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5. Four friends (from 6th grade, 1 point). Masha, Nina, Lena, and Olya are friends. They are all of different heights, but the difference is very small - it's hard to tell by eye. One day they decided to find out who is taller and who is shorter. It turned out that Nina is shorter than Masha, and Lena is taller than Ol... | Solution. The following equally possible distributions by height in descending order are possible:
## MNLO, MLNO, MLOL, LOMN, LMON, and LMNO.
Out of these six elementary outcomes, only one - MNLO - is favorable to the event "Nina is taller than Lena."
Answer: $1 / 6$. | \frac{1}{6} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,741 |
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. Let's assume for simplicity 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). We will combine both tables, but we will write the data not in percentages, but in ab... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,742 |
7. Anchurian Checkers ${ }^{1}$ (from 7th grade, 1 point). A tournament of Anchurian Checkers is held in several rounds. If an even number of players participate in a round, they are divided into random pairs. If the number of players is odd, random pairs are selected by lot, and one player remains without a pair and d... | Solution. Let there be $n$ participants in the tournament. As a result of each match, one participant is eliminated, so there will be a total of $n-1$ matches. At the same time, no pair can play two matches, since the loser is eliminated. Therefore, there will also be $n-1$ game pairs throughout the tournament. In tota... | \frac{1}{13} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,743 |
8. The King's Path (from 7th grade, 2 points). A chess king is on the a1 square of a chessboard and wants to move to the h8 square, moving right, up, or diagonally up-right. In how many ways can he do this? | Solution. Instead of letter designations for the columns, we use numerical ones. Then, instead of al, we can write $(1,1)$. Let $S_{m, n}$ be the number of ways to get from the cell $(1,1)$ to the cell $(m, n)$. We are interested in $S_{8,8}$.
Obviously, $S_{1, n}=1$ and $S_{m, 1}=1$ for all $m$ and $n$. In particular... | 48639 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,744 |
9. Buratino the Statistician (from 7th grade, 2 points). Every month, Buratino plays in the "6 out of 45" lottery organized by Karabas-Barabas. In the lottery, there are 45 numbered balls, and in each draw, 6 random winning balls are drawn.
Buratino noticed that in each subsequent draw, there are no balls that appeare... | Solution. The probability that no numbers from the first draw will be repeated in the second draw is $a=\frac{39 \cdot 38 \cdot 37 \cdot \ldots \cdot 34}{45 \cdot 44 \cdot 43 \cdot \ldots 40}=0.40056 \ldots$, which is slightly more than 0.4. The probability that there will be no repetitions from the previous draw in bo... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,745 |
10. Product of Digits (7th grade, 3 points). Nезнайка (lit. "Don't-know") was solving 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 ... | Solution. There is an error in Nезнayka's reasoning: if there is an even digit in the set, it does not mean that it is the 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 many even d... | 0.988 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,746 |
11. The Ace of Diamonds (From 8th grade, 2 points). Thirty-six players are playing a game: from a deck of 36 cards, they take turns drawing a random card. If a player draws the ace of diamonds, the player wins; if another card is drawn, the player returns it to the deck, and the next player draws a card. They draw card... | Solution. Let the probability of the first player winning be $p_{1}$. Suppose the first player does not draw the diamond ace on the first try. The probability of this is 35/36. At this point, the second player becomes the first, and the probability that he wins becomes equal to $p_{1}$. Returning to the beginning of th... | Thebetofeachsubsequentplayershouldbeintheof35:36tothebetofthepreviousone. | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,747 |
12. Winter Camp. In the winter camp, Vanya and Grisha live in a room. Every evening they draw lots to decide who will turn off the light before going to bed: the switch is near the door, and the loser has to go to bed in complete darkness, bumping into chairs.
Usually, Vanya and Grisha draw lots without any complicati... | Solution. a) Let's assume that heads in a coin toss give a one, and tails give a zero in the fractional part of a binary fraction. This results in some number $x$ represented by a binary fraction. For example, if the sequence of tosses starts with HTH, then the binary fraction is 0.101.
Obviously, $0 \leq x \leq 1$, a... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,748 |
13. Short algorithm (from 8th grade, 2 points). There are 4 coins of different weights. In one action, you can weigh two coins on a balance scale without weights and find out which one is heavier. The task is to order the coins by weight. Show that there is a way to order the coins such that the expected number of weig... | Solution. We will construct an algorithm that gives an average number of weighings of $4 \frac{2}{3}$. Let the coins be denoted as $A, B, C$, and $D$. Fewer than three weighings are not sufficient. The first two weighings will compare $A$ with $B$ and $C$ with $D$. For definiteness, assume that $A<B$ and $C<D$, meaning... | 4\frac{2}{3} | Combinatorics | proof | Yes | Yes | olympiads | false | 10,749 |
14. Intern's Mistake (from 8th grade, 3 points). At a bakery, a dosing machine measures out portions of dough weighing 400 g. The rule is: if the standard deviation of ten random measurements exceeds 5% of the nominal portion mass, the machine requires repair.
To check the machine, young intern Ivanov weighed 10 rando... | Solution. The master could reason as follows.
The root mean square deviation from 400 g is
$$
\sqrt{\frac{12^{2}+32^{2}+18^{2}+\ldots+21^{2}}{10}}=\sqrt{398}
$$
The standard deviation, which is the root mean square deviation of the numbers in the array from the arithmetic mean, is no greater than the root mean squar... | \sqrt{398}<20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,750 |
15. Red and Black (from 8th grade, 3 points). The dealer ${ }^{2}$ deals one card at a time to the player from a well-shuffled standard deck of cards. At any moment, while the dealer still has cards left, the player can say "Stop". After this, the dealer reveals one more card. If it is of a red suit, the player wins; i... | Solution. Suppose such a strategy exists, meaning that based on the sequence of already opened cards, the player can determine the optimal moment to say "Stop."
Let at this moment there remain $r$ red and $b$ black cards in the deck. Then the probability of success, i.e., the event "the next card is red," is $\frac{r}... | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,751 |
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 | 10,752 |
17. Happy Sums (from 8th grade, 4 points). In the "Happy Sum" lottery, there are $N$ balls numbered from 1 to $N$. During the main draw, 10 balls are randomly drawn. During the additional draw, 8 balls are randomly selected from the same set of balls. The sum of the numbers on the drawn balls in each draw is announced ... | Solution. In the main draw, there are $C_{N}^{10}$ possible combinations, and in the additional draw, there are $C_{N}^{8}$ combinations. Let $S_{63,10}$ be the number of combinations in the first draw where the sum is 63. The addends are different natural numbers from 1 to $N$. If the smallest number is 2 or more, the... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,753 |
18. Lucky Spy (from 9th grade, 2 points). A spy with a radio transmitter parachutes into a completely random point in a perfectly square forest measuring $10 \times 10$ km. Upon landing, he immediately starts a radio transmission, revealing important information to the enemy.
At each of the four corners of the forest,... | Solution. Suppose, for definiteness, that the direction finder in the northeast corner is not working (Fig. 4). The diagram shows how many direction finders "see" each of the six resulting regions. The location of the spy cannot be accurately determined if he is in one of the two regions visible to only one direction f... | 0.087 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,754 |
# 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 | 10,756 |
6. Probabilistic Voting (6-9). In the final of a play competition for March 8, two plays made it to the finals. In the first play, $n$ students from 5th grade A participated, and in the second play, $n$ students from 5th grade B participated. At the play, $2n$ mothers of all $2n$ students were present. The best play is... | # Solution.
a) Let's call a mother confident if her child is playing in the best play. A confident mother will vote for the best play with a probability of 1. There are exactly $n$ confident mothers, so the best play will receive at least half of the votes - it will get, at the very least, the votes of all confident m... | 1-(\frac{1}{2})^{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,757 |
18. The Absent-Minded Scholar on the Bus (4 p. 6-11). In the bus, there are $n$ seats, and all tickets have been sold to $n$ passengers. The first to enter the bus is the Absent-Minded Scholar, who, without looking at the ticket, takes the first available seat. Subsequently, passengers enter one by one. If an entering ... | Solution. First method. Let's number all passengers, starting with the Scientist, in the order in which they entered the bus. The last passenger has the number $n$. For simplicity, let's number the seats in the same way. Suppose all passengers except the last one have already entered and taken their seats. There is one... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,760 |
6. In the village of Znoynoye, there are exactly 1000 residents, which exceeds the average population of villages in the valley by 90 people.
How many residents are there in the village of Raduzhny, which is also located in Sunny Valley? | Solution. Let $x$ be the total number of residents in all other villages except Znoynoye. Then the average population is
$$
\frac{1000+x}{10}=100+0.1 x=910, \text{ hence } x=8100
$$
Thus, the average population in 9 villages, except Znoynoye, is 900 people. If there are more than 900 residents in Raduzhny, there must... | 900 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,761 |
5. Multicolored Balls (from 6th grade, 1 point). In the red box, there are 100 red balls, and in the green box, there are 100 green balls. Eight red balls were moved to the green box, and then the same number of balls were moved from the green box to the red box. The balls in the boxes were thoroughly mixed. What is no... | Solution. After all the balls have been moved, there are still 100 balls in each box. This means that the number of red balls in the green box is the same as the number of green balls in the red box.
## Answer: the probabilities are equal.
$\mathbf{6}^{1}$. Tournament. In a tournament, 100 wrestlers of different stre... | 33.5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,762 |
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 | 10,763 |
11. Polar Friends (from 8th grade, 2 points). At an Antarctic station, there are $n$ polar explorers, all of different ages. With a probability $p$, any two polar explorers will develop a friendly relationship, independently of other affinities or antipathies. When the wintering period ends and it's time to return home... | Solution. We will assign the polar explorers numbers from 1 to $n$ according to their age - the older they are, the smaller the number. The polar explorer with number $j$ can receive advice from $j-1$ older polar explorers. Therefore, he will not receive any advice only if he is not friends with any of them. The probab... | \frac{1-(1-p)^{n}}{p} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,765 |
12. Ball (from 8th grade. 3 points). At the ball, $n$ family couples arrived. In each couple, the husband and wife are of exactly the same height, but there are no two couples of the same height. A waltz begins, and all the attendees randomly form pairs: each gentleman dances with a randomly chosen lady. Find the mathe... | Solution. Let's assign the cavaliers numbers from 1 to $n$. Let $I_{j}$ be the indicator of the event "cavalier number $j$ is shorter than his lady."
$$
\mathrm{P}\left(I_{j}=1\right)=\frac{j-1}{n} . \text { Then } \mathrm{E} X=\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{n}=\frac{1}{n}(0+1+\ldots+n-1)=\frac... | \frac{n-1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,766 |
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 | 10,767 |
14. First group (from 8th grade. 3 points). The sequence consists of 19 ones and 49 zeros, arranged in a random order. We will call a group the maximal subsequence of identical symbols. For example, in the sequence 110001001111 there are five groups: two ones, then three zeros, then one one, then two zeros, and finally... | Solution. Let's separately number the ones from 1 to 19 and the zeros from 1 to 49. Let $I_{k}$ be the indicator of the event "there are no zeros before the $k$-th one" and $J_{m}$ be the indicator of the event "there are no ones before the $m$-th zero", where $1 \leq k \leq 19$ and $1 \leq m \leq 49$. If at least one ... | 2.83 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,768 |
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 resulting random vector is $\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 | 10,769 |
16. New Year's Tradition (from 9th grade. 3 points). On one island tribe, there is a tradition - during a ritual dance, the shaman throws up three thin straight rods of the same length, tied in a shape resembling the letter П. Adjacent rods are connected by a short thread and can freely rotate relative to each other. T... | Solution. Let the broken line formed by the rods be $A B C D$. Let $\alpha=\angle A B C$ be the angle between the first
link and the second, and $\beta=\angle B C D$ be the angle between th... | \frac{11}{12} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,770 |
17. Rabbits (from 9th grade, 3 points). Rabbits can be smooth-haired and fluffy. The type of fur is determined by genes inherited from parents. Each rabbit has two alleles of the same gene responsible for the type of fur. Let's denote the allele for fluffiness as $H$ (or $h$), and the allele for smooth fur as $S$ (or $... | Solution. The allele $S$ (or $s$) occurs with probability $q=1-p$. Let's list the possible variants. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,771 |
18. Stirling's Problem (from 9th grade, 5 points). An incredible legend says that one day Stirling ${ }^{4}$ was contemplating Stirling numbers of the second kind and, in a pensive mood, was tossing 10 fair dice onto the table. After one of the tosses, he suddenly noticed that the combination of points that came up inc... | Solution. Let's find the probability of event $A$ "some number is missing in the rolled combination". Denote $A_{k}$ as the event "number $k$ is missing". Then
$$
A=A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5} \cup A_{6}
$$
Using the inclusion-exclusion principle, we get:
$$
\begin{gathered}
\mathrm{P}(A)=\math... | 0.272 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,772 |
19. Darts (from 9th grade. 5 points). According to one implausible legend, Cauchy ${ }^{5}$ and Bunyakovsky ${ }^{6}$ loved to play darts in the evenings. But their target was unusual - the sectors on it were unequal, so the probabilities of hitting different sectors were not the same. One day, Cauchy threw a dart and ... | Solution. Let the probabilities of hitting the sectors be $p_{1}, p_{2}, \ldots, p_{n}$ (indices indicate the sector numbers). The probability that Bunyakovsky will hit the same sector as Cauchy is
$$
p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\ldots+p_{n}^{2}
$$
and the probability that Bunyakovsky will hit the next sector clock... | p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\ldots+p_{n}^{2}>p_{1}p_{2}+p_{2}p_{3}+\ldots+p_{n-1}p_{n}+p_{n}p_{1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,773 |
7. Solution. Let the random variables $\xi$ and $\eta$ be the lifetimes of the blue and red bulbs, respectively. The lifetime of the flashlight is equal to the minimum of these two values. It is clear that $\min (\xi, \eta) \leq \xi$. Let's move on to the expectations: $\operatorname{E} \min (\xi, \eta) \leq \mathrm{E}... | Answer: I made a mistake.
| Evaluation Criteria | Score |
| :--- | :--- |
| Solution is complete and correct | 2 points |
| It is stated that the service life of the device is exactly 2 years | 1 point |
| No solution, or incorrect, or only the answer | 0 points | | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 10,774 |
8. Solution. We will show by contradiction that it is impossible to make such coins. Suppose the minters succeeded. Let the probability of heads on the first coin be $p_{1}$, and on the second coin be $p_{2}$. Then we get:
$$
\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: will not be able to.
IX Internet Olympiad in Probability Theory and Statistics. Qualifying Round
© Laboratory of Probability Theory and Statistics of the Moscow Center for Continuous Mathematical Education, 2015
| Evaluation Criteria | Score |
| :--- | :--- |
| Complete and correct solution | 3 points |
| Th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 10,775 |
7. Solution. Let $\xi$ and $\eta$ be the lifespans of the blue and red bulbs, respectively. The flashlight fails when the last bulb burns out, which means the lifespan of the flashlight is the maximum of $\xi$ and $\eta$. Clearly, $\max (\xi, \eta) \geq \eta$. Let's move to the expectations: $\operatorname{E} \max (\xi... | Answer: I made a mistake.
| Evaluation Criteria | Score |
| :--- | :--- |
| Solution is complete and correct | 2 points |
| The solution states that the average term is exactly 4 years | 1 point |
| No solution, or incorrect, or only the answer | 0 points | | notfound | Other | proof | Yes | Yes | olympiads | false | 10,777 |
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 | 10,778 |
4. Neighbors (from 6th grade. 1 point). There are cards with natural numbers from 1 to 5. The cards are well shuffled and two are randomly selected. Find the probability that the numbers on the selected cards will be two consecutive numbers. | Solution. The total number of ways to choose two cards is $C_{5}^{2}=10$. There are 4 pairs of cards with two consecutive numbers. All combinations are equally likely, so the desired probability is 0.4. The problem can also be solved by enumeration.
Answer: 0.4.
. A tic-tac-toe board (three by three cells) is randomly filled with six X's and three O's. Find the probability that all three O's will end up in a "winning position," that is, in a row either vertically, horizontally, or diagonally. | Solution. The total number of ways to randomly place three noughts and six crosses is $C_{9}^{3}=84$. The number of winning positions for noughts is eight: three verticals, three horizontals, and two diagonals. Therefore, the desired probability is $\frac{8}{84}=\frac{2}{21}$.
Answer: $\frac{2}{21}$. | \frac{2}{21} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,781 |
6. Drawing Lots. In a deck of 36 playing cards of four suits, from six to ace. Vasya and Masha each draw one card. Vasya draws first, and then Masha. Masha wins if she draws a card higher than Vasya's. Otherwise, Vasya wins.
a) (for 6th grade. 1 point). Prove that the lot is unfair.
b) (for 6th grade. 1 point). How c... | Solution: a) The draw is unfair because Vasya's probability of winning is higher. He wins not only when his card is higher than Masha's but also when the cards are of the same rank - since there are four cards of each rank in the deck.
b) For example, we can order the suits. The usual order is: spades, clubs, diamonds... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,782 |
7. Promotion. The store "Crocus and Cactus" announced a promotion: if a customer buys three items, the cheapest one of them is free. The customer took ten items, all of different prices: 100 rubles, 200 rubles, 300 rubles, and so on - the most expensive item cost 1000 rubles.
a) (from 6th grade. 1 point). What is the ... | Solution. a) The cheapest item in each trio should be as expensive as possible. If the buyer distributes the purchases into trios as follows (indicating only the prices):
$(1000,900,800),(700,600,500),(400,300,200)$ and a separate item for $100 \mathrm{p}.$, then he will pay $1000+900+700+600+400+300+100=4000$ rubles.... | 4583 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,783 |
8. Shortening. There is a sequence of 2015 digits. All digits were chosen randomly and independently from the set from 0 to 9. In the obtained sequence, the following operation is performed. If several identical digits occur consecutively, they are replaced by one such digit. For example, if there was a fragment ... $0... | Solution. The first digit does not change, and the transformation of each subsequent digit represents a Bernoulli trial. With a probability of 0.1, a digit disappears, and the sequence shortens by 1. We will consider this a success. With a probability of 0.9, the digit remains unchanged. This results in a series of 201... | 1.564\cdot10^{-90} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,784 |
9. Stubborn Squares (from 7th grade. 2 points). Given 100 numbers. 2 was added to each of them. The sum of the squares of the numbers did not change. 2 was added to each of the resulting numbers again. How did the sum of the squares change now? | Solution. Let the given numbers be $x_{1}, x_{2}, \ldots, x_{100}$. By adding 2 to each number, we get the set $y_{j}=x_{j}+2$. Adding 2 again, we get the set $z_{j}=y_{j}+2=x_{j}+4$. However, the variances of all three sets are equal, that is,
$$
\overline{x^{2}}-\bar{x}^{2}=\overline{y^{2}}-\bar{y}^{2}=\overline{z^{... | 800 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,785 |
10. Football (from 7th grade. 4 points). A football match lasts 90 minutes. A draw is possible. What is more likely in such a match: that the total number of goals scored will be even or that it will be odd? | Solution. Let $p-$ be the probability of an even number of goals. Then $q=1-p-$ is the probability of an odd number of goals.
Consider the first and second halves of the match separately. Let $p_{1}$ be the probability that the number of goals scored during the first half is even. Then $q_{1}=1-p_{1}$ is the probabili... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,786 |
11. Similar events. If a die is thrown, then the event $A=\{1,2,3\}$ consists of one of the faces 1, 2, or 3 appearing. Similarly, the event $B=\{1,2,4\}$ is favorable if the faces 1, 2, or 4 appear.
A die is thrown 10 times. It is known that the event $A$ occurred exactly 6 times.
a) (from 8th grade. 2 points) Find ... | Solution. a) The probability of the event "A happened six times" is $\frac{C_{10}^{6}}{2^{10}}$. The event "A happened six times and B never happened" will occur only if a three is rolled six times, and in the other cases, a 5 or 6 is rolled. The probability of this is $C_{10}^{6} \cdot\left(\frac{1}{6}\right)^{6} \cdo... | \frac{16}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,787 |
12. Decimal fraction (from 8th grade. 2 points) There is a sequence of a thousand random digits. Before this sequence, a zero and a decimal point were placed. As a result, a random number $X$ was obtained, written as a decimal fraction: $X=0, x_{1} x_{2} x_{3} \ldots x_{1000}$. Find $\mathrm{E} X$. | Solution. $\quad \mathrm{E} X=0+0.1 \mathrm{E} x_{1}+0.01 \mathrm{E} x_{2}+\ldots+0.1^{k} \mathrm{E} x_{k}+\ldots+0.1^{1000} \mathrm{E} x_{1000} . \quad$ All digits take values from 0 to 9 with equal probabilities, so $\mathrm{Ex}_{k}=4.5$ for any $k$. Therefore,
$$
\mathrm{E} X=0.45+0.045+0.0045+\ldots+0. \underbrace... | 0.4\underbrace{4999}_{9999nines}5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,788 |
13. Test (from 8th grade. 3 points) Vasya and Misha were taking a test, which had 20 questions of the same difficulty, and for each question, answer options were provided. Since neither Vasya nor Misha knew anything, they only guessed the answers. Vasya guessed the correct answers to 6 questions, and Misha to 8 questio... | Solution. Let the random variable $I_{k}$ be the indicator of a match in the $k-\mathrm{th}$ question, that is, $I_{k}=1$ if both Vasya and Misha either guessed or did not guess the correct answer to question $k$, and $I_{k}=0$ if one guessed and the other did not.
According to the problem, Vasya will guess the correc... | 10.8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,789 |
14. Pills (from 9th grade. 2 points). The Absent-Minded Scientist had a sore knee. The doctor prescribed him 10 pills for his knee: to take one pill daily. The pills help in $90 \%$ of cases, and in $2 \%$ of cases, there is a side effect - absent-mindedness disappears if it was present.
Another doctor prescribed the ... | Solution. Let's consider the events $R$ "The scientist took the pills for absent-mindedness", $A$ "The knee stopped hurting" and $B$ "Absent-mindedness disappeared".
We need to find the conditional probability $\mathrm{P}(R \mid A \cap B)$:
$$
\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 | 10,790 |
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 | 10,791 |
16. Alternating Groups (from 9th grade, 3 points) A sequence consists of $k$ zeros and $m$ ones, arranged in a random order. Let's divide the sequence into alternating groups of zeros and ones (a group is a segment consisting of all identical digits standing consecutively) ${ }^{3}$. The total number of groups is a ran... | Solution. For each element of the sequence, we define a random variable $I_{j}$ - an indicator that equals 1 if this element is the first in its group. Otherwise, $I_{j}=0$. Clearly, $I_{1}=1$. For all other indicators $I_{j}$, the probability of the event $I_{j}=1$ is the probability that the symbols at the $(j-1)$-th... | 1+\frac{2k}{k+} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,792 |
17. Streak of Luck (from 9th grade, 4 points). A coin is tossed $n$ times. A streak of luck of length $k$ is a sequence of $k$ consecutive heads. Find the probability that the longest streak of luck in 10 coin tosses will be of length 4. | Solution. For brevity, let's call the length of the maximum lucky strip the "lucky value" of the sequence. Let $S_{n}^{k}$ be the number of different sequences of length $n$ with a lucky value of $k$. Each such sequence ends either with a tail (T) or a group of THH...H, where the number of heads can be any from 1 to $k... | \frac{139}{1024} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,793 |
19. Football-2. (from 10th grade. 6 points) A regular football match is in progress. A draw is possible. The waiting time for the next goal does not depend on previous events in the match. It is known that the expected number of goals in football matches between these teams is 2.8. Find the probability that an even num... | Solution. Under the given conditions, the waiting time for the next goal is a random variable with an exponential distribution, and the total number of goals in the match is distributed according to the Poisson law with parameter $\lambda=2.8$, that is, the probability that $n$ goals will be scored is $p_{n}=\frac{\lam... | 0.502 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,795 |
4. Extreme set (from 6th grade, $\mathbf{1}$ point). From the digits 1 to 9, three single-digit and three two-digit numbers are formed, with no digits repeating. Find the smallest possible arithmetic mean of the resulting set of numbers. | Solution. Let the sum of the first digits of two-digit numbers be $n$, and the sum of the remaining digits be $k$. Then the arithmetic mean is $\frac{10 n+k}{6}$, and this number will be the smallest only if the sum $n$ includes the digits 1, 2, and 3, and the sum $k$ includes the digits from 4 to 9, regardless of the ... | 16.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,796 |
5. Arithmetic density. It is clear that if we take a random natural number from a sufficiently long segment of the natural number series, the probability that this number will be divisible by 10 will be closer to 0.1, the longer the segment.
Let's ask a different question: how much can this probability differ from 0.1... | Solution. a) The greatest probability is 1. For this, it is sufficient to take the single number 10 as the segment.
b) Let there be $k \geq 1$ numbers divisible by 10 in the chosen segment. The greatest length of such a segment is $10 k+9$. Therefore, the smallest probability of the event "the chosen number is divisib... | \frac{1}{19} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,797 |
6. Average Weight (recommended from 6th grade, 2 points) There are fewer than 35 students in the class. The average weight of all students is 53.5 kg. The average weight of girls is 47 kg, and the average weight of boys is 60 kg. Could it be that the lightest student in this class is a boy, and the heaviest is a girl? ... | Solution. This is possible. Example: 15 boys each weighing 61 kg, 15 girls each weighing 46 kg, one boy weighing 45 kg, and one girl weighing 62 kg. The average weight of the boys is $\frac{15 \cdot 61+45}{16}=60$ kg, the average weight of the girls is $\frac{15 \cdot 46+62}{16}=47$ kg, and the overall average weight i... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,798 |
7. Three Punctures. On a rocky road, the Absent-Minded Scholar punctured the car's tire in three completely random and independent points. Serious repair is impossible, but fortunately, an old and rather torn inner tube was found in the trunk. The Scholar managed to cut out a large undamaged piece from it, which can co... | Solution. For simplicity, let's imagine that the tire is a circle of length 3, and the punctures are random points on this circle. The case where two punctures fall into the same point can be neglected, as the probability of such an event is zero.
a) There will definitely be two points out of three where the arc betwe... | \frac{2}{3} | Combinatorics | proof | Yes | Yes | olympiads | false | 10,799 |
8. Colorful Ribbons (from 6th grade, 3 points). Vanya had five colorful ribbons of lengths 15, 20, 24, 26, and 30 cm in her jewelry box. Kolya cut off pieces from some (possibly all) of the ribbons. The average length of the ribbons decreased by 5 cm, while the median and range of the lengths remained unchanged. What a... | Solution. Let the shortest ribbon have a length of $x \, \text{cm}$ after Colin's action. Then the length of the longest ribbon is $x+15$ cm, since the range has not changed. Another ribbon has a length of 24 cm, as the median has remained the same, and the two unknown lengths are denoted as $y$ cm and $z$ cm, with $y ... | 9,9,24,24,24 | Other | math-word-problem | Yes | Yes | olympiads | false | 10,800 |
9. Mini-tournament (recommended from 7th grade, 1 point). Alyosha, Borya, and Vasya are holding a mini-tournament in tennis: each plays with each other once. The winner gets 1 point, the loser gets 0, and there are no draws in tennis. The absolute winner of the mini-tournament is the one who has a total of 2 points. It... | Solution. Let the event “$X$ won against $Y$” be denoted as $X \square Y$. The event “no winner” will occur only if everyone has one point. This can happen in two cases:
$$
A \square B \square B \square A \text { and } B \square \delta \square A \square B .
$$
Let the unknown probability of the event $B \square A$ be... | 0.24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,801 |
10. Namesakes (recommended from 7th grade, 1 point) Let's assume that among eighth-graders, the name Alexander is three times more common than the name Alexandra, Yevgenyev is three times less common than Yevgeny, Valentinov is one and a half times more common than Valentin, and Vasilyev is 49 times more common than Va... | # Solution.
$$
\frac{1}{4} \cdot \frac{1}{4}+\frac{1}{4} \cdot \frac{3}{4}+\frac{1}{4} \cdot \frac{2}{5}+\frac{1}{4} \cdot \frac{1}{50}=0.355
$$
Answer: 0.355 . | 0.355 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,802 |
11. Magic Pen (recommended for 8th grade, 1 point). Katya correctly solves a problem with a probability of $4 / 5$, while the magic pen correctly solves a problem without Katya's help with a probability of $1 / 2$. In the test, there are 20 problems, and to get a B, one needs to solve at least 13 of them correctly. How... | Solution. Let $x$ be the number of examples Katya solves herself, and $20-x$ be the number of examples solved by the pen. Then the expected number of correctly solved problems is
$$
\frac{4}{5} x+\frac{1}{2}(20-x)=0.3 x+10
$$
From the inequality $0.3 x+10 \geq 13$ we get that $x \geq 10$. Therefore, Katya needs to tr... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,803 |
12. Retro Collection (recommended from 8th grade, 2 points). Vitya collects toy cars from the "Retro" series. The problem is that the total number of different models in the series is unknown - it is the biggest commercial secret, but it is known that different cars are produced in the same edition, and therefore it ca... | Solution. Let there be a series of $n \geq 13$ cars. The probability that among the next $k$ offers there will only be the 12 models that Vitya has already seen is
$$
\left(\frac{12}{n}\right)^{k} \leq\left(\frac{12}{13}\right)^{k}
$$
We form the inequality: $\left(\frac{12}{13}\right)^{k}\log _{12 / 13} 0.01=\frac{\... | 58 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,804 |
13. Young Shooters. Vanya and Kolya went to the shooting range. Kolya suggested:
- Vanya, let's shoot in turns. Whoever hits the target first wins.
- That's not fair! You shoot better. Remember, the Absent-Minded Scientist calculated exactly that on average you spend one shot less for one hit than I do.
- Alright, - a... | Solution. Let Kolya hit the target with probability $p$ on each separate shot, and Varya with probability $r$. The probabilities of missing the target for them are respectively $q=1-p$ and $s=1-r$.
When shooting in turns, the probability of event $A$ "Varya wins" (Varya shoots first) is
$$
\mathrm{P}(A)=r+s q r+s q s... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,805 |
14. New Year's Problem (recommended for 8th grade, 4 points). On the New Year's table, there are 4 glasses in a row: the first and third are filled with orange juice, while the second and fourth are empty. While waiting for guests, Vanya absent-mindedly and randomly pours the juice from one glass to another. In one mov... | Solution. We will encode full glasses with the digit 1 and empty ones with the digit 0. We will construct a graph of possible pourings (Fig. 4). This graph turns out to be the graph of an octahedron. From each state to any adjacent one, one can move with a probability of $1 / 4$, and each edge is "traversable" in both ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,806 |
15. Messengers (recommended for 9th grade, 3 points). Once, the beautiful Queen Guinevere, while staying at her parental castle, asked King Arthur to send her 20 pearls. The roads are not safe, and Arthur, just in case, decided to send 40 pearls, with different messengers, ordering them to ride on different roads. Band... | Solution. The probability of not saving at least 20 pearls if there are two messengers:
$$
\mathrm{P}_{2}=p^{2}
$$
The probability of not saving at least 20 pearls if there are three messengers:
$$
\mathrm{P}_{3}=p^{3}+2 p^{2}(1-p)=p^{2}(2-p) .
$$
The probability of not saving at least 20 pearls if there are four m... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,807 |
16. Technical Break (recommended for 9th grade, 3 points). A woman working as a cashier at a railway ticket office saw that there were no customers in front of the window, sighed, hung a sign saying "Technical Break 15 minutes," and left for exactly 15 minutes. When the Absent-Minded Scientist approached the ticket win... | Solution. We will measure time in minutes from the moment the cashier leaves. Let $x_{1}, x_{2}, \ldots, x_{5}, x_{6}$ be the moments when five people and the Scholar approached the window. All these numbers are greater than zero but less than 15, and it is known that the sixth number is greater than the others. We wil... | 0.738 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,808 |
17. The Capricious Princess (recommended for 9th grade, 4 points). A capricious princess is choosing a groom. 100 grooms are courting her, each better than the last, and there are no two equal among them. However, they court her in a random order. We will call a groom prominent if he pleases the princess more than all ... | Solution. Let's introduce indicators $I_{k}$ of the events "the $k$-th groom belongs to the first chain." This event is equivalent to the event "the 1st groom is the best among the first $k$ grooms." If this event occurs, then $I_{k}=1$, and if not, then $I_{k}=0$. Then the length of the first chain $X$ is equal to the... | 5.187 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,809 |
4. Bus (6 - 11 grades, $\mathbf{1}$ point). Anya is waiting for a bus. Which event has the highest probability
\[
\begin{gathered}
A=\{\text { Anya waits for the bus for no less than one minute }\} \\
B=\{\text { Anya waits for the bus for no less than two minutes. }\} \\
C=\{\text { Anya waits for the bus for no less... | Solution. Arrange the events on a timeline.
From the diagram, it is clear that event $A$ includes event $B$, and event $B$ includes event $C$, meaning that event $A$ is the most extensive:
... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,811 |
5. A row of dice. (6 - $\mathbf{1 1}$ grades, $\mathbf{1}$ point). 2012 fair dice (cubes) are arranged in a row such that each pair of adjacent dice touch each other with the same faces (domino principle). Otherwise, the position of the dice is random. Find the sum of the points that ended up on the surface of the resu... | Solution. The sum of the points on a die is $1+2+3+4+5+6=21$. The total points on all dice is $2012 \cdot 21$. For a fair die, the sum of the points on opposite faces is 7. Let the number of points on the end face of the first die be $X$ - this is a random variable taking natural values from 1 to 6. Then the number of ... | 28170,28172,28174,28176,28178,28180 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,812 |
6. Invariant variance (6th - 11th grades, 1 point). In the set
$$
-5,-4,-3,-2,-1,0,1,2,3,4,5
$$
replace one number with two other integers so that the variance of the set and its mean do not change. | Solution. As is known, the variance of a set is the difference between the mean of the squares and the square of the mean:
$$
S^{2}=\overline{x^{2}}-\bar{x}^{2}
$$
Therefore, the problem can be reformulated: we need to replace one number with two others so that the arithmetic mean and the mean of the squares of the n... | Weneedtoreplace-4withthe1-5orreplace4withthe-15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,813 |
8. Magic Pies (6th - 11th grades, 1 point). Alice has six magic pies in her pocket - two are 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... | 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 | 10,815 |
9. Round Table. Petr Ivanovich, 49 other men, and 50 women sit around a round table in a random order. We will call a man satisfied if a woman sits next to him. Find:
a) (6-11 grades, 1 point) the probability that Petr Ivanovich is satisfied;
b) (8-11 grades, 1 point) the expected number of satisfied men. | Solution. a) Pyotr Ivanovich is dissatisfied only if men are sitting on both sides of him. The probability of this is $\frac{49}{99} \cdot \frac{48}{98}=\frac{8}{33}$. Therefore, the probability of the opposite event “Pyotr Ivanovich is satisfied” is $\frac{25}{33}$.
b) The probability that one man is satisfied is $\f... | \frac{25}{33} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,816 |
10. Sequence of medians. Vanya wrote the number 1 on the board, and then several more numbers. As soon as Vanya writes the next number, Mitya calculates the median of the existing set of numbers and writes it down in his notebook. At some point, the numbers in Mitya's notebook are: $1 ; 2 ; 3 ; 2.5 ; 3 ; 2.5 ; 2 ; 2 ; ... | Solution. a) The first number is obviously one. Then, for the median of the first two numbers to be 2, the second number must be 3.
For clarity, we will write down Vanya's numbers. Vanya wrote the third number $a_{3}$:
$$
\begin{array}{lll}
1 & 3 & a_{3} .
\end{array}
$$
The median is now 3, which can only be if $a_... | )2;b)2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,817 |
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 | 10,818 |
13. The game "Unicum". Petya and 9 other people are playing such a game: each throws a die. A player wins a prize if they roll a number of points that no one else has rolled.
a) (6th - 11th grades, 1 point) What is the probability that Petya will win a prize?
b) (8th - 11th grades, 3 points) What is the probability t... | Solution. a) Suppose (for clarity only) that Petya throws before everyone else. He scores some number of points. The probability that each subsequent player will score a different number of points is $\frac{5}{6}$. Since the throws are independent, the probability that the nine remaining players will all score differen... | )(\frac{5}{6})^{9}\approx0.194;b)\approx0.919 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,820 |
14. What? Where? When? In the game "What? Where? When?", 13 envelopes with questions from the audience are played. The envelopes are chosen in a random order using a spinner. If the Experts answer correctly, they earn a point; if incorrectly, one point goes to the audience. The game ends as soon as one of the teams sco... | Solution. a) Suppose that the losing team earned $k$ points in one game. The winning team earned 6 points, so a total of $6+k$ envelopes were played. The point for the question from the last envelope was definitely earned by the winning team, while for the remaining $5+k$ questions, points could be earned by either tea... | )600\cdot(1-\frac{C_{12}^{6}}{2^{12}})\approx465;b)\frac{12}{13}\cdot(1-\frac{C_{12}^{6}}{2^{12}})\approx0.715 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,821 |
16. Unified State Exam in Anchuria. (7-11 grades, 2 points). In Anchuria, the Unified State Exam is taking place. The probability of guessing the correct answer to each question on the exam is 0.25. In 2011, to receive a certificate, one needed to answer correctly[^0]
3 questions out of 20. In 2012, the School Adminis... | Solution. If a graduate guesses the answers, the Unified State Exam (EGE) can be considered as a Bernoulli scheme with a success probability of $p=0.25$ and a failure probability of $q=0.75$. In 2011, to pass the exam, one needed to answer at least 3 questions correctly. It is more convenient to find the probability of... | In2012,theprobabilityofobtainingcertificateishigher | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,823 |
17. Successful Pairs. Vasya bought $n$ pairs of identical socks. For $n$ days, Vasya had no problems: every morning he took a new pair from the closet and wore it all day. After $n$ days, Vasya's mother washed all the socks in the washing machine and paired them up as they came out, since, to reiterate, the socks are i... | Solution. a) The probability that the first pair turned out to be a successful one is $\frac{1}{2 n-1}$, since when the mother took the first sock in her hand, there were $2 n-1$ socks left, and only one of them forms a successful pair with the first. Further - the same: the probability that the second pair is successf... | \frac{n}{2n-1}>\frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,824 |
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 | 10,825 |
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. No two pairs are the same. 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 with all... | 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 | 10,826 |
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 | 10,827 |
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 certain only that it could have happened in any year with equal probability... | 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 | 10,828 |
6. Nine cards (6th grade, 2 points). On 9 cards, the digits from 1 to 9 are written - one on each card. Three random cards are chosen and placed in a row. What is the probability that the resulting three-digit number is divisible by 3? | Solution. From the condition, it follows that the selected digits do not repeat - there is only one of each card. We will only talk about the digits from now on. Let's write them in three rows depending on the remainder when divided by 3.
$$
\begin{array}{llll}
3 & 6 & 9 & \text { - remainder } 0 \\
1 & 4 & 7 & \text ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,829 |
7. Ice Cream (from 6th grade, 2 points.) Anya wants to buy an ice cream that costs 19 rubles. In her pocket, she has two 10-ruble coins, two 5-ruble coins, and one 2-ruble coin. Anya takes out three coins without looking. Find the probability that the coins she took out will be enough to pay for the ice cream. | Solution. There are 5 coins in the pocket, and 3 of them are drawn. Therefore, this experiment has $C_{5}^{3}=10$ equally likely outcomes - the drawn sets of three coins. The event "Anya drew no less than 19 rubles" is favorable in five of them: three outcomes when two ten-ruble coins and one other coin are drawn, and ... | 0.5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,830 |
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 | 10,831 |
9. Bags in the Shed. A truck delivered 4 bags of cement. They are stacked in the truck bed. The worker can carry one bag at a time from the truck to the gate, or from the gate to the shed. The worker can carry the bags in any order, each time taking the top bag, carrying it where needed, and placing it on top of the st... | Solution. a) This is possible only if the worker each time carries a bag from the machine to the gate and immediately takes it to the shed. The probability of this is $(1 / 2)^{3}=1 / 8$.
b) This will happen only if the worker first brings three bags in a row to the gate (the probability of this is $(1 / 2)^{2}=1 / 4$... | \frac{1}{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,832 |
10. The Right Stars (from 8th grade, 3 points). Let natural numbers $k$ and $n$ be coprime, with $n \geq 5$ and $k < n / 2$. A proper $(n ; k)$-star is defined as a closed broken line that results from replacing every $k$ consecutive sides in a regular $n$-gon with a diagonal having the same endpoints. For example, the... | Solution. We will solve the problem in general for the $(n ; k)$-star. Let the vertices of the $n$-gon be denoted by $A_{0}, A_{1}$, and so on up to $A_{n-1}$. Take two consecutive segments of the star: $A_{0} A_{k}$ and $A_{1} A_{k+1}$. They intersect at a circle of radius $r_{1}$ with the center at the center of the ... | 48432 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,833 |
11. One Match (from 8th grade, 2 points). In the lottery ticket "6 out of 45," there are a total of 45 natural numbers. From these 45 numbers, a lottery participant must choose a combination that contains exactly 6 numbers. Two participants independently choose their combinations. What is the probability that their com... | Solution. Let's mark with some secret sign the six numbers chosen by the first participant. This results in 6 marked numbers and 39 unmarked numbers. We need to find the probability that the combination chosen by the second participant includes exactly 1 marked number and exactly 5 unmarked numbers. The probability of ... | 0.424 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,834 |
12. Target (from 8th grade, 2 points). A target is hanging on the wall, consisting of five zones: a central circle (the bullseye) and four colored rings (see figure). The width of each ring is equal to the radius of the bullseye. It is known that the number of points for hitting each zone is inversely proportional to t... | Solution. Suppose, for definiteness, that the apple has a radius of 1. Then the blue zone is enclosed between circles with radii 3 and 4. The probability of hitting the blue zone relative to the probability of hitting the apple is the ratio of the areas of these zones:
$$
\frac{p_{g}}{p_{c}}=\frac{1}{4^{2}-3^{2}}=\fra... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,835 |
13. Salvo-control. The artillery system of a missile cruiser fires at a target. If the target is not hit, the system automatically makes a second shot at the same target. A third shot at this target is not made. The cruiser has 10 missiles on board. During exercises, it conducts firings at several targets under identic... | Solution. a) Exactly three missiles will remain only in one case: the cruiser fired twice at two targets, and exactly three out of five targets were hit by the first salvo. Thus, the desired probability is that in 5 trials with a success probability of $p$, exactly 3 successes occur:
$$
C_{5}^{3} p^{3}(1-p)^{2}
$$
b)... | 10p-p^{10} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,836 |
14. Luggage at the airport. 200 suitcases were loaded onto the plane. After the flight, passengers meet their suitcases on the conveyor belt. As soon as the conveyor belt started, the loader began placing suitcases on it one by one every two seconds in a random order (as they were brought). Among the passengers is a gr... | Solution. a) To have exactly two minutes elapse, the last suitcase must be the 60th in the count. Among the first 59 suitcases, we select 9 spots for the 9 suitcases from the group we need, and place the 10th suitcase from this group in the 60th position. This results in exactly $C_{59}^{9}$ ways to choose 10 spots for... | \frac{C_{59}^{9}}{C_{200}^{10}};\frac{4020}{11} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,837 |
15. Two Umbrellas ${ }^{2}$ (from 8th grade, 4 points). Every morning, the Absent-Minded Scholar goes to work, and in the evening - home. The Scholar has two umbrellas in total, and he takes one with him only in one of two cases: either it is raining outside, or there is no umbrella where he is going. After some time, ... | Solution. It is assumed that the probability of rain is the same every day. Let's denote this probability by $x$.
We will consider the situation where the Scientist is traveling from point A to point B (it doesn't matter which is work and which is home). Let's go back one day and consider three possibilities: there we... | \frac{1}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,838 |
2. The Chairman is selected by a fair draw from the delegates of the chosen city each day.
In a private conversation with the prime minister, the president explained that the new procedure would increase the probability that both days' Chairmen would be loyal members of the ARP. The prime minister was against the chan... | Solution. Let $p_{k}$ be the probability that a member of the APP will be chosen from the delegation of the $k$-th city. Under the first method, the probability that the Chairperson will be a member of the APP on both days is $p=\left(\frac{p_{1}+\ldots+p_{5}}{5}\right)^{2}$. Under the second method, it is $p^{\prime}=... | both\\correct | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,839 |
17. The Paradox of Three Coins ${ }^{5}$ (from 9th grade, 4 points). Mitya, Vanya, and Dima each flipped a coin $2 n$ times. It is known that Mitya and Vanya got the same number of heads. What is more likely: event $A$ "Dima got exactly $n$ heads" or event $B$ "Mitya got exactly $n$ heads"? If the events are not equall... | Solution. The probability that Mitya and Vanya have the same number of heads is ${ }^{6} C_{4 n}^{2 n} / 2^{4 n}$. Clearly, $\mathrm{P}(A)=C_{2 n}^{n} / 2^{2 n}$. Let's find $\mathrm{P}(B)$:
$$
\mathrm{P}(B)=\mathrm{P}(\text { Mitya has } n \text { heads } \mid \text { Mitya and Vanya have the same number of heads })=... | \frac{C_{2n}^{n}\cdot2^{2n}}{C_{4n}^{2n}}(\approx\sqrt{2}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,840 |
18. Lonely Cars. (From 9th grade, 4 points.) On a very long narrow highway, where overtaking is impossible, $n$ cars are driving in a random order, each with its own preferred speed. If a fast car catches up to a slower one, the fast car has to slow down and drive at the same speed as the slower one. Thus, the cars for... | Solution. Let $I_{k}$ be the indicator of the event "the $k$-th car in line is alone." For $k \leq n$, this event consists of the slowest car among the first $k+1$ cars being the last, and the second slowest being the second to last. The probability of this is $\frac{1}{(k+1) k}$. If $k=n$, then this event consists of ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,841 |
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 | 10,842 |
5. The smallest set. (From 6th grade, 2 points) In a numerical set, the median is 3, the arithmetic mean is 5, and the only mode of the set is 6. What is the smallest number of numbers that can be in a set with the given properties? | Solution. It is clear that the number 6 appears at least twice in the set, and in addition, there are at least two more numbers.
If the set contains exactly four numbers $a, b, 6, 6$, then we can assume that $a \leq b \leq 3$ and, moreover, the sum of all numbers is 20, so $a+b=20-6-6=8$. Contradiction. The set cannot... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,844 |
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\}
$$
The first set has $n$ numbers, and the second set has $m$ numbers. The direct sum or simply the sum $X \oplus Y$ of these sets is the s... | Solution. a) For example, two identical sets: $X=\{0,0,1\}$ and $Y=\{0,0,1\}$. The median of each is 0, and the sum of these sets $\{0,0,0,0,1,1,1,1,2\}$ has a median of 1.
b) We will show that such sets do not exist. Let the first set $X$ consist of numbers $x_{1}, x_{2}, \ldots, x_{n}$, and the second set $Y$ consis... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,845 |
7. Easter Problem ${ }^{3}$. (From 6th grade, 2 points) Two players are playing an "Easter egg fight." In front of them is a large basket of eggs. They randomly take an egg each and hit them against each other. One of the eggs breaks, the loser takes a new egg, while the winner keeps their egg for the next round (the o... | Solution. Regardless of how events unfolded in the first ten rounds, in the 10th round, the egg that is the strongest among the first 11 eggs randomly drawn from the basket will be the winner. This means that the strongest egg will break in the 11th round only if the 12th egg drawn is stronger, that is, it is stronger ... | \frac{11}{12} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,846 |
8. Chairs. (From 7th grade, 1 point) A furniture store received a batch of office chairs. The chairs were identical in every way except for their colors: 15 chairs were black and 18 were brown. The chairs were in demand and were sold in random order. At some point, a customer on the store's website noticed that only tw... | Solution. We can assume that two random seats remain. Let's construct the experiment tree (Fig. 6). Event $A$ "the remaining seats are of the same color" is shown by the shaded oval. Its probability is
$$
\mathrm{P}(A)=\frac{15}{33} \cdot \frac{14}{32}+\frac{18}{33} \cdot \frac{17}{32}=\frac{43}{88} \approx 0.489
$$
... | 0.489 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,847 |
9. Card Drawing. Thirty-six players play a game: from a deck of 36 playing cards, players take turns to randomly draw a card. If a player draws the ace of diamonds, the player wins; if another card is drawn, the player returns it to the deck, and the next player draws a random card. They draw cards in a circle: first t... | Solution. a) Let the probability of the first player winning be $p_{1}$. Suppose the first player did not draw the diamond ace on the first try. The probability of this is $35 / 36$. At this point, the second player becomes the first, and the probability that he wins becomes equal to $p_{1}$. Returning to the beginning... | )no;b)\approx0.044 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,848 |
10. Lottery. (From 7th grade, 3 points) It so happened that the Absent-Minded Scholar had only 20 rubles left, and he needed to buy a bus ticket to go home. The ticket costs 45 rubles. Near the bus stop, instant lottery tickets are sold for exactly 10 rubles each. With a probability of \( p = 0.1 \), a ticket contains ... | Solution. As a result of one or several consecutive games, the Scientist may end up with no money at all (state 0), 10 rubles, 20, 30, 40 rubles, or he may win enough money to buy a bus ticket (state A). The possible transitions between states are schematically represented by a graph (Fig. 7). The probabilities of tran... | \frac{14}{419} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,849 |
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 | 10,850 |
12. Buses. (From 8th grade, 2 points) At the stop near the Absent-Minded Scholar's house, buses of two routes stop: No. 152 and No. 251. Both of them head to the metro station. The interval between buses No. 152 is exactly 5 minutes, and the interval between buses of route No. 251 is exactly 7 minutes. The intervals ar... | Solution. Let's assume that the Absent-Minded Scholar arrived at the bus stop at time 0 minutes. Let $x-5$ and $x$ be the moments when buses of route 152 arrived at the stop, and the Scholar did not catch the first of these buses but could have caught the second, i.e., $x-5<0 \leq x$. From this inequality, we get that ... | \frac{5}{14} | Other | math-word-problem | Yes | Yes | olympiads | false | 10,851 |
13. Line-up. (From 8th grade, 3 points) Twenty kindergarten children, including 11 girls and 9 boys, stand in a line-up at random. Find the probability that there will be no more than five girls between the first and the last boy. | Solution. Let's number the places in the row from left to right. The condition of the problem is satisfied by rows in which all boys are concentrated in the first 14 places. There are exactly $C_{14}^{9}$ such combinations, and thus such rows.
In addition to these rows, the condition is satisfied by all rows where the... | 0.058 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,852 |
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