (at least one automaton is operational) is equal to. Answer: .

21. The room is illuminated by a lantern with two lamps. The probability of one lamp burning out in a year is 0.3. Find the probability that at least one lamp does not burn out within a year. Solution: Both will burn out (the events are independent and we use the formula for the product of probabilities) with the probability p1=0.3⋅0.3=0.09
Opposite event(NOT both will burn out = at least ONE will not burn out)
will happen with probability p=1-p1=1-0.09=0.91
ANSWER: 0.91

22. Probability that new Electric kettle will serve more than a year, is equal to 0.97. The probability that it will last more than two years is 0.89. Find the probability that it lasts less than two years but more than a year.

Decision.

Let A = "the kettle will last more than a year, but less than two years", B = "the kettle will last more than two years", then A + B = "the kettle will last more than a year."

Events A and B are joint, the probability of their sum is equal to the sum of the probabilities of these events, reduced by the probability of their product. The probability of the product of these events, consisting in the fact that the kettle will fail in exactly two years - exactly on the same day, hour and second - is equal to zero. Then:

P(A + B) = P(A) + P(B) − P(A B) = P(A) + P(B),

whence, using the data from the condition, we obtain 0.97 = P(A) + 0.89.

Thus, for the desired probability we have: P(A) = 0.97 − 0.89 = 0.08.

23. Agrofirm purchases chicken eggs in two households. 40% of the eggs from the first farm are eggs of the highest category, and from the second farm - 20% of the eggs of the highest category. In total, 35% of eggs receive the highest category. Find the probability that the egg purchased from this farm will be from the first farm. Decision: Let in the first farm, the agricultural firm buys eggs, including eggs of the highest category, and in the second farm - eggs, including eggs of the highest category. Thus, in total, the agroform buys eggs, including eggs of the highest category. By condition, 35% of eggs have the highest category, then:

Therefore, the probability that the purchased egg will be from the first farm is equal to =0,75

24. There are 10 digits on the telephone keypad, from 0 to 9. What is the probability that a randomly pressed number will be even?

25. What is the probability that a randomly chosen natural number from 10 to 19 is divisible by three?

26. Cowboy John hits a fly on the wall with a probability of 0.9 if he shoots from a shot revolver. If John fires an unshot revolver, he hits a fly with a probability of 0.2. There are 10 revolvers on the table, of which only 4 are shot. Cowboy John sees a fly on the wall, randomly grabs the first revolver he comes across and shoots at the fly. Find the probability that John misses. Solution: John hits a fly if he grabs a sighted revolver and hits from it, or if he grabs an unfired revolver and hits from it. According to the conditional probability formula, the probabilities of these events are 0.4 0.9 = 0.36 and 0.6 0.2 = 0.12, respectively. These events are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: 0.36 + 0.12 = 0.48. The event that John misses is the opposite. Its probability is 1 − 0.48 = 0.52.

27. There are 5 people in a group of tourists. With the help of lots, they choose two people who must go to the village for food. Tourist A. would like to go to the store, but he submits to the lot. What is the probability that A will go to the store? Decision: There are five tourists in total, two of them are randomly selected. The probability of being selected is 2:5 = 0.4. Answer: 0.4.

28.Before you start football match The referee tosses a coin to determine which team will start the ball game. The Physicist team plays three matches with different teams. Find the probability that in these games "Physicist" wins the lot exactly twice. Solution: Let's denote by "1" the side of the coin that is responsible for winning the lot by "Physicist", the other side of the coin will be denoted by "0". Then there are three favorable combinations: 110, 101, 011, and there are 2 combinations in total 3 = 8: 000, 001, 010, 011, 100, 101, 110, 111. Thus, the desired probability is:

29. A dice is thrown twice. How many elementary outcomes of experience favor the event "A = sum of points equals 5"? Solution: The sum of points can be equal to 5 in four cases: "3 + 2", "2 + 3", "1 + 4", "4 + 1". Answer: 4.

30. In a random experiment, a symmetrical coin is tossed twice. Find the probability that the outcome of the OR will come (heads the first time, tails the second). Decision: There are four possible outcomes: heads-heads, heads-tails, tails-heads, tails-tails. Favorable is one: heads-tails. Therefore, the desired probability is 1: 4 = 0.25. Answer: 0.25.

31. Groups perform at the rock festival - one from each of the declared countries. The order of performance is determined by lot. What is the probability that a band from Denmark will perform after a band from Sweden and after a band from Norway? Round the result to the nearest hundredth. Decision: The total number of groups performing at the festival does not matter to answer the question. No matter how many there are, there are 6 ways for the indicated countries relative position among speakers (D - Denmark, S - Sweden, N - Norway):

L...S...N..., ...D...N...Sh..., ...Sh...N...L..., ...Sh. ..L...N..., ...N...D...W..., ...N...S...L...

Denmark comes after Sweden and Norway on two occasions. Therefore, the probability that the groups will be randomly distributed in this way is equal to Answer: 0.33.

32. When firing artillery automatic system makes a shot at the target. If the target is not destroyed, the system fires again. The shots are repeated until the target is destroyed. The probability of destroying a certain target with the first shot is 0.4, and with each subsequent shot - 0.6. How many shots will be required to ensure that the probability of destroying the target is at least 0.98? Decision: You can solve the problem "by actions", calculating the probability of surviving after a series of successive misses: P(1) = 0.6. P(2) = P(1) 0.4 = 0.24. P(3) = P(2) 0.4 = 0.096. P(4) = P(3) 0.4 = 0.0384; P(5) = P(4) 0.4 = 0.01536. The last probability is less than 0.02, so five shots on the target are sufficient.

33. To advance to the next round of the competition, a football team needs to score at least 4 points in two games. If a team wins it gets 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will be able to advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4. Decision : A team can get at least 4 points in two games in three ways: 3+1, 1+3, 3+3. These events are incompatible, the probability of their sum is equal to the sum of their probabilities. Each of these events is a product of two independent events - the result in the first and second game. Hence we have:

34. In a certain city, out of 5000 babies born, 2512 are boys. Find the birth frequency of girls in this city. Round the result to thousandths. Decision: 5000 – 2512 = 2488; 2488: 5000 = 0,4976 ≈0,498

35. There are 12 seats on board the aircraft next to the emergency exits and 18 seats behind the partitions separating the cabins. The rest of the seats are inconvenient for the passenger tall. Passenger V. is tall. Find the probability that at check-in, with a random choice of seat, passenger B. will get a comfortable seat if there are 300 seats on the plane. Decision : On the plane, 12 + 18 = 30 seats are convenient for passenger V., and in total there are 300 seats on the plane. Therefore, the probability that passenger B. will get a comfortable seat is 30: 300 = 0.1. Answer: 0.1.

36. At the Olympiad at the university, participants are seated in three classrooms. In the first two, 120 people each, the rest are taken to a reserve auditorium in another building. When counting, it turned out that there were 250 participants in total. Find the probability that a randomly selected participant wrote the Olympiad in the spare room. Decision: In total, 250 − 120 − 120 = 10 people were sent to the reserve audience. Therefore, the probability that a randomly selected participant wrote the Olympiad in the spare room is 10: 250 = 0.04. Answer: 0.04.

37. There are 26 people in the class, among them two twins - Andrey and Sergey. The class is randomly divided into two groups of 13 people each. Find the probability that Andrey and Sergey will be in the same group. Decision: Let one of the twins be in some group. Together with him, 12 people out of the 25 remaining classmates will be in the group. The probability that the second twin will be among these 12 people is 12:25 = 0.48.

38. There are 50 cars in the taxi company; 27 of them are black with yellow inscriptions on the sides, the rest are yellow with black inscriptions. Find the probability that a yellow car with black inscriptions will arrive at a random call. Solution: 23:50=0.46

39. There are 30 people in a group of tourists. They are thrown by helicopter in several steps into a remote area, 6 people per flight. The order in which the helicopter transports tourists is random. Find the probability that tourist P. will take the first helicopter flight. Decision: There are 6 seats on the first flight, a total of 30 seats. Then the probability that tourist P. will fly on the first helicopter flight is: 6:30 \u003d 0.2

40. The probability that a new DVD player will be repaired within a year is 0.045. In a certain city, out of 1,000 DVD players sold during the year, 51 pieces arrived at the warranty workshop. How different is the frequency of the “warranty repair” event from its probability in this city? Solution: The frequency (relative frequency) of the "warranty repair" event is 51: 1000 = 0.051. It differs from the predicted probability by 0.006.

41. In the manufacture of bearings with a diameter of 67 mm, the probability that the diameter will differ from the specified one by no more than 0.01 mm is 0.965. Find the probability that a random bearing will have a diameter less than 66.99 mm or greater than 67.01 mm. Decision. According to the condition, the bearing diameter will be in the range from 66.99 to 67.01 mm with a probability of 0.965. Therefore, the desired probability of the opposite event is 1 − 0.965 = 0.035.

42. The probability that student O. correctly solves more than 11 tasks on a biology test is 0.67. The probability that O. will correctly solve more than 10 problems is 0.74. Find the probability that O. correctly solves exactly 11 problems. Decision: Consider the events A = “the student will solve 11 problems” and B = “the student will solve more than 11 problems”. Their sum is the event A + B = "the student will solve more than 10 problems." Events A and B are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: P(A + B) = P(A) + P(B). Then, using the data of the problem, we get: 0.74 = P(A) + 0.67, whence P(A) = 0.74 − 0.67 = 0.07. Answer: 0.07.

43. To enter the Institute for the specialty "Linguistics", the applicant must score at least 70 points on the Unified State Examination in each of the three subjects - mathematics, Russian language and a foreign language. To enter the specialty "Commerce", you need to score at least 70 points in each of the three subjects - mathematics, Russian language and social studies. The probability that the applicant Z. will receive at least 70 points in mathematics is 0.6, in the Russian language - 0.8, in foreign language- 0.7 and in social studies - 0.5. Find the probability that Z. will be able to enter at least one of the two specialties mentioned. Solution: In order to enter at least somewhere, Z. needs to pass both Russian and mathematics with at least 70 points, and in addition to that, pass a foreign language or social studies with at least 70 points. Let be A , B , C and D - these are events in which Z. passes mathematics, Russian, foreign and social studies, respectively, with at least 70 points. Then since

For the probability of arrival, we have:

44. At the factory of ceramic dishes, 10% of the produced plates have a defect. During product quality control, 80% of defective plates are detected. The rest of the plates are for sale. Find the probability that a plate randomly selected at the time of purchase has no defects. Round your answer to the nearest hundredth. Decision : Let the factory producedplates. All high-quality cymbals and 20% of undetected defective cymbals will go on sale:plates. Because the quality ones, the probability of buying a quality plate is 0.9p:0.92p=0.978 Answer: 0.978.

45. There are three sellers in the store. Each of them is busy with a client with a probability of 0.3. Find the probability that, at a random time, all three salespeople are busy at the same time (assume that the customers enter independently of each other). Decision : The probability of producing independent events is equal to the product of the probabilities of these events. Therefore, the probability that all three sellers are busy is

46. Based on customer reviews, Ivan Ivanovich assessed the reliability of two online stores. The probability that desired product delivered from store A is 0.8. The probability that this product will be delivered from store B is 0.9. Ivan Ivanovich ordered the goods at once in both stores. Assuming that online stores operate independently of each other, find the probability that none of the stores will deliver the goods. Decision: The probability that the first store will not deliver the goods is 1 − 0.9 = 0.1. The probability that the second store will not deliver the goods is 1 − 0.8 = 0.2. Since these events are independent, the probability of their product (both stores will not deliver the goods) is equal to the product of the probabilities of these events: 0.1 0.2 = 0.02

47. A bus runs daily from the district center to the village. The probability that on Monday there will be less than 20 passengers on the bus is 0.94. The probability that there will be less than 15 passengers is 0.56. Find the probability that the number of passengers will be between 15 and 19. Solution: Consider the events A = “there are less than 15 passengers on the bus” and B = “there are between 15 and 19 passengers on the bus”. Their sum is the event A + B = "less than 20 passengers on the bus". Events A and B are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: P(A + B) = P(A) + P(B). Then, using the data of the problem, we obtain: 0.94 = 0.56 + P(B), whence P(B) = 0.94 − 0.56 = 0.38. Answer: 0.38.

48. Before the start of a volleyball match, the team captains draw fair lots to determine which team will start the ball game. The Stator team takes turns playing with the Rotor, Motor and Starter teams. Find the probability that Stator will start only the first and last games. Decision. It is required to find the probability of product of three events: "Stator" starts the first game, does not start the second game, starts the third game. The probability of producing independent events is equal to the product of the probabilities of these events. The probability of each of them is equal to 0.5, whence we find: 0.5 0.5 0.5 = 0.125. Answer: 0.125.

49. There are two types of weather in Fairyland: good and excellent, and the weather, having settled in the morning, remains unchanged all day. It is known that with a probability of 0.8 the weather tomorrow will be the same as today. Today is July 3rd, the weather in Fairyland is fine. Find the probability that there will be great weather in Magicland on July 6th. Decision. For the weather on July 4, 5 and 6, there are 4 options: XXO, XOO, OXO, LLC (here X is good, O is excellent weather). Let's find the probabilities of such weather: P(XXO) = 0.8 0.8 0.2 = 0.128; P(XOO) = 0.8 0.2 0.8 = 0.128; P(OXO) = 0.2 0.2 0.2 = 0.008; P(OOO) = 0.2 0.8 0.8 = 0.128. These events are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: P(XXO) + P(XXO) + P(XXO) + P(OOO) = 0.128 + 0.128 + 0.008 + 0.128 = 0.392.

50. All patients with suspected hepatitis are given a blood test. If the analysis reveals hepatitis, then the result of the analysis is called positive . In hepatitis patients, the analysis gives positive result with a probability of 0.9. If the patient does not have hepatitis, then the test may give a false positive result with a probability of 0.01. It is known that 5% of patients admitted with suspected hepatitis actually have hepatitis. Find the probability that the test result of a patient admitted to the clinic with suspected hepatitis will be positive. Decision . The analysis of the patient can be positive for two reasons: A) the patient has hepatitis, his analysis is correct; B) the patient does not have hepatitis, his analysis is false. These are incompatible events, the probability of their sum is equal to the sum of the probabilities of these events. We have: p(A)=0.9 0.05=0.045; p(B)=0.01 0.95=0.0095; p(A+B)=P(A)+p(B)=0.045+0.0095=0.0545.

51. In Misha's pocket there were four sweets - Grillage, Squirrel, Cow and Swallow, as well as the keys to the apartment. Taking out the keys, Misha accidentally dropped one piece of candy from his pocket. Find the probability that the Grillage candy is lost.

52.Mechanical watches with a twelve-hour dial at some point broke and stopped walking. Find the probability that hour hand froze, reaching the 10 mark, but not reaching the 1 hour mark. Solution: 3: 12=0.25

53. The probability that the battery is defective is 0.06. The customer in the store selects a random package containing two of these batteries. Find the probability that both batteries are good. Solution: The probability that the battery is good is 0.94. The probability of producing independent events (both batteries will be good) is equal to the product of the probabilities of these events: 0.94 0.94 \u003d 0.8836. Answer: 0.8836.

54. An automatic line makes batteries. The probability that a finished battery is defective is 0.02. Before packaging, each battery goes through a control system. The probability that the system will reject a bad battery is 0.99. The probability that the system will mistakenly reject a good battery is 0.01. Find the probability that a randomly selected manufactured battery will be rejected by the control system. Decision. The situation in which the battery will be rejected can be the result of the following events: A = the battery is really bad and was rejected fairly, or B = the battery is good, but rejected by mistake. These are incompatible events, the probability of their sum is equal to the sum of the probabilities of these events. We have:

55. The figure shows a labyrinth. The spider crawls into the maze at the "Entrance" point. The spider cannot turn around and crawl back, therefore, at each fork, the spider chooses one of the paths that it has not crawled yet. Assuming that the choice of the further path is purely random, determine with what probability the spider will come to the exit.

Decision.

At each of the four marked forks, the spider can choose either the path leading to exit D or another path with a probability of 0.5. These are independent events, the probability of their product (the spider reaches exit D) is equal to the product of the probabilities of these events. Therefore, the probability of coming to the output D is (0.5) 4 = 0,0625.

Plan for a workshop for mathematics teachers of the educational institution of the city of Tula on the topic “Solving USE tasks in mathematics from the sections: combinatorics, probability theory. Teaching Methods»

Time spending: 12 00 ; 15 00

Location: MBOU "Lyceum No. 1", room. No. 8

I. Problem solving for probability

1. Solving problems on the classical definition of probability

We, as teachers, already know that the main types of tasks in the USE in probability theory are based on the classical definition of probability. Recall what is called the probability of an event?

Probability of an event is the ratio of the number of outcomes favorable to a given event to total number outcomes.

In our scientific and methodological association of teachers of mathematics, a general scheme problem solving for probability. I would like to present it to your attention. By the way, we shared our work experience, and in the materials that we gave to your attention for a joint discussion of solving problems, we gave this scheme. However, I want to voice it.

In our opinion, this scheme helps to quickly logically put everything on the shelves, and after that the task can be solved much easier for both the teacher and the students.

So, I want to analyze in detail the problem of the following content.

I wanted to talk with you in order to explain the methodology of how to convey such a solution to the guys, during which the guys would understand this typical task, and later they would understand these tasks themselves.

What is a random experiment in this problem? Now we need to isolate the elementary event in this experiment. What is this elementary event? Let's list them.

Issue questions?

Dear colleagues, you, too, have obviously considered probability problems with dice. I think we need to disassemble it, because there are some nuances. Let's analyze this problem according to the scheme that we proposed to you. Since there is a number from 1 to 6 on each face of the cube, the elementary events are the numbers 1, 2, 3, 4, 5, 6. We found that the total number of elementary events is 6. Let us determine which elementary events favor the event. Only two events favor this event - 5 and 6 (since it follows from the condition that 5 and 6 points should fall out).

Explain that all elementary events are equally possible. What will be the questions on the task?

How do you understand that the coin is symmetrical? Let's get this straight, sometimes certain phrases cause misunderstandings. Let's understand this problem conceptually. Let's deal with you in that experiment, which is described, what elementary outcomes can be. Can you imagine where is the head, where is the tail? What are the fallout options? Are there other events? What is the total number of events? According to the problem, it is known that the heads fell out exactly once. So this eventelementary events from these four OR and RO favor, this cannot happen twice already. We use the formula by which the probability of an event is found. Recall that the answers in Part B must be either an integer or a decimal.

Show on the interactive whiteboard. We read the task. What is the elementary outcome in this experience? Clarify that the pair is ordered - that is, the number fell on the first die, and on the second die. In any task, there are moments when you need to choose rational methods, forms and present the solution in the form of tables, diagrams, etc. In this problem, it is convenient to use such a table. I already give you turnkey solution, but in the course of solving it turns out that in this problem it is rational to use the solution in the form of a table. Explain what the table means. You understand why the columns say 1, 2, 3, 4, 5, 6.

Let's draw a square. The lines correspond to the results of the first roll - there are six of them, because the die has six faces. As are the columns. In each cell we write the sum of the dropped points. Show the completed table. Let's color the cells where the sum is equal to eight (as it is required in the condition).

I believe that the next problem, after analyzing the previous ones, can be given to the guys to solve on their own.

In the following problems, there is no need to write down all the elementary outcomes. It is enough just to count their number.

(Without solution) I gave the guys to solve this problem on their own. Algorithm for solving the problem

1. Determine what a random experiment is and what is a random event.

2. Find the total number of elementary events.

3. We find the number of events that favor the event specified in the condition of the problem.

4. Find the probability of an event using the formula.

Students can be asked a question, if 1000 batteries went on sale, and among them 6 are faulty, then the selected battery is determined as? What is it in our task? Next, I ask a question about finding what is used here as a numberand I propose to find itnumber. Then I ask, what is the event here? How many accumulators favor the completion of the event? Next, using the formula, we calculate this probability.

Here the children can be offered a second solution. Let's discuss what this method can be?

1. What event can be considered now?

2. How to find the probability of a given event?

The children need to be told about these formulas. They are next

The eighth task can be offered to the children on their own, since it is similar to the sixth task. It can be offered to them as independent work, or on a card at the board.

This problem can be solved in relation to the Olympiad, which is currently taking place. Despite the fact that different events participate in the tasks, however, the tasks are typical.

2. The simplest rules and formulas for calculating probabilities (opposite events, sum of events, product of events)

This is a task from USE collection. We put the solution on the board. What questions should we put before the students in order to analyze this problem.

1. How many machine guns were there? Once two automata, then there are already two events. I ask the children what the event will be? What will be the second event?

2. is the probability of the event. We do not need to calculate it, since it is given in the condition. According to the condition of the problem, the probability that "coffee runs out in both machines" is 0.12. There was an event A, there was an event B. And a new event appears? I ask the children the question - what? This is an event when both vending machines run out of coffee. In this case, in the theory of probability, this is a new event, which is called the intersection of two events A and B and is denoted in this way.

Let's use the probability addition formula. The formula is as follows

We give it to you in the reference material and the guys can give this formula. It allows you to find the probability of the sum of events. We were asked the probability of the opposite event, the probability of which is found by the formula.

Problem 13 uses the concept of a product of events, the formula for finding the probability of which is given in the Appendix.

3. Tasks for the application of the tree of possible options

According to the condition of the problem, it is easy to draw up a diagram and find the indicated probabilities.

With what help theoretical material Have you worked with students to solve problems of this kind? Did you use a tree of possibilities or did you use other methods for solving such problems? Did you give the concept of graphs? In the fifth or sixth grade, the guys have such problems, the analysis of which gives the concept of graphs.

I would like to ask you, have you and your students considered using a tree of possibilities when solving probability problems? The fact is that not only do the USE have such tasks, but rather complex tasks have appeared, which we will now solve.

Let's discuss with you the methodology for solving such problems - if it coincides with my methodology, as I explain to the guys, then it will be easier for me to work with you, if not, then I will help you deal with this problem.

Let's discuss the events. What events in problem 17 can be identified?

When constructing a tree on a plane, a point is designated, which is called the root of the tree. Next, we begin to consider the eventsand. We will construct a segment (in probability theory it is called a branch). The condition says that the first factory produces 30% mobile phones this brand (what? The one they produce), so in this moment I ask students, what is the probability of the first factory producing phones of this brand, those that they produce? Since the event is the release of the phone at the first factory, the probability of this event is 30% or 0.3. The remaining phones are produced at the second factory - we are building the second segment, and the probability of this event is 0.7.

Students are asked the question - what type of phone can be produced by the first factory? With or without defect. What is the probability that the phone produced by the first factory has a defect? According to the condition, it is said that it is equal to 0.01. Question: What is the probability that the phone produced by the first factory does not have a defect? Since this event is opposite to the given one, its probability is equal.

It is required to find the probability that the phone is defective. It may be from the first factory, or it may be from the second. Then we use the formula for adding probabilities and we get that the whole probability is the sum of the probabilities that the phone is defective from the first factory, and that the phone is defective from the second factory. The probability that the phone has a defect and was produced at the first factory is found by the formula for the product of probabilities, which is given in the appendix.

4. One of the most challenging tasks from the USE bank for probability

Let's analyze, for example, No. 320199 from the FIPI Task Bank. This is one of the most difficult tasks in B6.

In order to enter the institute for the specialty "Linguistics", the applicant Z. must score at least 70 points on the Unified State Examination in each of the three subjects - mathematics, Russian language and a foreign language. To enter the specialty "Commerce", you need to score at least 70 points in each of the three subjects - mathematics, Russian language and social studies.

The probability that applicant Z. will receive at least 70 points in mathematics is 0.6, in Russian - 0.8, in a foreign language - 0.7 and in social studies - 0.5.

Find the probability that Z. will be able to enter at least one of the two specialties mentioned.

Note that the problem does not ask whether an applicant named Z. will study both linguistics and commerce at the same time and receive two diplomas. Here we need to find the probability that Z. will be able to enter at least one of these two specialties - that is, he will gain required amount points.

In order to enter at least one of the two specialties, Z. must score at least 70 points in mathematics. And in Russian. And yet - social science or foreign.

The probability of scoring 70 points in mathematics for him is 0.6.

The probability of scoring points in mathematics and Russian is equal.

Let's deal with foreign and social studies. The options are suitable for us when the applicant scored points in social studies, in a foreign language, or in both. The option is not suitable when he did not score points either in language or in “society”. This means that the probability of passing social studies or a foreign one is at least 70 points equal. As a result, the probability of passing mathematics, Russian and social studies or a foreign one is equal to

This is the answer.

II . Solving combinatorial problems

1. Number of combinations and factorials

Let's briefly analyze the theoretical material.

Expressionn ! reads "en-factorial" and denotes the product of all natural numbers from 1 ton inclusive:n ! = 1 2 3 ...n .

In addition, in mathematics, by definition, it is considered that 0! = 1. Such an expression is rare, but still occurs in problems in probability theory.

Definition

Let there be objects (pencils, sweets, whatever) from which it is required to choose exactly different objects. Then the number of options for such a choice is callednumber of combinations from the elements. This number is indicated and calculated according to a special formula.

Designation

What does this formula give us? In fact, almost no serious task can be solved without it.

For a better understanding, let's analyze a few simple combinatorial problems:

Task

The bartender has 6 varieties of green tea. For the tea ceremony, you need to submit green tea exactly 3 different varieties. In how many ways can a bartender complete an order?

Decision

Everything is simple here: there isn = 6 varieties to choose fromk = 3 varieties. The number of combinations can be found by the formula:

Answer

Substitute in the formula. We cannot solve all problems, but typical tasks we have written, they are presented to your attention.

Task

In a group of 20 students, 2 representatives must be selected to speak at the conference. In how many ways can this be done?

Decision

Again, all we haven = 20 students, but you have to choosek = 2 students. Finding the number of combinations:

Please note that the factors included in different factorials are marked in red. These multipliers can be painlessly reduced and thereby significantly reduce the total amount of calculations.

Answer

190

Task

17 servers with various defects were brought to the warehouse, which cost 2 times cheaper than normal servers. The director bought 14 such servers for the school, and spent the saved money in the amount of 200,000 rubles on the purchase of other equipment. In how many ways can a director choose defective servers?

Decision

There is quite a lot of extra data in the task, which can be confusing. Most important facts: there is everythingn = 17 servers, and the director needsk = 14 servers. We count the number of combinations:

The red color again indicates the multipliers that are being reduced. In total, it turned out 680 combinations. In general, the director has plenty to choose from.

Answer

680

This task is capricious, as there is extra data in this task. They confuse many students right decision. There were 17 servers in total, and the director needed to choose 14. Substituting into the formula, we get 680 combinations.

2. Law of multiplication

Definition

multiplication law in combinatorics: the number of combinations (ways, combinations) in independent sets is multiplied.

In other words, let there beA ways to perform one action andB ways to perform another action. The path also these actions are independent, i.e. not related in any way. Then you can find the number of ways to perform the first and second action by the formula:C = A · B .

Task

Petya has 4 coins of 1 ruble each and 2 coins of 10 rubles each. Petya, without looking, took out of his pocket 1 coin with a face value of 1 ruble and another 1 coin with a face value of 10 rubles to buy a pen for 11 rubles. In how many ways can he choose these coins?

Decision

So, first Petya getsk = 1 coin fromn = 4 available coins with a face value of 1 ruble. The number of ways to do this isC 4 1 = ... = 4.

Then Petya reaches into his pocket again and takes outk = 1 coin fromn = 2 available coins with a face value of 10 rubles. Here the number of combinations isC 2 1 = ... = 2.

Since these actions are independent, the total number of options isC = 4 2 = 8.

Answer

Task

There are 8 white balls and 12 black ones in a basket. In how many ways can you get 2 white balls and 2 black balls from this basket?

Decision

Total in cartn = 8 white balls to choose fromk = 2 balls. It can be doneC 8 2 = ... = 28 different ways.

In addition, the cart containsn = 12 black balls to choose from againk = 2 balls. The number of ways to do this isC 12 2 = ... = 66.

Since the choice of the white ball and the choice of the black one are independent events, the total number of combinations is calculated according to the multiplication law:C = 28 66 = 1848. As you can see, there can be quite a few options.

Answer

1848

The law of multiplication shows how many ways you can perform a complex action that consists of two or more simple ones - provided that they are all independent.

3. Law of addition

If the law of multiplication operates on "isolated" events that do not depend on each other, then in the law of addition the opposite is true. It deals with mutually exclusive events that never happen at the same time.

For example, “Peter took out 1 coin from his pocket” and “Peter did not take out a single coin from his pocket” are mutually exclusive events, since it is impossible to take out one coin without taking out any.

Similarly, the events "Randomly selected ball - white" and "Randomly selected ball - black" are also mutually exclusive.

Definition

Addition Law in combinatorics: if two mutually exclusive actions can be performedA andB ways, respectively, these events can be combined. This will generate a new event that can be executedX = A + B ways.

In other words, when combining mutually exclusive actions (events, options), the number of their combinations is added up.

We can say that the law of addition is a logical "OR" in combinatorics, when any of the mutually exclusive options suits us. Conversely, the law of multiplication is a logical "AND", in which we are interested in the simultaneous execution of both the first and second actions.

Task

There are 9 black balls and 7 red balls in a basket. The boy takes out 2 balls of the same color. In how many ways can he do this?

Decision

If the balls are the same color, then there are few options: both of them are either black or red. Obviously, these options are mutually exclusive.

In the first case, the boy has to choosek = 2 black balls fromn = 9 available. The number of ways to do this isC 9 2 = ... = 36.

Similarly, in the second case we choosek = 2 red balls fromn = 7 possible. The number of ways isC 7 2 = ... = 21.

It remains to find the total number of ways. Since the variants with black and red balls are mutually exclusive, according to the law of addition we have:X = 36 + 21 = 57.

Answer57

Task

The stall sells 15 roses and 18 tulips. A 9th grade student wants to buy 3 flowers for his classmate, and all flowers must be the same. In how many ways can he make such a bouquet?

Decision

According to the condition, all flowers must be the same. So, we will buy either 3 roses or 3 tulips. Anyway,k = 3.

In the case of roses, you will have to choose fromn = 15 options, so the number of combinations isC 15 3 = ... = 455. For tulipsn = 18, and the number of combinations -C 18 3 = ... = 816.

Since roses and tulips are mutually exclusive options, we work according to the law of addition. Get the total number of optionsX = 455 + 816 = 1271. This is the answer.

Answer

1271

Additional terms and restrictions

Very often in the text of the problem there are additional conditions that impose significant restrictions on the combinations of interest to us. Compare two sentences:

There is a set of 5 pens different colors. In how many ways can 3 stroke handles be selected?

There is a set of 5 pens in different colors. In how many ways can 3 stroke handles be chosen if one of them must be red?

In the first case, we have the right to take any colors that we like - there are no additional restrictions. In the second case, everything is more complicated, since we must choose a red handle (it is assumed that it is in the original set).

Obviously, any restrictions drastically reduce the total number of options. So how do you find the number of combinations in this case? Just remember next rule:

Let there be a set ofn elements to choose fromk elements. With the introduction of additional restrictions on the numbern andk decrease by the same amount.

In other words, if you need to choose 3 out of 5 pens, and one of them must be red, then you will have to choose fromn = 5 − 1 = 4 elements byk = 3 − 1 = 2 elements. Thus, instead ofC 5 3 must be consideredC 4 2 .

Now let's see how this rule works for concrete examples:

Task

In a group of 20 students, including 2 excellent students, you need to choose 4 people to participate in the conference. In how many ways can these four be chosen if the excellent students must get to the conference?

Decision

So there is a group ofn = 20 students. But you just have to choosek = 4 of them. If there were no additional restrictions, then the number of options was equal to the number of combinationsC 20 4 .

However, we were given additional condition: 2 honors must be among these four. Thus, according to the above rule, we reduce the numbersn andk by 2. We have:

Answer

153

Task

Petya has 8 coins in his pocket, of which 6 are ruble coins and 2 are 10 ruble coins. Petya shifts some three coins into another pocket. In how many ways can Petya do this if it is known that both 10-ruble coins ended up in another pocket?

Decision

So there isn = 8 coins. Petya shiftsk = 3 coins, of which 2 are ten rubles. It turns out that out of 3 coins that will be transferred, 2 are already fixed, so the numbersn andk must be reduced by 2. We have:

Answer

III . Solving combined problems on the use of formulas of combinatorics and probability theory

Task

Petya had 4 ruble coins and 2 2 ruble coins in his pocket. Petya, without looking, shifted some three coins into another pocket. Find the probability that both two-ruble coins are in the same pocket.

Decision

Suppose that both two-ruble coins really ended up in the same pocket, then 2 options are possible: either Petya did not shift them at all, or he shifted both at once.

In the first case, when two-ruble coins were not transferred, 3 ruble coins would have to be transferred. Since there are 4 such coins in total, the number of ways to do this is equal to the number of combinations of 4 by 3:C 4 3 .

In the second case, when both two-ruble coins have been transferred, one more ruble coin will have to be transferred. It must be chosen from 4 existing ones, and the number of ways to do this is equal to the number of combinations from 4 to 1:C 4 1 .

Now let's find the total number of ways to shift the coins. Since there are 4 + 2 = 6 coins in total, and only 3 of them need to be chosen, the total number of options is equal to the number of combinations from 6 to 3:C 6 3 .

It remains to find the probability:

Answer

0,4

Show on the interactive whiteboard. Pay attention to the fact that, according to the condition of the problem, Petya, without looking, shifted three coins into one pocket. In answering this question, we can assume that two two-ruble coins really remained in one pocket. Refer to the formula for adding probabilities. Show the formula again.

Task

Petya had 2 coins of 5 rubles and 4 coins of 10 rubles in his pocket. Petya, without looking, shifted some 3 coins into another pocket. Find the probability that five-ruble coins are now in different pockets.

Decision

In order for five-ruble coins to lie in different pockets, you need to shift only one of them. The number of ways to do this is equal to the number of combinations of 2 by 1:C 2 1 .

Since Petya transferred 3 coins in total, he will have to transfer 2 more coins of 10 rubles each. Petya has 4 such coins, so the number of ways is equal to the number of combinations from 4 to 2:C 4 2 .

It remains to find how many options there are to shift 3 coins out of 6 available. This number, as in the previous problem, is equal to the number of combinations from 6 to 3:C 6 3 .

Finding the probability:

In the last step, we multiplied the number of ways to choose two-ruble coins and the number of ways to choose ten-ruble coins, since these events are independent.

Answer

0,6

So, problems with coins have their own probability formula. It is so simple and important that it can be formulated as a theorem.

Theorem

Let the coin be tossedn once. Then the probability that heads will land exactlyk times can be found using the formula:

WhereC n k - number of combinations ofn elements byk , which is calculated by the formula:

Thus, to solve the problem with coins, two numbers are needed: the number of tosses and the number of heads. Most often, these numbers are given directly in the text of the problem. Moreover, it does not matter what exactly to count: tails or eagles. The answer will be the same.

At first glance, the theorem seems too cumbersome. But it's worth a little practice - and you no longer want to return to the standard algorithm described above.

The coin is tossed four times. Find the probability that heads will come up exactly three times.

Decision

According to the condition of the problem, the total number of throws wasn = 4. Required number of heads:k = 3. Substituten andk into the formula:

With the same success, you can count the number of tails:k = 4 − 3 = 1. The answer will be the same.

Answer

0,25

Task [ Workbook“USE 2012 in mathematics. Tasks B6»]

The coin is tossed three times. Find the probability that it never comes up tails.

Decision

Writing out the numbers againn andk . Since the coin is tossed 3 times,n = 3. And since there should be no tails,k = 0. It remains to substitute the numbersn andk into the formula:

Let me remind you that 0! = 1 by definition. SoC 3 0 = 1.

Answer

0,125

Task [ Trial USE in mathematics 2012. Irkutsk]

In a random experiment, a symmetrical coin is tossed 4 times. Find the probability that heads will come up more times than tails.

Decision

In order for there to be more heads than tails, they must fall out either 3 times (then there will be 1 tails) or 4 (then there will be no tails at all). Let's find the probability of each of these events.

Let bep 1 - the probability that heads will fall out 3 times. Thenn = 4, k = 3. We have:

Now let's findp 2 - the probability that heads will fall out all 4 times. In this casen = 4, k = 4. We have:

To get the answer, it remains to add the probabilitiesp 1 andp 2 . Remember: you can only add probabilities for mutually exclusive events. We have:

p = p 1 + p 2 = 0,25 + 0,0625 = 0,3125

Answer

0,3125

In order to save your time when preparing with the guys for the Unified State Exam and the GIA, we have presented solutions to many more tasks that you can choose and solve with the guys.

Materials of the GIA, Unified State Examination of various years, textbooks and sites.

IV. Reference material

The theory of probability on the exam in mathematics can be represented as in the form simple tasks on the classical definition of probability, and in the form of rather complex ones, on the application of the corresponding theorems.

In this part, we consider problems for which it is sufficient to use the definition of probability. Sometimes here we will also apply a formula for calculating the probability of the opposite event. Although this formula can be dispensed with here, it will still be needed when solving the following problems.

Theoretical part

A random event is an event that may or may not occur (it is impossible to predict in advance) during an observation or test.

Let during the test (tossing a coin or a die, pulling examination card etc.) equally possible outcomes are possible. For example, when tossing a coin, the number of all outcomes is 2, since there can be no other outcomes except for the loss of “tails” or “eagles”. When throwing a dice, 6 outcomes are possible, since any of the numbers from 1 to 6 can appear on the upper face of the dice. Let also some event A be favored by outcomes.

The probability of an event A is the ratio of the number of outcomes favorable for this event to the total number of equally possible outcomes (this is the classical definition of probability). We write

For example, let event A consist of getting an odd number of points on a roll of a die. There are 6 possible outcomes in total: 1, 2, 3, 4, 5, 6 on the top face of the die. At the same time, outcomes with 1, 3, 5 rolls are favorable for event A. Thus, .

Note that the double inequality always holds, so the probability of any event A lies on the interval, that is . If in your answer the probability is greater than one, then you made a mistake somewhere and you need to double-check the solution.

Events A and B are called opposite each other if any outcome is favorable for exactly one of them.

For example, when a die is rolled, the event "rolled odd number” is the opposite of the “even number rolled” event.

The event opposite to event A is denoted. From the definition of opposite events it follows
, means,
.

Problems about selecting objects from a set

Task 1. 24 teams participate in the World Championship. With the help of lots, they need to be divided into four groups of six teams each. In the box are mixed cards with group numbers:

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4.

Team captains draw one card each. What is the probability that the Russian team will be in the third group?

The total number of outcomes is equal to the number of cards - there are 24 of them. There are 6 favorable outcomes (since the number 3 is written on six cards). The desired probability is equal to .

Answer: 0.25.

Task 2. An urn contains 14 red, 9 yellow and 7 green balls. One ball is drawn at random from the urn. What is the probability that this ball is yellow?

The total number of outcomes is equal to the number of balls: 14 + 9 + 7 = 30. The number of outcomes favorable to this event is 9. The desired probability is equal to .

Task 3. There are 10 numbers on the phone keypad, from 0 to 9. What is the probability that a randomly pressed number will be even and greater than 5?

The outcome here is pressing a certain key, so there are 10 equally possible outcomes in total. The indicated event is favored by the outcomes, which mean pressing the key 6 or 8. There are two such outcomes. The required probability is .

Answer: 0.2.

Task 4. What is the probability that a randomly chosen natural number from 4 to 23 is divisible by 3?

There are 23 - 4 + 1 = 20 natural numbers on the interval from 4 to 23, which means that there are 20 possible outcomes in total. On this segment, the following numbers are multiples of three: 6, 9, 12, 15, 18, 21. There are 6 such numbers in total, so 6 outcomes favor the event in question. The desired probability is equal to .

Answer: 0.3.

Task 5. Of the 20 tickets offered in the exam, the student can answer only 17. What is the probability that the student will not be able to answer the ticket chosen at random?

1st way.

Since the student can answer 17 tickets, he cannot answer 3 tickets. The probability of getting one of these tickets is, by definition, .

2nd way.

Denote by A the event "the student can answer the ticket". Then . The probability of the opposite event is =1 - 0.85 = 0.15.

Answer: 0.15.

Task 6. In the championship rhythmic gymnastics 20 athletes participate: 6 from Russia, 5 from Germany, the rest from France. The order in which the gymnasts perform is determined by lot. Find the probability that the seventh athlete is from France.

There are 20 athletes in total, all of them have equal chances to perform seventh. Therefore, there are 20 equally likely outcomes. From France 20 - 6 - 5 = 9 athletes, so there are 9 favorable outcomes for this event. The required probability is .

Answer: 0.45.

Task 7. The scientific conference is held in 5 days. A total of 50 reports are planned - the first three days, 12 reports each, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by a draw. What is the probability that Professor N.'s report will be scheduled for the last day of the conference?

First, let's find how many reports are scheduled for the last day. Reports are scheduled for the first three days. There are still 50 - 36 = 14 reports that are distributed equally between the remaining two days, so the reports are scheduled for the last day.

We will consider as the outcome the serial number of the report of Professor N. There are 50 such equally possible outcomes. There are 7 outcomes that favor the indicated event (the last 7 numbers in the list of reports). The required probability is .

Answer: 0.14.

Task 8. There are 10 seats on board the aircraft next to the emergency exits and 15 seats behind the partitions separating the cabins. The rest of the seats are inconvenient for tall passengers. Passenger K. is tall. Find the probability that at check-in, with a random choice of seat, passenger K. will get a comfortable seat if there are 200 seats on the plane.

The outcome in this problem is the choice of location. In total there are 200 equally possible outcomes. Favor the event "the chosen place is convenient" 15 + 10 = 25 outcomes. The required probability is .

Answer: 0.125.

Task 9. Of the 1000 coffee grinders assembled at the factory, 7 pieces are defective. The expert checks one randomly selected coffee grinder out of this 1000. Find the probability that the coffee grinder being checked is defective.

When choosing a coffee grinder at random, 1000 outcomes are possible, the event A "the selected coffee grinder is defective" is favorable for 7 outcomes. By definition of probability.

Answer: 0.007.

Task 10. The plant produces refrigerators. On average, for every 100 high-quality refrigerators, there are 15 refrigerators with hidden defects. Find the probability that the purchased refrigerator will be of high quality. Round the result to the nearest hundredth.

This task is similar to the previous one. However, the wording “for every 100 quality refrigerators, there are 15 with defects” tells us that defective 15 pieces are not included in the 100 quality. Therefore, the total number of outcomes is 100 + 15 = 115 (equal to the total number of refrigerators), favorable outcomes are 100. The required probability is . To calculate the approximate value of a fraction, it is convenient to use division by a corner. We get 0.869 ... which is 0.87.

Answer: 0.87.

Task 11. Before the start of the first round of the tennis championship, participants are randomly divided into game pairs by drawing lots. In total, 16 tennis players participate in the championship, including 7 participants from Russia, including Maxim Zaitsev. Find the probability that in the first round Maxim Zaitsev will play any tennis player from Russia.

As in the previous task, you need to carefully read the condition and understand what is the outcome and what is the favorable outcome (for example, the thoughtless application of the probability formula leads to the wrong answer).

Here the outcome is the rival of Maxim Zaitsev. Since there are 16 tennis players in total, and Maxim cannot play with himself, there are 16 - 1 = 15 equally probable outcomes. A favorable outcome is a rival from Russia. There are 7 such favorable outcomes - 1 = 6 (we exclude Maxim himself from among the Russians). The required probability is .

Answer: 0.4.

Task 12. The football section is attended by 33 people, among them two brothers - Anton and Dmitry. Those attending the section are randomly divided into three teams of 11 people each. Find the probability that Anton and Dmitry will be on the same team.

Let's form teams by sequentially placing the players in empty places, while starting with Anton and Dmitry. First, let's place Anton on a randomly selected place out of 33 free places. Now we put Dmitry on an empty place (we will consider the choice of a place for him as the outcome). There are 32 free places in total (one has already been taken by Anton), so there are 32 possible outcomes in total. There are 10 free places left in the same team with Anton, so the event "Anton and Dmitry in the same team" is favored by 10 outcomes. The probability of this event is .

Answer: 0.3125.

Task 13. The mechanical clock with a twelve-hour dial broke at some point and stopped working. Find the probability that the hour hand is frozen when it reaches 11 but does not reach 2 o'clock.

Conventionally, the dial can be divided into 12 sectors located between the marks of neighboring numbers (between 12 and 1, 1 and 2, 2 and 3, ..., 11 and 12). We will consider the stop of the hour hand in one of the indicated sectors as the outcome. In total there are 12 equally possible outcomes. This event is favored by three outcomes (sectors between 11 and 12, 12 and 1, 1 and 2). The desired probability is equal to .

Answer: 0.25.

Summarize

After studying the material on solving simple problems in probability theory, I recommend completing tasks for an independent solution, which we publish on our Telegram channel. You can also check the correctness of their implementation by entering your answers in the proposed form.