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14. Asymmetric coin. (From 9th grade, 2 points) Billy Bones has two coins - a gold one and a silver one. One of them is symmetric, and the other is not. It is unknown which coin is asymmetric, but it is known that the asymmetric coin lands heads with a probability of \( p = 0.6 \).
Billy Bones tossed the gold coin, an... | Solution. Let's introduce notations for the events:
$$
A=\{\text { the gold coin is biased }\},
$$
$B=\left\{\begin{array}{l}\text { when the gold coin is tossed, heads appear immediately, } \\ \text { and when the silver coin is tossed, heads appear on the second attempt. }\end{array}\right\}$
We need to find the c... | \frac{5}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,969 |
19. How many airplanes? (11th grade, 8 points) The work of the Absent-Minded Scientist involves long business trips, and therefore he often flies on the same airline. This airline has many identical airplanes, and all of them have names. Since the Scientist does not fly every day or even every week, it can be assumed t... | Solution. Consider the random variable $X$ "the ordinal number of the flight when the Scientist first gets a plane he has flown before". We will make a point estimate of the number of planes using the method of moments. For this, we need to solve the equation $\mathrm{E} X=15$ at least approximately.
Let's find $\math... | 134 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,970 |
7. Solution. To increase the average score in both groups, it is necessary to transfer students from group A to group B whose average score is higher than the average in group B but lower than the average in group A. Two students have a score higher than the average in B but lower than the average in A: Lopatin and Fil... | Answer: It is necessary to transfer Lapatin and Fylin from Group A to Group B.
Note: The solution may not be the only one.
| Evaluation Criterion | Score |
| :--- | :---: |
| The solution is complete and correct | 2 |
| The solution is basically correct, but it is not shown that both students need to be transferred, ... | ItisnecessarytotransferLapatinFylinfromGroupAtoGroupB | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,971 |
7. Boy and girl. Let's assume that the birth of a girl and a boy is equally likely. It is known that in a certain family there are two children.
a) What is the probability that one is a boy and one is a girl
b) It is additionally known that one of the children is a boy. What is the probability now that there is one b... | Solution. a) Children appear in a certain sequence (MM, MD, DM, or DD). All sequences are equally likely, and the probability of each is $\frac{1}{4}$. The condition "Boy and girl" is favorable to two outcomes, MD and DM, so the probability of this is $\frac{2}{4}=\frac{1}{2}$.
b) Now it is known that one of the child... | \frac{1}{2};\frac{2}{3};\frac{14}{27} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,972 |
4. For a procrastinator, the waiting time will increase by $a$ minutes.
Therefore, the total wasted time of all standing in the queue will decrease by $b-a$ minutes. We will thus swap people in pairs of "Procrastinator-Hurry" until we get a queue where all the hurries stand before all the procrastinators. In this queu... | Answer: a) 40 minutes and 100 minutes; b) 70 minutes.
Comment. The expected time turned out to be the arithmetic mean between the maximum and minimum possible. Try to prove the fact $\mathrm{E} T=\frac{T_{\min }+T_{\max }}{2}$ in general, that is, prove the combinatorial identity
$$
C_{n+m}^{2} \cdot \frac{b m+a n}{m... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,974 |
14. Pills (from 9th grade. 2 points). The Absent-Minded Scientist had a sore knee. The doctor prescribed him 10 pills for his knee: to take one pill daily. The pills help in $90 \%$ of cases, and in $2 \%$ of cases, there is a side effect - absent-mindedness disappears if it was present.
Another doctor prescribed the ... | Solution. Let's consider the events $R$ "The scientist took the pills for absent-mindedness", $A$ "The knee stopped hurting" and $B$ "Absent-mindedness disappeared".
We need to find the conditional probability $\mathrm{P}(R \mid A \cap B)$:
$$
\begin{gathered}
\mathrm{P}(R \mid A \cap B)=\frac{\mathrm{P}(R \cap A \ca... | 0.69 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,975 |
15. Mysterious Animal (from 9th grade. 2 points). Exactly half of the population of the Island of Misfortune are rabbits, and all the others are hares. If a resident of the Island of Misfortune asserts something, they always sincerely believe what they say. However, hares are honestly mistaken on average in one out of ... | Solution. Let's consider the events $A$ "The animal is a hare", $B$ "The animal claimed it is not a hare", $C$ "The animal claimed it is not a rabbit".
We need to find the conditional probability $\mathrm{P}(A \mid B \cap C)$:
$$
\begin{aligned}
\mathrm{P}(A \mid B \cap C) & =\frac{\mathrm{P}(A \cap B \cap C)}{\mathr... | 0.458 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,976 |
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
, and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\ba... | 40 | Other | math-word-problem | Yes | Yes | olympiads | false | 10,977 |
7. What? Where? When? The game wheel in the TV quiz is
divided into 13 identical sectors. The sectors are numbered from 1 to 13. At the beginning of the game, each sector contains an envelo... | Solution. a) The probabilities are equal, since no sector has an advantage over another.
b) Consider the sequence of sectors $n_{1}, n_{2}, \ldots, n_{6}$ that fell when the spinner was spun. The event $A$ "sectors №1 - №6 have been played" is favorable to all sequences where $n_{k} \leq k$, and all permutations of ea... | 0.00348 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,978 |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,980 |
15. Random Vectors. There are $n$ random vectors of the form $\left(y_{1}, y_{2}, y_{3}\right)$, where exactly one random coordinate is equal to 1, and the others are 0. They are added together. The result is a random vector $\vec{a}$ with coordinates $\left(Y_{1}, Y_{2}, Y_{3}\right)$.
a) (from 9th grade. 2 points). ... | Solution. a) Let's find the expected value $\mathrm{E} Y_{j}^{2}$. The quantity $Y_{j}$ is the number of ones among the numbers $y_{j}$, that is, this quantity is distributed according to the binomial law with probability $p=1 / 3$ and the number of trials $n$. Therefore,
$$
\mathrm{E} Y_{j}=n \cdot \frac{1}{3}=\frac{... | \frac{2n+n^{2}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,981 |
9. Solution. Suppose Olga Pavlovna has \( x \) liters of jam left, and Maria Petrovna has \( y \) liters of jam left. The numbers \( x \) and \( y \) are randomly and independently chosen from the interval from 0 to 1. We will consider that a random point with coordinates \((x; y)\) is selected from the unit square \( ... | Answer: 0.375.
Note. Other solution methods are possible.
## Grading Criteria
| Solution is complete and correct | 3 points |
| :--- | :--- |
| The solution contains correct reasoning for individual cases and an enumeration of these cases, and the formula for total probability is applied to them. However, the answer... | 0.375 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,982 |
3. Overcrowded carriages. The histogram (Fig. 1) shows the distribution of passenger carriages by the number of passengers (in $4 \%$ of carriages, passengers are from 10 to 19, in $6 \%$ of carriages - from 20 to 29, etc.). If a carriage has 60 passengers or more, it is called overcrowded.
a) (for 6th grade, 1 point)... | Answer: a) $40 \%$; b) approx. $49 \%$; c) cannot.
Solution. b) To make the share of passengers in overcrowded cars as small as possible, there should be as few passengers as possible in them, and as many as possible in non-overcrowded cars. Let the total number of cars be $N$. We find the maximum possible number of p... | )40;b)\approx49;) | Other | math-word-problem | Yes | Yes | olympiads | false | 10,984 |
7. Highway (from 7th grade, 3 points). A highway running from west to east intersects with $n$ equal roads, numbered from 1 to $n$ in order. Cars travel on these roads from south to north and from north to south. The probability that a car will approach the highway from each of these roads is $\frac{1}{n}$. Similarly, ... | Answer: $\frac{2 k n-2 k^{2}+2 k-1}{n^{2}}$.
Solution. The specified event will occur in one of three cases. | \frac{2kn-2k^{2}+2k-1}{n^{2}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,986 |
12. Stefan Banach ${ }^{1}$ and matches (8th grade, 4 points). An incredible legend says that one day in a small shop in the center of Lviv, Stefan Banach bought two boxes of matches and put them in his jacket pocket. Each box contained 60 matches.
When Banach needed a match, he would randomly pick a box and take a ma... | Answer: approximately 7.795.
Solution proposed by Olympiad participant Alexandra Nesterenko. Let there be $n$ matches in the box. Banach took a box (let's call it red for definiteness) and found it to be empty. Banach knows that he took the red box $n$ times, and the second (blue) box he could have taken $k$ times, wh... | 7.795 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,987 |
7. What? Where? When? The game wheel in the TV quiz is
divided into 13 identical sectors. The sectors are numbered from 1 to 13. At the beginning of the game, each sector contains an envelo... | Solution. a) The probabilities are equal, since no sector has an advantage over another.
b) Consider the sequence of sectors $n_{1}, n_{2}, \ldots, n_{6}$ that fell when the spinner was spun. Event $A$ "sectors №1 - №6 have been played" is favorable to all sequences where $n_{k} \leq k$, and all permutations of each s... | 0.00348 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,988 |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,990 |
4. Cities of Anchuria (from 6th grade, 1 point). In Anchuria, there is only one river, Rio Blanco, which originates somewhere in the mountains and flows into the ocean, and there are only five cities: San Mateo, Alasan, Coralio, Alforan, and Solitas.
In the map of Anchuria, the city names are missing, and the cities a... | Solution. Alasan is the only city not located on the banks of the Rio Blanco, but gets its water from wells. Therefore, Alasan has the number 2.
The lower a city is along the river, the lower its elevation above sea level. Only the "Elevation above sea level" value is required from all available statistics.
Answer: 1... | 1-Alforan,2-Alasan,3-Solitas,4-San-Mateo,5-Coralio | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,991 |
6. Experimental Textbook (from 6th grade, 2 points). In the testing of a new mathematics textbook, parallel seventh-grade classes from two schools, approximately equal in number, participated in the experiment. In each school, part of the seventh-graders studied using the experimental textbook, while the other part use... | Solution. The Absent-Minded Scientist is right. For simplicity, let's assume that each school has the same number of seventh graders, specifically $100 x$ people (the multiplier 100 is taken to make it easier to calculate percentages). Let's combine both tables, but write the data not in percentages, but in absolute nu... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 10,992 |
16. Mailboxes. In a newly inhabited apartment building, workers installed a block of 80 mailboxes. Inside each mailbox, they placed a ring with a key and a tag with the apartment number, but they did not bother to ensure that each mailbox contained the key to that mailbox. They simply threw the keys into the mailboxes ... | Solution. We will number the mailboxes and keys in the same way as the apartments. We need to show that sooner or later the Scholar will take out key 37 from mailbox 37.
The sequence of the numbers of the extracted keys cannot be infinite, since there are only 80 apartments. The sequence must end with the number 37. I... | 4.965 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,993 |
8. Magic Pies (6th - 11th grades, 1 point). Alice has six magic pies in her pocket - two enlarging (eat one and you grow), and the rest are shrinking (eat one and you shrink). When Alice met Mary Ann, she took out three pies from her pocket without looking and gave them to Mary. Find the probability that one of the gir... | Solution. One of the increasing pies (IncP), undoubtedly, ended up with one of the girls. We need to find the probability that the second IncP also ended up with her. This girl, after the distribution, has two more pies out of the remaining five, in addition to the first IncP. Therefore, the probability that the second... | 0.4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,995 |
11. Large Cube. A cube is assembled from 27 playing dice. a) (6 - 11 grades, 1 point) Find the probability that exactly 25 sixes are on the surface of the cube.
b) (7 - 11 grades, 1 point) Find the probability that at least one one is on the surface of the cube.
c) (8 - 11 grades, **1** point) Find the expected value... | Solution. a) One of the 27 cubes is in the center and therefore not visible at all. The other 26 cubes are visible. Thus, the required event
$$
A=\{25 \text { sixes }\}
$$
consists of all the cubes showing sixes outward, except for one - let's call it the special cube.
Consider all the cubes. If a cube is in the cen... | )\frac{31}{2^{13}\cdot3^{18}}\approx9.77\cdot10^{-12};b)1-\frac{5^{6}}{2^{2}\cdot3^{18}}\approx0.99998992;)9;)189;e)6-\frac{5^{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,996 |
18. (10-11 grades, 5 points) Favorite Pair. On the drying rack, in a random order (as taken out of the washing machine), there are p socks. Among them are two favorite socks of the Absent-Minded Scientist. The socks are hidden by a drying sheet, so the Scientist cannot see them and takes one sock at a time by feel. Fin... | Solution. It is convenient to form a triangular table. The shaded cells correspond to pairs of favorite socks. For example, the pair $(2; 4)$, marked with an "X", corresponds to the case where the first favorite sock was the second one, and the second one was the fourth in the sequence. All pairs are equally likely, an... | \frac{2(n+1)}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,997 |
19. First pair. (10-11 grades, 6 points). On the drying line, in a random order (as taken out of the washing machine), p pairs of socks are hanging. There are no two identical pairs. The socks are hanging behind a drying sheet, so the Absent-Minded Scientist takes one sock at a time by feel and compares each new sock w... | Solution. Let $\xi_{n}$ be a random variable equal to the number of socks removed under the condition that $n$ pairs are hanging on the line. Obviously, $E \xi_{1}=2$. Now let $n>1$. Let's try to establish a recurrence relation between $\xi_{n}$ and $\xi_{n-1}$. Number the socks in the order in which the Absent-Minded ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,998 |
4. Contest (from 6th grade, 1 point). In one social network, a photo contest was held. Several photos were submitted to the contest, and each participant could rate each photo by giving it either 0 (do not like), 1 (not very like), or 2 (like very much).
Two categories were announced: the most attractive photo, which ... | Solution. We will build a table showing how this can happen with just two photos and three voters.
| | Photo 1 | Photo 2 |
| :--- | :---: | :---: |
| 1st voter | 2 | 1 |
| 2nd voter | 2 | 1 |
| 3rd voter | 0 | 1 |
Answer: yes, this is possible | yes,thisispossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,999 |
5. Anniversary of the Meeting (from 6th grade, 2 points). As is known, Robinson Crusoe's friend and companion was named Friday because they met on a Friday. This happened in January, but it is impossible to determine the exact year. Robinson is only sure that it could have happened in any year with equal probability, s... | Solution. It is necessary to take into account that years can be common and leap years (if the year number is divisible by 4). A common year has 365 days, and a leap year has 366. When dividing by 7, the numbers 365 and 366 give remainders of 1 and 2, respectively. Therefore, if an event falls on a Friday in January of... | on\\Friday\with\\probability\of\0.75\or\on\\Thursday\with\\probability\of\0.25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,000 |
8. Cards and PIN codes. Once, a pickpocket named Brick ${ }^{1}$ stole a wallet, which contained four credit cards and a note with four PIN codes for these cards. Brick does not know which PIN code corresponds to which card. If the wrong PIN code is entered three times for any card, the card will be blocked.
a) (from ... | Solution. a) Taking the first code and trying it on four cards in turn, Brick will find the card to which this code fits. Taking the second code and trying it on the three remaining cards, he will find the second match. Then - the third. Thus, he will find codes for three cards. The fourth one may turn out to be blocke... | \frac{23}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,001 |
19. Essay on Suitcases (from 10th grade, 6 points). In the book by Szczepan Jelenski ${ }^{7}$, there is a small essay about suitcases.
A department store received 10 suitcases and 10 keys in a separate envelope, with the warning that each key opens only one suitcase and that a suitable key can be found for each suitc... | Solution. We will solve the problem in a general form, assuming that there are $n$ suitcases. Let $Y$ be a random variable equal to the number of attempts to open the first of $n$ suitcases. Obviously, $Y=1$ with probability $1 / n$, $Y=2$ with probability $\frac{n-1}{n} \cdot \frac{1}{n-1}=\frac{1}{n}$, and so on, up ... | 29.62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,002 |
# 9. Solution.
1st method. An elementary outcome in the random experiment is a triplet of positions where the children in green caps stand. Consider the event $A$ "all three green caps are together." This event is favorable in 9 elementary outcomes. The event $B$ "two green caps are together, and the third is separate... | Answer: $\frac{5}{14}$.
| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| The chosen method is clear from the solution, but the probability of the event "not all three together" is correctly found | 2 |
| The chosen method is clear from the solution, but only the probability ... | \frac{5}{14} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,003 |
14. Pills (from 9th grade. 2 points). The Absent-Minded Scientist had a sore knee. The doctor prescribed him 10 pills for his knee: to take one pill daily. The pills help in $90 \%$ of cases, and in $2 \%$ of cases, there is a side effect - absent-mindedness disappears if it was present.
Another doctor prescribed the ... | Solution. Let's consider the events $R$ "The scientist took the pills for absent-mindedness", $A$ "The knee stopped hurting" and $B$ "Absent-mindedness disappeared".
We need to find the conditional probability $\mathrm{P}(R \mid A \cap B)$:
$$
\begin{gathered}
\mathrm{P}(R \mid A \cap B)=\frac{\mathrm{P}(R \cap A \ca... | 0.69 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,004 |
15. Mysterious Animal (from 9th grade. 2 points). Exactly half of the population of the Island of Misfortune are rabbits, and all the others are hares. If a resident of the Island of Misfortune asserts something, they always sincerely believe what they say. However, hares are honestly mistaken on average in one out of ... | Solution. Let's consider the events $A$ "The animal is a hare", $B$ "The animal claimed it is not a hare", $C$ "The animal claimed it is not a rabbit".
We need to find the conditional probability $\mathrm{P}(A \mid B \cap C)$:
$$
\begin{aligned}
\mathrm{P}(A \mid B \cap C) & =\frac{\mathrm{P}(A \cap B \cap C)}{\mathr... | 0.458 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,005 |
8. Solution. We will show by contradiction that it is impossible to make such coins. Suppose the minters succeeded. Let the probability of heads on the first coin be $p_{1}$, and on the second coin be $p_{2}$. Then we get:
$$
\left(1-p_{1}\right)\left(1-p_{2}\right)=p_{1} p_{2}=p_{1}\left(1-p_{2}\right)+p_{2}\left(1-p... | Answer: cannot.
| Evaluation Criteria | Score |
| :--- | :--- |
| Complete and correct solution | 3 points |
| The solution assumes that $p_{1}=p_{2}$. The solution is correct under this assumption | 2 points |
| The system of equations is correctly set up, but the solution is not completed or contains an error | 1 po... | proof | Algebra | proof | Yes | Yes | olympiads | false | 11,008 |
7. Boy and girl. Let's assume that the birth of a girl and a boy is equally likely. It is known that in a certain family there are two children.
a) What is the probability that one is a boy and one is a girl
b) It is additionally known that one of the children is a boy. What is the probability now that there is one b... | Solution. a) Children appear in a certain sequence (MM, MD, DM, or DD). All sequences are equally likely, and the probability of each is $\frac{1}{4}$. The condition "Boy and girl" is favorable to two outcomes, MD and DM, so the probability of this is $\frac{2}{4}=\frac{1}{2}$.
b) Now it is known that one of the child... | \frac{1}{2};\frac{2}{3};\frac{14}{27} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,009 |
18. Queue at the bank. If one person spends one minute waiting in the queue, we will say that one person-minute is wasted. In the bank queue, there are eight people, five of whom plan to perform simple operations taking 1 minute, while the rest plan to perform long operations taking 5 minutes. Find:
a) the minimum and... | Solution. Again, we will solve the problem in a general case. Let a short operation take $a$ minutes, and a long one - $b$ minutes, with $a<b$. For brevity, we will call a customer planning a simple operation a "hurry-up," and one who intends to take a long time - a "slacker." Suppose there are $n$ hurry-ups and $m$ sl... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,011 |
7. What? Where? When? The game wheel in the TV quiz is
divided into 13 identical sectors. The sectors are numbered from 1 to 13. At the beginning of the game, each sector contains an envelo... | Solution. a) The probabilities are equal, since no sector has an advantage over another.
b) Consider the sequence of sectors $n_{1}, n_{2}, \ldots, n_{6}$ that fell when the spinner was spun. Event $A$ "sectors №1 - №6 have been played" is favorable to all sequences where $n_{k} \leq k$, and all permutations of each s... | 0.00348 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,012 |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,014 |
14. Pills (from 9th grade. 2 points). The Absent-Minded Scientist had a sore knee. The doctor prescribed him 10 pills for his knee: to take one pill daily. The pills help in $90 \%$ of cases, and in $2 \%$ of cases, there is a side effect - absent-mindedness disappears if it was present.
Another doctor prescribed the ... | Solution. Let's consider the events $R$ "The scientist took the pills for absent-mindedness", $A$ "The knee stopped hurting" and $B$ "Absent-mindedness disappeared".
We need to find the conditional probability $\mathrm{P}(R \mid A \cap B)$:
$$
\begin{gathered}
\mathrm{P}(R \mid A \cap B)=\frac{\mathrm{P}(R \cap A \ca... | 0.69 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,015 |
15. Mysterious Animal (from 9th grade. 2 points). Exactly half of the population of the Island of Misfortune are rabbits, and all the others are hares. If a resident of the Island of Misfortune asserts something, they always sincerely believe what they say. However, hares are honestly mistaken on average in one out of ... | Solution. Let's consider the events $A$ "The animal is a hare", $B$ "The animal claimed it is not a hare", $C$ "The animal claimed it is not a rabbit".
We need to find the conditional probability $\mathrm{P}(A \mid B \cap C)$:
$$
\begin{aligned}
\mathrm{P}(A \mid B \cap C) & =\frac{\mathrm{P}(A \cap B \cap C)}{\mathr... | 0.458 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,016 |
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
, and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\ba... | 40 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,017 |
9. Solution. Suppose Olga Pavlovna has \( x \) liters of jam left, and Maria Petrovna has \( y \) liters of jam left. The numbers \( x \) and \( y \) are randomly and independently chosen from the interval from 0 to 1. We will consider that a random point with coordinates \((x; y)\) is selected from the unit square \( ... | Answer: 0.375.
Note. Other solution methods are possible.
## Grading Criteria
| Solution is complete and correct | 3 points |
| :--- | :--- |
| The solution contains correct reasoning for individual cases and an enumeration of these cases, and the formula for total probability is applied to them. However, the answer... | 0.375 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,018 |
4. Stem-and-leaf plot. (From 6th grade, 2 points). To represent whole numbers or decimal fractions, a special type of diagram called a "stem-and-leaf plot" is often used. Such diagrams are convenient for representing people's ages. Suppose that in the studied group, there are 5 people aged 19, 34, 37, 42, and 48. For t... | Solution. The digits from 0 to 5, representing decades of years, can be placed immediately (Fig. 4a). It is clear that less than 10 years have passed, otherwise there would be no digits in line "1".
If 7 or more years had passed, then the person who is 13 years old would have moved to line "2", and there would be fewe... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,020 |
6. Median of the sum of numerical sets. Let there be two numerical sets:
$$
X=\left\{x_{1}, x_{2}, \ldots, x_{n}\right\} \text { and } Y=\left\{y_{1}, y_{2}, \ldots, y_{m}\right\}
$$
The first set has $n$ numbers, and the second set has $m$ numbers. The direct sum or simply the sum $X \oplus Y$ of these sets is the s... | Solution. a) For example, two identical sets: $X=\{0,0,1\}$ and $Y=\{0,0,1\}$. The median of each is 0, and the sum of these sets $\{0,0,0,0,1,1,1,1,2\}$ has a median of 1.
b) We will show that such sets do not exist. Let the first set $X$ consist of numbers $x_{1}, x_{2}, \ldots, x_{n}$, and the second set $Y$ consis... | ){0,0,1} | Other | math-word-problem | Yes | Yes | olympiads | false | 11,021 |
11. The Collector's Two Tasks. A chocolate egg manufacturer with a toy inside announced the release of a new collection called "The Nile Family," featuring ten different charming crocodiles. The crocodiles are evenly and randomly distributed among the chocolate eggs, meaning that in a randomly selected egg, each crocod... | Solution. a) Let $B_{k}$ be the event "at the moment when the last crocodile is acquired for the first collection, the second collection is missing exactly $k$ crocodiles." We need to show that
$$
p_{1}=\mathrm{P}\left(B_{1}\right)=\mathrm{P}\left(B_{2}\right)=p_{2}
$$
Let $A_{j, k}$ be the event "at some point, the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,022 |
14. Asymmetric coin. (From 9th grade, 2 points) Billy Bones has two coins - a gold one and a silver one. One of them is symmetric, and the other is not. It is unknown which coin is asymmetric, but it is known that the asymmetric coin lands heads with a probability of \( p = 0.6 \).
Billy Bones tossed the gold coin, an... | Solution. Let's introduce notations for the events:
$$
A=\{\text { the gold coin is biased }\},
$$
$B=\left\{\begin{array}{l}\text { when the gold coin is tossed, heads appear immediately, } \\ \text { and when the silver coin is tossed, heads appear on the second attempt. }\end{array}\right\}$
We need to find the c... | \frac{5}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,023 |
3. Overcrowded carriages. The histogram (Fig. 1) shows the distribution of passenger carriages by the number of passengers (in $4 \%$ of carriages, passengers are from 10 to 19, in $6 \%$ of carriages - from 20 to 29, etc.). If a carriage has 60 passengers or more, it is called overcrowded.
a) (for 6th grade, 1 point)... | Answer: a) $40 \%$; b) approx. $49 \%$; c) cannot.
Solution. b) To make the share of passengers in overcrowded cars as small as possible, there should be as few passengers as possible in them, and as many as possible in non-overcrowded cars. Let the total number of cars be $N$. We find the maximum possible number of p... | )40;b)\approx49;) | Other | math-word-problem | Yes | Yes | olympiads | false | 11,024 |
7. Highway (from 7th grade, 3 points). A highway running from west to east intersects with $n$ equal roads, numbered from 1 to $n$ in order. Cars travel on these roads from south to north and from north to south. The probability that a car will approach the highway from each of these roads is $\frac{1}{n}$. Similarly, ... | Answer: $\frac{2 k n-2 k^{2}+2 k-1}{n^{2}}$.
Solution. The specified event will occur in one of three cases. | \frac{2kn-2k^{2}+2k-1}{n^{2}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,026 |
12. Stefan Banach ${ }^{1}$ and matches (8th grade, 4 points). An incredible legend says that one day in a small shop in the center of Lviv, Stefan Banach bought two boxes of matches and put them in his jacket pocket. Each box contained 60 matches.
When Banach needed a match, he would randomly pick a box and take a ma... | Answer: approximately 7.795.
Solution proposed by Olympiad participant Alexandra Nesterenko. Let there be $n$ matches in the box. Banach took a box (let's call it red for definiteness) and found it to be empty. Banach knows that he took the red box $n$ times, and the second (blue) box he could have taken $k$ times, wh... | 7.795 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,027 |
7. Boy and girl. Let's assume that the birth of a girl and a boy is equally likely. It is known that in a certain family there are two children.
a) What is the probability that one is a boy and one is a girl
b) It is additionally known that one of the children is a boy. What is the probability now that there is one b... | Solution. a) Children appear in a certain sequence (MM, MD, DM, or DD). All sequences are equally likely, and the probability of each is $\frac{1}{4}$. The condition "Boy and girl" is favorable to two outcomes, MD and DM, so the probability of this is $\frac{2}{4}=\frac{1}{2}$.
b) Now it is known that one of the child... | \frac{1}{2};\frac{2}{3};\frac{14}{27} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,028 |
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young dr... | # Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,030 |
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean ... | # Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now give... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,031 |
9. Steel doors (from 9th grade. 2 points). At the factory named after Sailor Zheleznyak, they manufacture rectangles with a length of 2 m and a width of 1 m. The length is measured by worker Ivanov, and the width, independently of Ivanov, is measured by worker Petrov. Both have an average error of zero, but Ivanov has ... | # Solution.
a) Let $X$ be the width and $Y$ be the length of the cut rectangle in meters. According to the problem, $\mathrm{E} X=2$, $\mathrm{E} Y=1$. Since the measurements are independent, $\mathrm{E}(X Y)=\mathrm{E} X \cdot \mathrm{E} Y=2$ (sq.m.).
b) From the condition, it follows that $\mathrm{D} X=0.003^{2}=9 ... | )2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,032 |
15. English Club (from 9th grade. 6 points). Every Friday, ten gentlemen come to the club, and each hands their hat to the doorman. Each hat fits its owner perfectly, but no two hats are the same size. The gentlemen leave one by one in a random order.
When seeing off each gentleman, the doorman tries to put on the fir... | Solution. Let the number of gentlemen be $n$. We will number them in the order of increasing sizes of their hats from 1 to $n$. No hats will be left only if each took his own hat. Let the probability of this be $p_{n}$. If the $k$-th gentleman leaves first (the probability of this is $\frac{1}{n}$), then the probabilit... | 0.000516 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,033 |
# 9. Solution.
1st method. An elementary outcome in a random experiment is a triplet of positions where children in red caps stand. Consider the event $A$ "all three red caps are next to each other." This event is favorable in 10 elementary outcomes. The event $B$ "two red caps are next to each other, and the third is... | Answer: $\frac{7}{12}$.
| Evaluation Criteria | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| The chosen method is clear from the solution, but only the probability of the event "exactly two together" is correctly found | 2 |
| The chosen method is clear from the solution, but only the probabilit... | \frac{7}{12} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,034 |
# 9. Solution.
1st method. An elementary outcome in the random experiment is a triplet of positions where the children in green caps stand. Consider the event $A$ "all three green caps are together." This event is favorable in 9 elementary outcomes. The event $B$ "two green caps are together, and the third is separate... | Answer: $\frac{5}{14}$.
| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| The chosen method is clear from the solution, but the probability of the event "not all three together" is correctly found | 2 |
| The chosen method is clear from the solution, but only the probability ... | \frac{5}{14} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,035 |
4. Contest (from 6th grade, 1 point). In one social network, a photo contest was held. Several photos were submitted to the contest, and each participant could rate each photo by giving it either 0 (dislike), 1 (somewhat dislike), or 2 (like very much).
Two categories were announced: the most attractive photo, which r... | Solution. We will build a table showing how this can happen with just two photos and three voters.
| | Photo 1 | Photo 2 |
| :--- | :---: | :---: |
| 1st voter | 2 | 1 |
| 2nd voter | 2 | 1 |
| 3rd voter | 0 | 1 |
Answer: yes, this is possible | yes,thisispossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,036 |
8. Cards and PIN codes. Once, a pickpocket named Brick ${ }^{1}$ stole a wallet, which contained four credit cards and a note with four PIN codes for these cards. Brick does not know which PIN code corresponds to which card. If the wrong PIN code is entered three times for any card, the card will be blocked.
a) (from ... | Solution. a) Taking the first code and trying it on four cards in turn, Brick will find the card to which this code fits. Taking the second code and trying it on the three remaining cards, he will find the second match. Then - the third. Thus, he will find codes for three cards. The fourth one may turn out to be blocke... | \frac{23}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,037 |
11. Large Cube. A cube is assembled from 27 playing dice. a) (6 - 11 grades, 1 point) Find the probability that exactly 25 sixes are on the surface of the cube.
b) (7 - 11 grades, 1 point) Find the probability that at least one one is on the surface of the cube.
c) (8 - 11 grades, **1** point) Find the expected numbe... | Solution. a) One of the 27 cubes is in the center and therefore not visible at all. The other 26 cubes are visible. Thus, the required event
$$
A=\{25 \text { sixes }\}
$$
consists of all the cubes showing sixes outward, except for one - let's call it the special cube.
Consider all the cubes. If a cube is in the cen... | )\frac{31}{2^{13}\cdot3^{18}}\approx9.77\cdot10^{-12};b)1-\frac{5^{6}}{2^{2}\cdot3^{18}}\approx0.99998992;)9;)189;e)6-\frac{5^{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,041 |
18. (10-11 grades, 5 points) Favorite Pair. On the drying rack, in a random order (as taken out of the washing machine), there are p socks. Among them are two favorite socks of the Absent-Minded Scientist. The socks are hidden by a drying sheet, so the Scientist cannot see them and takes one sock at a time by feel. Fin... | Solution. It is convenient to form a triangular table. The shaded cells correspond to pairs of favorite socks. For example, the pair $(2; 4)$, marked with an "X", corresponds to the case where the first favorite sock was the second one, and the second one was the fourth in the sequence. All pairs are equally likely, an... | \frac{2(n+1)}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,042 |
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean ... | # Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now give... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,043 |
9. Steel doors (from 9th grade. 2 points). At the factory named after Sailor Zheleznyak, they manufacture rectangles with a length of 2 m and a width of 1 m. The length is measured by worker Ivanov, and the width, independently of Ivanov, is measured by worker Petrov. Both have an average error of zero, but Ivanov has ... | # Solution.
a) Let $X$ be the width and $Y$ be the length of the cut rectangle in meters. According to the problem, $\mathrm{E} X=2$, $\mathrm{E} Y=1$. Since the measurements are independent, $\mathrm{E}(X Y)=\mathrm{E} X \cdot \mathrm{E} Y=2$ (sq.m.).
b) From the condition, it follows that $\mathrm{D} X=0.003^{2}=9 ... | )2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,044 |
15. English Club (from 9th grade. 6 points). Every Friday, ten gentlemen come to the club, and each hands their hat to the doorman. Each hat fits its owner perfectly, but no two hats are the same size. The gentlemen leave one by one in a random order.
When seeing off each gentleman, the doorman tries to put on the fir... | Solution. Let the number of gentlemen be $n$. We will number them in the order of increasing sizes of their hats from 1 to $n$. No hats will be left only if each took his own hat. Let the probability of this be $p_{n}$. If the $k$-th gentleman leaves first (the probability of this is $\frac{1}{n}$), then the probabilit... | 0.000516 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,045 |
4. Cities of Anchuria (from 6th grade, 1 point). In Anchuria, there is only one river, Rio-Blanco, which originates somewhere in the mountains and flows into the ocean, and there are only five cities: San-Mateo, Alasan, Coralio, Alforan, and Solitas.
In the map of Anchuria, the city names are not shown, and the cities... | Solution. Alasan is the only city not located on the banks of the Rio Blanco, but gets its water from wells. Therefore, Alasan has the number 2.
The lower a city is along the river, the lower its elevation above sea level. Only the "Elevation above sea level" value is required from all available statistics.
Answer: 1... | 1-Alforan,2-Alasan,3-Solitas,4-San-Mateo,5-Coralio | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,046 |
10. Product of Digits (7th grade, 3 points). Nезнайка (Nезнayka) solved the following problem. "Given a set of 20 random digits, find the probability that the product of these digits ends in 0." Nезнайка reasoned as follows.
If the set contains the digit 0, then the product of the digits will definitely end in 0. The ... | Solution. There is an error in Little Don't-Know's reasoning: if there is an even digit in the set, it does not mean that there is only one. If there are several even digits, then a five cannot stand on any of the positions occupied by these digits, so the conditional probability that there is a five depends on how man... | 0.988 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,047 |
4. Contest (from 6th grade, 1 point). In one social network, a photo contest was held. Several photos were submitted to the contest, and each participant could rate each photo by giving it either 0 (do not like), 1 (not very like), or 2 (like very much).
Two categories were announced: the most attractive photo, which ... | Solution. We will build a table showing how this can happen with just two photos and three voters.
| | Photo 1 | Photo 2 |
| :--- | :---: | :---: |
| 1st voter | 2 | 1 |
| 2nd voter | 2 | 1 |
| 3rd voter | 0 | 1 |
Answer: yes, this is possible | yes,thisispossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,048 |
8. Cards and PIN codes. Once, a pickpocket named Brick ${ }^{1}$ stole a wallet, which contained four credit cards and a note with four PIN codes for these cards. Brick does not know which PIN code corresponds to which card. If the wrong PIN code is entered three times for any card, the card will be blocked.
a) (from ... | Solution. a) Taking the first code and trying it on four cards in turn, Brick will find the card to which this code fits. Taking the second code and trying it on the three remaining cards, he will find the second match. Then - the third. Thus, he will find codes for three cards. The fourth one may turn out to be blocke... | \frac{23}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,049 |
19. Essay on Suitcases (from 10th grade, 6 points). In the book by Szczepan Jelenski ${ }^{7}$, there is a small essay about suitcases.
A department store received 10 suitcases and 10 keys in a separate envelope, with the warning that each key opens only one suitcase and that a suitable key can be found for each suitc... | Solution. We will solve the problem in a general form, assuming that there are $n$ suitcases. Let $Y$ be a random variable equal to the number of attempts to open the first of $n$ suitcases. Obviously, $Y=1$ with probability $1 / n$, $Y=2$ with probability $\frac{n-1}{n} \cdot \frac{1}{n-1}=\frac{1}{n}$, and so on, up ... | 29.62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,050 |
4. Three Rolls (from 6th grade. 1 point). A die is rolled three times. Which event is more likely:
$A$ "some number of points will fall at least twice" or
$B$ "three different numbers of points will fall on three rolls." | Solution. The probability of event $B$ is
$$
\frac{6 \cdot 5 \cdot 4}{6^{3}}=\frac{5}{9}>\frac{1}{2}
$$
Therefore, event $B$ is more likely than event $A=\bar{B}$.
Answer: $B$. | B | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,051 |
6. Stopped Clock (from 6th grade. 1 point). On the clock, there are two hands: the hour hand and the minute hand. At a random moment in time, the clock stopped. Find the probability that the angle between the hands on the stopped clock is acute. | Solution. Suppose the clock face also rotates at the speed of the hour hand so that the hour hand always points to 12 o'clock. Then the minute hand will form an acute angle if it is in the interval from 12 to 3 o'clock or from 9 o'clock to 12, that is, within six of the twelve hour intervals. The desired probability is... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,053 |
7. Letters (from 6th grade. 1 point). At the post office, a courier sends 100 letters. Letters weighing more than 100 g are not accepted. The cost of sending depends on the weight of the letter. The minimum cost (for letters weighing up to 20 g inclusive) is 22 rubles, and for each subsequent full or partial 20 g, an a... | Solution. Not correct. Suppose that two letters weigh 100 g, and all the others weigh 20 g. Then the first operator will calculate that the courier should pay $98 \cdot 22 + 2 \cdot 32 = 2220 \text{ p}$.
The second operator will find that the average weight of a letter is
$$
\frac{20 \cdot 98 + 100 \cdot 2}{100} = 21... | no,notcorrect | Other | math-word-problem | Yes | Yes | olympiads | false | 11,054 |
8. Reviews (from 7th grade. 1 point). Angry reviews about the work of an online store are left by $80 \%$ of dissatisfied customers (those who were poorly served in the store). Only $15 \%$ of satisfied customers leave positive reviews. A certain online store has received 60 angry and 20 positive reviews. Using this st... | Solution. Let $p$ be the probability that a customer is served well, and $q=1-p$ be the probability that they are served poorly. Then the probability that a customer leaves a good review is $0.15 p$, and the probability that there will be a bad review is $0.8(1-p)$. Then
$$
\frac{0.15 p}{0.8(1-p)} \approx \frac{1}{3},... | 0.64 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,055 |
12. Each subsequent number is chosen independently of the previous ones. If it matches any of the already chosen numbers, move clockwise to the first unchosen number.
In the end, $k$ distinct numbers are obtained.
a) (From 6th grade. 1 point). Is it true that the appearance of each specific number in such a sample is... | Solution. a) Yes, the probability of each number appearing is $k / n$ due to symmetry.
b) No. We will provide an example showing that different samples are not equally probable. Suppose that $k=2$. The sample (1;3) will appear with a probability of $\frac{2}{n^{2}}$ (verify this). The sample (1;2) can appear in three ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,056 |
10. Length of the river (from 8th grade. 2 points). The Anchurian Geographical Society (GSA) sent an expedition with the goal of measuring the length of the great Anchurian river Rio-Coralio. It turned out that the length of the riverbed from the source to the mouth is 402 km plus or minus 500 m, with a probability of ... | Solution. The scientist could reason as follows. Measurement errors in either direction are equally likely. Therefore, according to GSA data, the probability that the river is longer than 402.5 km is 0.02. Similarly, AWRA results indicate that with a probability of 0.02, the river's length is less than 402.5 km. As is ... | 402.5withanerrorprobabilityof0.04 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,057 |
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entra... | Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
. Ivan the Tsarevich is learning to shoot a bow. He put 14 arrows in his quiver and went to shoot at pine cones in the forest. He knocks down a pine cone with a probability of 0.1, and for each pine cone he knocks down, the Frog-Princess gives him 3 more arrows.... | Solution. First method. Let Ivan have $n$ arrows at the present moment. Let $X_{0}$ be the random variable "the number of shots needed to reduce the number of arrows by one." Ivan makes a shot. Consider the random variable - the indicator $I$ of a successful shot. $I=0$, if the shot is unsuccessful (probability of this... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,059 |
13. Random graph (from 8th grade. 3 points). On a plane, 20 points were pairwise connected by segments (edges) ${ }^{3}$ - each point with each other. Then, 35 edges were randomly selected and removed. Find the probability that the resulting graph is connected (i.e., from any point, you can reach any other point by edg... | Solution. A graph is disconnected if it can be divided into two groups of points, such that no point in the first group is connected by an edge to any point in the second group. Suppose the first group has $n$ points, where $2 \leq n \leq 10$, and the second group has the remaining $20-n$ points. The remaining edges af... | 1-\frac{20C_{171}^{16}}{C_{190}^{35}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,060 |
4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the ... | Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,061 |
15. Who is richer? (9th grade, 4 points). Cat Basilio and Fox Alisa have no solido left. And then they got hired to set up and remove the scenery in Carabas-Barabas's theater. Every evening after the performance, Carabas gives them a one-soldo coin, and a fight immediately breaks out between the cat and the fox.
With ... | Solution. Let $X_{n}$ be the random variable "the absolute difference between the state of the cat and the state of the fox at the end of the $n$-th day of operation." It is convenient to track not the absolute difference, but the signed difference. For definiteness, we will track the random variable "the difference be... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,062 |
16. Fifty cards (from 9th grade, 3 points). On the table, in a row, 50 cards numbered from 1 to 50 from left to right were laid out. The bottom of the cards is blue, and the top is red. Vasya randomly selects 25 consecutive cards (a segment of the row) and flips them. Then Asey approaches the table and does the same - ... | Solution. Randomly selecting a fragment from 25 cards means randomly selecting the first (leftmost) card in this fragment. With equal chances, this card can have a number from 1 to 26.
Consider card No. 1. The probability that Vasya will flip it is \( \frac{1}{26} \) (only if the fragment starts with card No. 1). The ... | 1338 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,063 |
17. Three Forecasts (9th grade. 4 points). In Anchuria, days can be either clear or rainy, with the probability of a clear day being 0.74 regardless of the time of year. In the Committee on Climate and Forecasts of the Anchurian Parliament, there are two senators and one invited meteorologist. It is known that the prop... | Solution. Let event $G$ be "The VNL day will be clear", event $M_{1}$ "the first senator gave a forecast of 'clear'", event $M_{2}$ "the second senator gave a forecast of 'clear'", and event $S$ "the meteorologist gave a forecast of 'rain'". Let
$$
\mathrm{P}\left(M_{1} \mid G\right)=\mathrm{P}\left(M_{2} \mid G\right... | themeteorologist'forecast | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,064 |
18. The Paradox of the Last Roll (from 9th grade. 4 points). A fair die is rolled until the sum of the points that fall in sequence reaches the number 2019 (becomes equal to 2019 or more). Prove that the probability of the event “6 points fall on the last roll” is greater than $1 / 6$ by a factor of ${ }^{5}$.
Comment... | Solution. In the conditions of the experiment, before the last roll, the sum of the points that fell in sequence $Y$ became equal to one of the numbers
$$
n-6, n-5, \ldots, n-1
$$[^4]
Let's introduce brief notations for the probabilities of these six events: $p_{n-k}=\mathrm{P}(Y=n-k)$. Obviously, $p_{n-1}+p_{n-2}+\l... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,065 |
19. Monochromatic pairs. In a box, there are $n$ white and $n$ black balls. Balls are drawn from the box in pairs - random pairs.
a) (from 9th grade. 2 points). Find the expected number of mixed-color pairs drawn from the box by the time it is empty.
b) (from 9th grade. 5 points). Suppose that if a pair is mixed-colo... | Solution. a) Let us mentally number all the black balls. Introduce a random variable $I_{k}$ - the indicator of the event $A_{k}$ "the $k$-th black ball has a white one in its pair", that is,
$$
I_{k}=\left\{\begin{array}{l}
0, \text { if event } A_{k} \text { does not occur, } \\
1, \text { if event } A_{k} \text { o... | 2n-H_n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,066 |
18. The diagram shows a track layout for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they like, returning to point $A$.
The young driver, Yu... | # Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,067 |
8. Reviews (from 7th grade. 1 point). Angry reviews about the work of an online store are left by $80 \%$ of dissatisfied customers (those who were poorly served in the store). Only $15 \%$ of satisfied customers leave positive reviews. A certain online store has received 60 angry and 20 positive reviews. Using this st... | Solution. Let $p$ be the probability that a customer is served well, and $q=1-p$ be the probability that they are served poorly. Then the probability that a customer leaves a good review is $0.15 p$, and the probability that there will be a bad review is $0.8(1-p)$. Then
$$
\frac{0.15 p}{0.8(1-p)} \approx \frac{1}{3},... | 0.64 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,068 |
10. Length of the river (from 8th grade. 2 points). The Anchurian Geographical Society (GSA) sent an expedition with the goal of measuring the length of the great Anchurian river Rio-Coralio. It turned out that the length of the riverbed from the source to the mouth is 402 km plus or minus 500 m, with a probability of ... | Solution. The scientist could reason as follows. Measurement errors in either direction are equally likely. Therefore, according to GSA data, the probability that the river is longer than 402.5 km is 0.02. Similarly, AWRA results indicate that with a probability of 0.02, the river's length is less than 402.5 km. As is ... | 402.5withanerrorprobabilityof0.04 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,069 |
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entra... | Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
. Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the ... | Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,071 |
# 4. Calculator (6-9).
a) (1 pt.) On the calculator keyboard, there are digits from 0 to 9 and symbols for two operations (see figure). Initially, the display shows the number 0. You can press any keys. The calculator performs operations in the sequence of key presses. If the operation symbol is pressed several times ... | Solution. a) Note that at least one addition operation is performed, even if the Scholar entered only one number - thereby adding this number to zero.
Let $A$ be the event "the result is odd". The parity of the result is determined by the last addend. Let's explain this in more detail.
Suppose the penultimate number ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,072 |
6. Median of the sum of numerical sets. Let there be two numerical sets:
$$
X=\left\{x_{1}, x_{2}, \ldots, x_{n}\right\} \text { and } Y=\left\{y_{1}, y_{2}, \ldots, y_{m}\right\}
$$
The first set has $n$ numbers, and the second set has $m$ numbers. The direct sum or simply the sum $X \oplus Y$ of these sets is the s... | Solution. a) For example, two identical sets: $X=\{0,0,1\}$ and $Y=\{0,0,1\}$. The median of each is 0, and the sum of these sets $\{0,0,0,0,1,1,1,1,2\}$ has a median of 1.
b) We will show that such sets do not exist. Let the first set $X$ consist of numbers $x_{1}, x_{2}, \ldots, x_{n}$, and the second set $Y$ consis... | ){0,0,1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,076 |
11. The Collector's Two Tasks. A chocolate egg manufacturer with a toy inside announced the release of a new collection called "The Nile Family," featuring ten different charming crocodiles. The crocodiles are evenly and randomly distributed among the chocolate eggs, meaning that in a randomly selected egg, each crocod... | Solution. a) Let $B_{k}$ be the event "at the moment when the last crocodile is acquired for the first collection, the second collection is missing exactly $k$ crocodiles." We need to show that
$$
p_{1}=\mathrm{P}\left(B_{1}\right)=\mathrm{P}\left(B_{2}\right)=p_{2}
$$
Let $A_{j, k}$ be the event "at some point, the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,077 |
14. Asymmetric coin. (From 9th grade, 2 points) Billy Bones has two coins - a gold one and a silver one. One of them is symmetric, and the other is not. It is unknown which coin is asymmetric, but it is known that the asymmetric coin lands heads with a probability of \( p = 0.6 \).
Billy Bones tossed the gold coin, an... | Solution. Let's introduce notations for the events:
$$
A=\{\text { the gold coin is biased }\},
$$
$B=\left\{\begin{array}{l}\text { when the gold coin is tossed, heads appear immediately, } \\ \text { and when the silver coin is tossed, heads appear on the second attempt. }\end{array}\right\}$
We need to find the c... | \frac{5}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,078 |
11. Large Cube. A cube is assembled from 27 playing dice. a) (6 - 11 grades, 1 point) Find the probability that exactly 25 sixes are on the surface of the cube.
b) (7 - 11 grades, 1 point) Find the probability that at least one one is on the surface of the cube.
c) (8 - 11 grades, **1** point) Find the expected value... | Solution. a) One of the 27 cubes is in the center and therefore not visible at all. The other 26 cubes are visible. Thus, the required event
$$
A=\{25 \text { sixes }\}
$$
consists of all the cubes showing sixes outward, except for one - let's call it the special cube.
Consider all the cubes. If a cube is in the cen... | )\frac{31}{2^{13}\cdot3^{18}}\approx9.77\cdot10^{-12};b)1-\frac{5^{6}}{2^{2}\cdot3^{18}}\approx0.99998992;)9;)189;e)6-\frac{5^{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,080 |
18. (10-11 grades, 5 points) Favorite Pair. On the drying rack, in a random order (as taken out of the washing machine), there are p socks. Among them are two favorite socks of the Absent-Minded Scientist. The socks are hidden by a drying sheet, so the Scientist cannot see them and takes one sock at a time by feel. Fin... | Solution. It is convenient to form a triangular table. The shaded cells correspond to pairs of favorite socks. For example, the pair $(2; 4)$, marked with an "X", corresponds to the case where the first favorite sock was the second one, and the second one was the fourth in the sequence. All pairs are equally likely, an... | \frac{2(n+1)}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,081 |
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean ... | # Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now give... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,082 |
9. Steel doors (from 9th grade. 2 points). At the factory named after Seaman Zheleznyak, they manufacture rectangles with a length of 2 m and a width of 1 m. The length is measured by worker Ivanov, and the width, independently of Ivanov, is measured by worker Petrov. Both have an average error of zero, but Ivanov has ... | # Solution.
a) Let $X$ be the width and $Y$ be the length of the cut rectangle in meters. According to the problem, $\mathrm{E} X=2$, $\mathrm{E} Y=1$. Since the measurements are independent, $\mathrm{E}(X Y)=\mathrm{E} X \cdot \mathrm{E} Y=2$ (sq.m.).
b) From the condition, it follows that $\mathrm{D} X=0.003^{2}=9 ... | )2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,083 |
15. English Club (from 9th grade. 6 points). Every Friday, ten gentlemen come to the club, and each hands their hat to the doorman. Each hat fits its owner perfectly, but no two hats are the same size. The gentlemen leave one by one in a random order.
When seeing off each gentleman, the doorman tries to put on the fir... | Solution. Let the number of gentlemen be $n$. We will number them in the order of increasing sizes of their hats from 1 to $n$. No hats will be left only if each took his own hat. Let the probability of this be $p_{n}$. If the $k$-th gentleman leaves first (the probability of this is $\frac{1}{n}$), then the probabilit... | 0.000516 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,084 |
4. Solution. The first one can be right. For example, let there be 4 white chocolate candies out of 7 in the first box, and 3 white candies out of 5 in the second. Then, in the bag, there are 7 white candies out of 12 in total. The second one can also be right: if there were 8 white candies out of 14 in the first box, ... | Answer: the first and second mathematicians may be right; the third is wrong.
Comment. Other methods of solving are possible. For example, the problem can be solved using a system of equations.
Evaluation criteria
| Justification of the correctness/incorrectness of all three | 3 points |
| :--- | :---: |
| Justifica... | thefirstmathematiciansmayberight;thethirdiswrong | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,086 |
5. Solution. Let the probabilities of heads and tails be $p$ and $q=1-p$ respectively. We form the equation
$$
C_{10}^{7} p^{7} q^{3}=C_{10}^{6} p^{6} q^{4}
$$
from which it follows: $120 p=210 q ; \frac{p}{q}=\frac{7}{4} ; p=\frac{7}{11}$. | Answer: $\frac{7}{11}$.
Grading Criteria
| Correct and justified solution | 2 points |
| :--- | :---: |
| Correct equation is set up, but an error is made or the solution is not completed | 1 point |
| Solution is incorrect or missing (including only providing the answer) | 0 points | | \frac{7}{11} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,087 |
7. Solution. Suppose for clarity of reasoning that when a bite occurs, the Absent-Minded Scholar immediately pulls out and re-casts the fishing rod, and does so instantly. After this, he waits again. Consider a 6-minute time interval. During this time, on average, there are 3 bites on the first fishing rod and 2 bites ... | Answer: 1 minute 12 seconds.
Evaluation Criteria
| Correct and justified solution | 3 points |
| :--- | :---: |
| It is shown that on average there are 5 bites in 6 minutes, or an equivalent statement is proven | 1 point |
| The solution is incorrect or missing (in particular, only the answer is given) | 0 points | | 1 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,089 |
# 8. Solution.
a) Suppose there are 9 numbers in the set. Then five of them do not exceed the median, which is the number 2. Another four numbers do not exceed the number 13. Therefore, the sum of all numbers in the set does not exceed
$$
5 \cdot 2 + 4 \cdot 13 = 62
$$
Since the arithmetic mean is 7, the sum of the ... | Answer: a) no; b) 11.
Scoring criteria
| Both parts solved correctly or only part (b) | 3 points |
| :--- | :---: |
| The correct estimate of the number of numbers in part (b) is found, but no example is given | 2 points |
| Part (a) is solved correctly | 1 point |
| The solution is incorrect or missing (in particula... | 11 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,090 |
1. Solution. The first one can be right. For example, let there be 2 blue candies out of 5 in the first packet, and 3 blue candies out of 8 in the second. Then, in the box, there are 5 blue candies out of 13. The second one can also be right: for example, if there are 4 blue candies out of 10 in the first packet, and 3... | Answer: the first and second mathematicians may be right; the third is wrong.
Comment. Other methods of solving are possible. For example, the problem can be solved using a system of equations.
## Evaluation Criteria
| The correctness/incorrectness of all three is justified | 3 points |
| :--- | :---: |
| The correc... | thefirstmathematiciansmayberight;thethirdiswrong | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,091 |
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