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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 | 10,853 |
16. Bridges. (From 9th grade, 3 points) Along the southern shore of an endless sea, there stretches an archipelago of an infinite number of islands. The islands are connected by an infinite chain of bridges, and each island is connected by a bridge to the shore. In the event of a strong earthquake, each bridge independ... | Solution. Let's number the bridges as shown in Figure 11. If bridge $b_{1}$ remains intact (the probability of this is $q=1-p$), then one can cross to the shore using it.
Fig. 11.
If brid... | \frac{2}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,855 |
17. Maximum Groups. (From 9th grade, 4 points) Imitating random processes is difficult for humans. Even a series of coin tosses is almost impossible to invent "from the head." One of the reasons is that in a long series, a person subconsciously tries to avoid "too many" heads or tails in a row. In a truly random series... | Solution. For brevity, we will call a $k$-series a series where there is a group of length $k$ or more. Let's count all $k$-series of length $n (n \geq k)$ that start with a head. Denote their number by $A_{n, k}\left(A_{n, k}=0\right.$ when $\left.n<k\right)$. It is clear that
$$
A_{n, k}=A_{n-1, k}+A_{n-2, k}+A_{n-3... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,856 |
18. Decreasing density. (From 10th grade, 3 points) A continuous random variable $x$ has a density function $y=f(x)$, which is zero for $x<a$ and for $x \geq b$, and on the interval $[a ; b)$ is continuous, positive, and monotonically decreasing. Prove that the mathematical expectation of the random variable $X$ is gre... | Solution. Without loss of generality, we can assume that the median is zero. Then we need to prove that the mathematical expectation is positive. Let's schematically depict the density function (Fig. 12).
Reflect the segment of the graph on the interval $[0; -a]$ symmetrically with respect to the y-axis and divide the... | proof | Calculus | proof | Yes | Yes | olympiads | false | 10,857 |
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 | 10,858 |
4. Ministers in Anchuria. In the government cabinet of ministers in Anchuria, there are 100 ministers. Among them, there are crooks and honest ministers. It is known that among any ten ministers, at least one minister is a crook. What is the smallest number of crook ministers that can be in the cabinet? | Solution. There are no more than nine honest ministers, otherwise, a group of ten honest ministers would be found, which contradicts the condition. Therefore, the number of minister-cheats is no less than $100-9=91$.
Answer: 91. | 91 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,859 |
5. Random walk of a point. A point exits from the origin on a line and makes $a$ steps one unit to the right, $b$ steps one unit to the left in some order, where $a>b$. The range of the point's walk is defined as the difference between the greatest and the smallest coordinates of the point over the entire walk.
a) Fin... | Solution. Let's solve all three problems at once. At the end of its path, the point inevitably (regardless of the order of steps to the right and to the left) has the coordinate $a-b$. Therefore, the greatest span can be $a$, and the smallest span can be $a-b$.
The greatest span is achieved if all steps to the right a... | )b)-)b+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,860 |
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 | 10,862 |
9. Patrick and Slippers. Every day, the dog Patrick gnaws one slipper from the existing supply in the house. With a probability of 0.5, Patrick wants to gnaw a left slipper, and with a probability of 0.5 - a right slipper. If the desired slipper is not available, Patrick gets upset. How many pairs of identical slippers... | Solution. It is clear that if 7 pairs are bought, Patrick will definitely have enough of the desired, even if he chooses only left slippers every day. The question is about the smallest number of slippers that need to be bought so that with a probability of 0.8 or higher, Patrick will not be disappointed. Probability t... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,864 |
10. Even number of successes. Find the probability that heads will appear an even number of times, in an experiment where:
a) a fair coin is tossed $n$ times;
b) a coin, for which the probability of heads on a single toss is $p(0<p<1)$, is tossed $n$ times. | Solution. We will solve problem b), thereby obtaining the answer to question a) as well. The probability that heads will appear exactly $k$ times, as is known, is
$$
C_{n}^{k} p^{k} q^{n-k}
$$
where $q=1-p$ is the probability of tails. Therefore, the event "Heads appeared an even number of times" has the probability
... | \frac{1+(1-2p)^{n}}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,865 |
11. Electoral Districts. In Anchuria, presidential elections are being prepared, in which President Miraflores wants to win. Exactly half of the numerous voters support Miraflores, and the other half support Dick Maloney. Miraflores is also a voter. By law, he has the right to divide all voters into two electoral distr... | Solution. Suppose there are $2 n$ voters in Anchuria in total. It may happen that in each district, exactly half of the voters are supporters of Miraflores. The probability of this $p$ depends on the method of dividing into districts devised by Miraflores, but in any case, $p1, q=1$ and $p=0$.
If Miraflores includes o... | \frac{n-1}{2n-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,866 |
12. Gnomes and the Thunderstorm. In a terrible thunderstorm, n gnomes are climbing up a rope ladder in a chain. If there is a thunderclap, each gnome, independently of the others, may fall with probability $p(0<p<1)$. If a gnome falls, they knock down all the gnomes below them. Find:
a) The probability that exactly $k... | Solution. Let $q=1-p$ be the probability that a gnome will not fall from fright (although, it might get scared and even fall, but not from fright, but because someone fell on top of it).
a) Exactly $k$ gnomes will fall only if the $k$-th gnome, counting from the bottom, falls from fright and none of the $n-k$ gnomes a... | )p(1-p)^{n-k};b)n+1-\frac{1}{p}+\frac{(1-p)^{n+1}}{p} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,867 |
13. Where is the deviation greater? We will toss a symmetric coin $n$ times. Suppose that heads appeared $m$ times. The number $m / n$ is called the frequency of heads. The number $\frac{m}{n}-0.5$ is called the deviation of the frequency from the probability, and the number $\left|\frac{m}{n}-0.5\right|$ is called the... | Solution. Let the number of heads in a series of 10 tosses be $m_{10}$. Then the deviation of the frequency from the probability is $|\alpha|$, where $\alpha=\frac{m_{10}}{10}-0.5$. Let $\mathrm{E}|\alpha|=a$.
The second sequence, consisting of 100 tosses, can be divided into 10 series of ten tosses each. Each of thes... | Theexpectedabsolutedeviationinseriesof10tossesisgreaterthaninseriesof100tosses | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,868 |
14. Concept. In Anchuria, there are $K$ laws and $N$ ministers. The probability that a randomly chosen minister knows a randomly chosen law is $p$. One day, the ministers gathered for a council to write the Concept. If at least one minister knows a law, then this law will be considered in the Concept; otherwise, this l... | Solution. a) Let's find the probability that no minister knows the first law: $(1-p)^{N}$. Therefore, the probability that the first law will be taken into account is $q=1-(1-p)^{N}$.
The probability that exactly $M$ of the $K$ laws will be taken into account in the Concept can be easily found using the binomial proba... | )C_{K}^{M}(1-(1-p)^{N})^{M}(1-p)^{N\cdot(K-M)};b)K(1-(1-p)^{N}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,869 |
15. Twins. The probability of twins being born in Shvambria is $p$, and triplets are not born in Shvambria.
a) Estimate the probability that a Shvambrian met on the street is one of a pair of twins?
b) In a certain Shvambrian family, there are three children. What is the probability that among them there is a pair of... | Solution. a) Let $M$ pairs of twins and $N$ "singletons" walk the streets of the Shvambrian towns and villages. Then $\frac{M}{M+N} \approx p$, from which $\frac{M}{N} \approx \frac{p}{1-p}$. If we meet a random Shvambrian, the probability that they are one of the twins is
$$
\frac{2 M}{2 M+N}=\frac{2 \frac{M}{N}}{2 \... | \frac{2p}{2p+q^{2}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,870 |
16. Variance of the number of matches. A deck of playing cards is laid out on the table (for example, in a row). On top of each card, a card from another deck is placed. Some of the cards may have matched. Find:
a) the expected value of the number of matches;
b) the variance of the number of matches. | Solution. Here, the plot of a well-known matching problem is used, but the question is not about probability, but about the expectation and variance of the number of matches. Surprisingly, this problem is easier to solve than finding the probability. Again, we will use indicators.
a) Let's number the pairs from 1 to $... | )1;b)1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,871 |
17. Three-sided roof. The tower in King Arthur's castle is topped with a roof that is a triangular pyramid, where all the plane angles at the apex are right angles. The three roof slopes are painted in different colors. The red roof slope is inclined to the horizontal at an angle $\alpha$, and the blue slope at an angl... | Solution. We will assume that the drop falls on the roof randomly in the sense that the probability of it hitting a certain area of the roof is proportional to the area of the projection of this area onto a horizontal surface. Then we should be interested in the areas of the projections of the slopes onto the triangula... | \cos^{2}\gamma=1-\cos^{2}\beta-\cos^{2}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,872 |
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 smallest an... | 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 | 10,873 |
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 | 10,874 |
4. Median. In a set of 100 numbers. If one number is removed, the median of the remaining numbers is 78. If another number is removed, the median of the remaining numbers is 66. Find the median of the entire set. | Solution. Arrange the numbers in ascending order. If we remove a number from the first half of the sequence (with a number up to 50), the median of the remaining numbers will be the $51-\mathrm{st}$ number in the sequence. If we remove a number from the second half, the median of the remaining numbers will be the numbe... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,876 |
5. Easy Numbers. In the city where the Absent-Minded Scientist lives, telephone numbers consist of 7 digits. The Scientist easily remembers a telephone number if the number is a palindrome, that is, it reads the same from left to right and from right to left.
For example, the number 4435344 the Scientist remembers eas... | Solution. As is known, the first digit of a phone number can't be just any digit. Let the number of allowed first digits be $m$. Then the total number of numbers is $m \cdot 10^{6}$. A palindromic number is uniquely determined by its first four digits. Therefore, the number of seven-digit palindromes is $m \cdot 10^{3}... | 0.001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,877 |
6. Origami. Mitya is going to fold a square sheet of paper $A B C D$. Mitya calls a fold beautiful if side $A B$ intersects side $C D$ and the four resulting right triangles are equal.
Before that, Vanya chooses a random point $F$ on the sheet. Find the probability that Mitya can make a beautiful fold passing through ... | Solution. Let's unfold a beautiful fold (see fig.). The diagonal $B D$ and the fold line $S T$ intersect at point $O$. Triangles $B S O$ and $D T O$ are equal (by side and two angles). Therefore, $B O=O D$, and thus $O$ is the center of the square. This means that the fold line $S T$ passes through the center of the sq... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,878 |
7. Voters. $40 \%$ of the supporters of a certain political party are women. $70 \%$ of the supporters of this party are urban residents. Of these urban supporters, $60 \%$ are men. Are the events "party supporter - urban resident" and "party supporter - woman" independent? | Solution. Let event $A$ be "supporter is a woman", and $B$ - "supporter is an urban resident". According to the condition, $\mathrm{P}(A)=0.4$.
In the city, $70\%$ of the party's voters live, of whom $60\%$ are men. Therefore, the remaining $40\%$ are women, which means the proportion of women among urban residents is... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 10,879 |
8. Reliability of the device. The Absent-Minded Scientist designed a device consisting of a sensor and a transmitter. The average (expected) service life of the sensor is 3 years, and the average service life of the transmitter is 5 years. Knowing the distributions of the service life of the sensor and the transmitter,... | Solution. Let $\xi$ and $\eta$ be the lifespans of the sensor and the transmitter, respectively. Clearly, $\min (\xi, \eta) \leq \xi$. Transitioning to expectations, we get that the average lifespan of the device is $\operatorname{Emin}(\xi, \eta) \leq \mathrm{E} \xi=3$. Therefore, the average lifespan of the device is... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 10,880 |
9. Fly on the Grid. A fly crawls from the origin. In doing so, the fly moves only along the lines of the integer grid to the right or up (monotonic walk). At each node of the grid, the fly randomly chooses the direction of further movement: up or to the right. Find the probability that at some point:
a) the fly will b... | Solution. a) The fly can reach the point $(8 ; 10)$ in exactly 18 steps - no more, no less. Therefore, it is sufficient to consider the first 18 steps of the fly. The total number of possible paths consisting of 18 steps is $2^{18}$.
To reach the point $(8 ; 10)$ from these 18 steps, the fly must take exactly 8 steps ... | )\frac{C_{18}^{8}}{2^{18}}\approx0.167;\quadb)\frac{C_{11}^{5}\cdotC_{6}^{2}}{2^{18}}\approx0.026;\quad)\frac{2C_{9}^{2}C_{9}^{6}+2C_{9}^{} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,881 |
10. Minimum Variance. In a set of $n$ numbers, one of the numbers is 0, and another is 1. What is the smallest possible variance of such a set? What should the set be for this to happen? | Solution. Let the numbers be $a_{k}$, where $k=1, \ldots, n$, and the arithmetic mean of all numbers $a$.
The variance of the set is
$$
S^{2}=\frac{(0-a)^{2}+\left(a_{2}-a\right)^{2}+\ldots+\left(a_{n-1}-a\right)^{2}+(1-a)^{2}}{n}
$$
We can split the sum in the numerator into two groups - the first term will be comb... | \frac{1}{2n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,882 |
11. What? Where? When? Experts and Viewers play "What, Where, When" until six wins - whoever wins six rounds first is the winner. The probability of the Experts winning in one round is 0.6, and there are no ties. Currently, the Experts are losing with a score of $3: 4$. Find the probability that the Experts will still ... | Solution. Let the Experts win one round with probability $p$, and the Viewers with probability $q=1-p$.
The Experts need 3 more winning rounds to victory, while the Viewers need 2 more rounds. Thus, there can be no more than 4 rounds in total. Consider the possible chains of wins for the Experts (3) and the Viewers (V... | 0.4752 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,883 |
14. King Arthur's Advisors. King Arthur has two equally wise advisors - Merlin and Percival. Each of them finds the correct answer to any question with probability $p$ or the wrong answer - with probability $q=1-p$.
If both advisors say the same thing, the king listens to them. If they say the opposite, the king makes... | Solution. Merlin should reason as follows.
"Let's figure out whether the advisor should be fired. If there is only one advisor, then the king, following his advice, will make the correct decision with probability $p$ on any issue.
What happens when there are two advisors? The probability that both of us give the corr... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,884 |
16. Line-up. At the physical education teacher's whistle, all 10 boys and 7 girls lined up in a random order - wherever they managed to get. Find the mathematical expectation of the quantity "The number of girls standing to the left of all boys." | Solution. Let's mentally add one more boy, placing him to the right of everyone. Then the entire row is divided into 11 groups, each ending with a boy. The smallest group consists of one boy, and the largest can consist of 8 people - 7 girls and one boy.
Let's number these groups from left to right and call $\xi_{k}$ ... | \frac{7}{11} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,885 |
17. Durability of Zmey Gorynych. Ilya Muromets meets a three-headed Zmey Gorynych. Every minute, Ilya cuts off one of Zmey's heads. Let $x$ be the durability of Zmey ($x > 0$). The probability that $s$ new heads will grow in place of the cut head $(s=0,1,2)$ is
$$
p_{s}=\frac{x^{s}}{1+x+x^{2}}
$$
During the first 10 ... | Solution. Due to the mutual independence of head regrowth (the number of heads that have grown back depends only on the survivability, but not on previous events), the probability of the vector $K$ is $\mathrm{P}(K)=p_{1} p_{2} p_{2} p_{1} p_{0} p_{2} p_{1} p_{0} p_{1} p_{2}=\frac{x^{12}}{\left(1+x+x^{2}\right)^{10}}$.... | \frac{\sqrt{97}+1}{8}\approx1.36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,886 |
18. Population ${ }^{10}$. The Absent-Minded Scientist in his laboratory has developed a single-celled organism that with a probability of 0.6 divides into two such organisms, and with a probability of 0.4 dies without leaving offspring. Find the probability that after some time the Absent-Minded Scientist will not hav... | Solution. It does not matter how much time will be spent. Therefore, for simplicity, we will assume that organisms divide or die every second, but strictly one at a time. When something happens to one of them, the others patiently wait their turn. By making this assumption, we get a standard random walk problem: every ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,887 |
19. The End of Zmey Gorynych. Ilya Muromets meets the three-headed Zmey Gorynych. And the battle begins. Every minute Ilya cuts off one of Zmey's heads. With a probability of $\frac{1}{4}$, two new heads grow in place of the severed one, with a probability of $\frac{1}{3}$, only one new head grows, and with a probabili... | Solution. The problem is slightly complicated by the case when, in place of the severed head, one head grows back, and thus the number of heads does not change. Let's remove this difficulty by considering only those strikes by Ilya where the number of heads changes. We will call such strikes by Ilya Muromets successful... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,889 |
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 | 10,891 |
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 | 10,892 |
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 | 10,893 |
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 | 10,896 |
4. Code Lock. Petya approaches the entrance door with a code lock, on which buttons with digits from 0 to 9 are located. To open the door, he needs to press three correct buttons simultaneously. Petya doesn't remember the code and tries combinations one by one. For each attempt, Petya spends 2 seconds.
a) (6th grade. ... | Solution: a) The total number of possible combinations of three digits is $C_{10}^{3}=\frac{10 \cdot 9 \cdot 8}{2 \cdot 3}=120$. Therefore, Petya will spend $120 \cdot 2=240$ seconds, which is 4 minutes.
b) In the best case, Petya will need exactly 1 attempt, and in the worst case, all 120. Since all combinations are ... | )4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,897 |
5. First-grader (6th grade. 2 points). Maksim will go to the first grade on September 1, 2014, at the age of 6 years, and this day will not be his birthday. What is the probability that he was born in 2008? | Solution. Maxim could have been born on any day starting from September 2, 2007, and ending on August 31, 2008. In total, there are 365 days, considering that 2008 was a leap year with February 29. Assuming that all days are equally likely (we have no information suggesting otherwise), we find the desired probability:
... | \frac{244}{365} | Other | math-word-problem | Yes | Yes | olympiads | false | 10,898 |
6. The Strategy of the Great Combinator (6th grade. 2 points). Ostap Bender is playing chess against Vasyukov, the Champion of Russia, and the World Champion. According to the rules, to become the Absolute Champion, Ostap needs to win two games in a row. Bender has the right to choose the order in which he plays agains... | Solution. We will assume that Ostap's chances of winning against the Russian Champion are lower than against the Champion of Vasyuki, but higher than against the World Champion.
Let $p_{1}, p_{2}$, and $p_{3}$ be the probabilities of Ostap winning against his opponents in the order they follow.
Then the probability o... | OstapshouldchoosetheChampionofVasyukitheopponent,itdoesnotmatterwhowillbethefirstthird | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,899 |
7. Multiplicative Dice. Fedia suggested Vasya play a game: roll two dice and multiply the numbers that come up.
- The largest product is 36, and the smallest is 1, - Fedia declared. Let's do this: if the product is from 1 to 18, I win, and if it's from 19 to 36, you win.
a) (6th grade. 2 points). Should Vasya agree t... | Solution. a) It is necessary to recalculate the combinations of two dice where the product is from 1 to 18 and separately the combinations where the product is from 19 to 36. This can be done, for example, using a table. It is clear that there are more of the former. All combinations are equally likely, so Fedya is in ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,900 |
11. Stocks of "Maloney & Co". On Monday, April 7, the stock of the "Maloney & Co" Trading House was worth 5 Anchurian dollars. Over the next six days, the stock did not fall in price or even rose, and by Sunday, April 13, the stock price reached 5 dollars and 14 cents. For the entire following week, the stocks did not ... | Solution. b) The lowest average stock price over the period from April 7 to April 13 is no less than $\frac{5+5+5+5+5+5+5.14}{7}=5.02$ dollars.
The highest average stock price over the period from April 14 to April 20 is no more than $\frac{5.14+5.14+5.14+5.14+5.14+5.14+5}{7}=5.12$ dollars. Thus, the difference betwee... | B | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,901 |
9. Sleepy Flies (7th grade. 1 point). One day in October, Nikita glanced at the clock hanging on the wall and saw that four flies had fallen asleep on the clock face. The first one was sleeping precisely at the 12 mark, while the others were also precisely positioned at the 3, 6, and 9 marks.
- How beautiful! - Nikita... | Solution. The event "exactly 2 flies are swatted" occurs only if at the moment when Petya looked at the clock, the minute hand was between 11 and 12 o'clock, 2 and 3 o'clock, 5 and 6 o'clock, or 8 and 9 o'clock. It is not hard to notice that in total these intervals amount to 4 hours, which covers exactly $\frac{1}{3}$... | \frac{1}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,902 |
10. Fyodorovna's Tea Party. (7th grade. 3 points). Fyodorovna has 3 teacups. One day, she washed them and started a new life. Each time she wants to drink tea, Fyodorovna randomly picks one of the cups, and then puts it back with the others, but does not wash it again. Find the probability that during the fifth tea par... | Solution. The specified event will occur only if the first four times Fedor used some two cups (with each being used at least once), and the fifth time - the last clean cup. For definiteness, let's number the cups. Then there are 16 ways to use cups 1 and 2 four times. Two of these ways are not suitable - these are the... | \frac{14}{81} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,903 |
11. The First Ace. In a certain card game, four players participate. To determine who starts the game, they usually draw lots "until the first ace." There are 32 cards in the deck, from sevens to aces. One of the players deals the cards face up to each player in turn, clockwise (to himself - last), until someone gets a... | Solution. a) It is naturally assumed that the deck is well shuffled, meaning each card can end up in any of the 32 possible positions with equal probability.
Let's assign the players numbers $1, 2, 3$, and 4. The dealer (4) starts with player (1). Player (1) will get the ace first if the first ace is in position $1, 5... | 0.3008,0.2647,0.2320,0.2024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,904 |
12. Predictions (8th grade. 2 points). On Monday, Misha has five lessons, and on Tuesday - six. To determine which lessons unpredictable teachers will ask him about his homework, Misha tosses a coin 11 times - for the number of possible troubles. If it lands on heads, Misha assumes he will be asked on that lesson, if t... | Solution. The probability of guessing or not guessing a particular lesson is 0.5, regardless of the probability that the teacher will ask for Mishin's homework.
We are dealing with the conditional probability of the intersection of events $A_{1}$ "3 correct predictions on Monday" and $A_{2}$ "4 correct predictions on ... | \frac{5}{11} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,905 |
13. Fox Alice and Cat Basilio (8th grade. 2 points). Every day, Cat Basilio and Fox Alice visit all 20 courtyards of the capital of the Country of Fools, and in each courtyard, they either receive or do not receive one gold coin with a probability of $\frac{1}{2}$. If the number of gold coins collected by the end of th... | Solution: Let $X$ be the revenue of the cat, and $Y$ be the revenue of the fox. Then the random variable $X+Y$ has an expected value of 10. The random variable $X-Y$ with equal probabilities takes the values 0 and 1. Therefore, $\mathrm{E}(X-Y)=0.5$.
Represent $X$ as $\frac{(X+Y)+(X-Y)}{2}$. We get:
$$
\mathrm{E} X=... | 5.25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,906 |
16. Check Digit (8th grade. 3 points). The Simple-Dairy Telephone Company “Simple-Telecom” uses three-digit telephone numbers. The equipment is old, so during connection, errors in individual digits of the transmitted subscriber number are possible - each digit, independently of the others, has a probability of $p=0.02... | Solution. Numbers that satisfy the rule will be called correct. The task is to ensure that, despite the erroneous digits, the number turns out to be correct.
If two digits are known, then the third digit is uniquely determined. Therefore, we will not consider the first digit separately, but will assume that all digits... | 0.000131 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,909 |
17. Closed Rural Store (8th grade. 3 points). At noon, Anna Kuzminichna looked out the window and saw Klava, the store clerk, leaving for a break. At two minutes past one, Anna Kuzminichna looked out the window again and saw that no one was in front of the closed store. Klava was absent for exactly 10 minutes, and when... | Solution. It is reasonable to assume that the arrival time of each customer is uniformly distributed over the permissible interval. This can be understood as: the probability of arrival at each minute (or second) is the same. More precisely - the probability of arrival in some time interval is proportional to the durat... | 0.75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,910 |
18. Weather Diary (8th grade. 5 points). Throughout March, 7th-grader Ivanov kept a weather observation diary. Every day at exactly 3 PM, he would peek out the kitchen window, look at the outdoor thermometer, and record the air temperature in his diary. After statistical processing, the results were:
- Average tempera... | Solution. Probably, the math teacher knew that for any set of numbers, the median is not more than one standard deviation away from the arithmetic mean:
$$
|\bar{x}-m| \leq \sqrt{S^{2}} \text {, hence }(\bar{x}-m)^{2} \leq S^{2} \text {. }
$$
Let's prove this. Suppose the set consists of numbers $x_{1}, x_{2}, \ldots... | (0-4)^2<15.917 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,911 |
19. Pills. On February 28, the doctor prescribed Pills for Absentmindedness to the Absent-Minded Mathematician, which he had to take once a day after lunch. The Mathematician bought two bottles, each containing 10 pills. Starting from March 1, the Mathematician takes one of the bottles to work with him each day (choosi... | Solution. a) The specified event will occur only if the Mathematician ate 10 pills from one bottle and 3 pills from the other during the first 13 days of March, and on March 14th, he accidentally took the empty bottle with him. The probability of this is
$$
2 \cdot C_{13}^{3} \frac{1}{2^{13}} \cdot \frac{1}{2}=C_{13}^... | )\frac{143}{4096}\approx0.035;b)17.3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,912 |
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 | 10,913 |
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 | 10,915 |
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 | 10,916 |
8. Solution. The greatest deviation from the average among the measurements that have been preserved is 39 g, which does not exceed $10 \%$ of the nominal mass of the portion. The deviations of the measurements that cannot be read are less than 39 g: for example, the number 38 ? is not less than 380, so the deviation o... | Answer: not required.
## Evaluation Criteria
The solution is complete and correct. Other reasoning is possible. For example, the participant added the missing values in such a way as to not reduce the standard deviation, found the standard deviation of the resulting set, and indicated that the deviation of this set i... | notrequired | Other | math-word-problem | Yes | Yes | olympiads | false | 10,917 |
8. Solution. The greatest deviation from the average among the measurements that have been preserved is 37 g, which does not exceed $10 \%$ of the nominal mass of the portion. The deviations of the measurements that cannot be read are less than 37 g: for example, the number 41 ? is less than 420, so the deviation of th... | Answer: not required.
## Evaluation Criteria
The solution is complete and correct. Other reasoning is possible. For example, the participant added the missing values in such a way as to not reduce the standard deviation, found the standard deviation of the resulting set, and indicated that the deviation of this set i... | notrequired | Other | math-word-problem | Yes | Yes | olympiads | false | 10,918 |
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 | 10,919 |
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 | 10,921 |
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 | 10,922 |
16. Extra Crocodiles (from 9th grade, 6 points). A chocolate egg manufacturer with a toy inside announced that a new collection is being released, featuring ten different crocodiles. The crocodiles are uniformly and randomly distributed among the chocolate eggs, meaning that in a randomly chosen egg, each crocodile can... | Answer: approximately 3.59.
Solution. Let's assume there are a total of $n$ crocodiles in the collection. We will number the exhibits in the order in which they appeared in the collection. For example, the crocodile with glasses was the first, so it is a crocodile of type 1. After some number of attempts, a crocodile ... | 3.59 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,923 |
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 | 10,924 |
14. Concept. In Anchuria, there are $K$ laws and $N$ ministers. The probability that a randomly chosen minister knows a randomly selected law is $p$. One day, the ministers gathered for a council to write the Concept. If at least one minister knows a law, then this law will be considered in the Concept; otherwise, this... | Solution. a) Let's find the probability that no minister knows the first law: $(1-p)^{N}$. Therefore, the probability that the first law will be taken into account is $q=1-(1-p)^{N}$.
The probability that exactly $M$ of the $K$ laws will be taken into account in the Concept can be easily found using the binomial proba... | )C_{K}^{M}(1-(1-p)^{N})^{M}(1-p)^{N\cdot(K-M)};b)K(1-(1-p)^{N}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,926 |
1. Conduct the following experiment 10 times: first, toss a coin 10 times in a row and record the number of heads, then toss the coin 9 times in a row and also record the number of heads. We will call the experiment successful if the number of heads in the first case is greater than in the second. After conducting a se... | # Solution.
a) The problem is a particular case of the more general problem from part b).
Answer: $\frac{1}{2}$.
b) Let Vanya first toss the coin $n$ times, just like Tanya. After this, the following events arise:
$$
\begin{aligned}
A= & \{y \text { Vanya has more heads }\} ; B=\{y \text { Tanya has more heads }\} ... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,927 |
2. In a school football tournament, 8 teams of equal skill are participating. Each game ends with one team winning. A randomly drawn number determines a team's position in the tournament table:
| Quarterfinals | | Semifinals | Final |
| :---: | :---: | :---: | :---: |
| 1 | ? | | |
| 2 | ? | 0 | |
| | | | |
| ... | # Solution.
a) For the teams to meet in the semifinal, they must end up in different but converging to one semifinal subgroups (Event $X$).
Note: On the diagram, one subgroup consists of positions "1" and "2", "3" and "4", etc.
Team "A" can end up in any subgroup. For Team "B" to meet "A" in the semifinal, it must e... | \frac{1}{14} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,928 |
3. Ivan Semenov is taking a Unified State Exam (USE) in mathematics. The exam consists of three types of tasks: A, B, and C. For each task of type A, there are four answer options, only one of which is correct. There are 10 such tasks in total. Tasks of types B and C require detailed answers. Since Ivan constantly skip... | # Solution.
Event $A_{2 n}=\{$ Ivan Semenov receives $n$ points or more $\}$.
Event $B=\{$ he solves a problem of type $B\}$.
Then $\mathrm{P}(\{$ he gets more than 5 points $\})=\mathrm{P}\left(A_{4}\right) \mathrm{P}(B)+\mathrm{P}\left(A_{6}\right) \mathrm{P}(\bar{B})$.
Obviously, the probability of correctly gue... | 0.088 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,929 |
4. A city is considered a millionaire city if it has a population of more than one million people. Which event has a higher probability:
$$
A=\{\text { a randomly chosen city dweller lives in a millionaire city }\}
$$
$$
B=\{\text { a randomly chosen city is a millionaire city }\} ?
$$ | Justify the answer
Take the statistics on the urban population of Russia from the website http://www.perepis2002.ru/ct/doc/1_TOM_01_05.xls. Check if your previous conclusion is valid for Russia. For this, calculate the probability that a randomly chosen urban resident lives in a city with a population of over a millio... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,930 |
5. The final score in figure skating is determined as follows. The judging panel consists of ten people. Each judge gives the athlete their own score for the performance. After that, seven of the ten received scores are randomly selected. The sum of these seven scores is the final score. Places among the athletes are d... | Solution.
a) a special case of point b).
Answer: Yes, it could.
b) Let's provide an example where such a situation is possible.
| Athlete | Scores | | | | | | | | | | Sum of 10 | Sum of 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 1... | Yes,itcould | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,931 |
6. Is it possible:
a) to load two coins so that the probabilities of getting "heads" and "tails" are different, but the probabilities of getting any of the combinations "tails, tails", "heads, tails", "heads, heads" are the same?
b) to load two dice so that the probability of getting any sum from 2 to 12 is the same? | # Solution.
a) We will show that it is impossible to load the coins in such a way. Immediately note that if both coins are symmetric, the probability of the combination "heads, tails" is higher than the probability of the other combinations. Now, without loss of generality, we can assume that the probability of "heads... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,932 |
7. Petya is playing a computer game called "Pile of Stones." Initially, there are 16 stones in the pile. Players take turns taking 1, 2, 3, or 4 stones from the pile. The player who takes the last stone wins. Petya is playing for the first time and therefore takes a random number of stones each time, without violating ... | # Solution.
Note that the player making the first move always has an advantage and wins with the correct strategy. Indeed, on the first step, one should take one stone from the pile, and on each subsequent step, take such a number of stones so that the remaining number of stones is divisible by 5. Since according to t... | \frac{1}{256} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,933 |
8. Petya suggests Vasya play the following game. Petya gives Vasya two boxes of candies. Each of the two boxes contains chocolate candies and caramels. In total, there are 25 candies in both boxes. Petya suggests Vasya take one candy from each box. If both candies turn out to be chocolate, Vasya wins. Otherwise, Petya ... | # Solution.
Since Vasya will pull out two caramels with a probability of 0.54, the probability that he will pull out two chocolate candies is certainly no more than $1-0.54=0.46$, i.e., the probability of winning for Vasya is less than $\frac{1}{2}$. Therefore, Petya has a higher chance of winning.
Answer: Petya. | Petya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,934 |
9. On each of four cards, a natural number is written. Two cards are randomly taken, and the numbers on them are added. With equal probability, this sum can be less than 9, equal to 9, and greater than 9. What numbers can be written on the cards?
# | # Solution.
Let the natural numbers $a, b, c, d$ be written on four cards. Two out of the four cards can be chosen in six different ways. Since, according to the problem, a pair of cards is chosen randomly, the probability of selecting a specific pair is $\frac{1}{6}$. Furthermore, since the events $\{s9\}$ (where $s$... | (1;2;7;8);(1;3;6;8);(1;4;5;8);(2;3;6;7);(2;4;5;7);(3;4;5;6) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,935 |
10. There are two symmetrical dice. Can the numbers on their faces be written in such a way that the sum of the points when rolled takes the values $1,2, \ldots, 36$ with equal probabilities
# | # Solution.
Let's provide an example of such dice. Suppose the first die has the numbers 1, 2, $3, 4, 5, 6$, and the second die has $0, 6, 12, 18, 24, 30$. Notice that any number from 1 to 36 can be uniquely represented as the sum of numbers written on the two dice. Thus, the sum of the points when rolling the dice ta... | Yes,itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,936 |
11. There are 25 children in the class. Two are chosen at random for duty. The probability that both duty students will be boys is $\frac{3}{25}$. How many girls are in the class? | # Solution.
Let there be $n$ boys in the class, then the number of ways to choose two duty students from them is $\frac{n(n-1)}{2}$, the number of ways to choose two duty students from the entire class is $25 \cdot 24$. The ratio of the two obtained fractions, i.e., $\frac{n(n-1)}{25 \cdot 24}$. It is also equal to $\... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,937 |
12. In the first quarter, Vasya had 5 math grades, with the most being fives. It turned out that the median of all grades was 4, and the arithmetic mean was 3.8. What grades could Vasya have had? | # Solution.
Let's write down Vasya's grades in ascending order. The grade in the third (middle) position will be 4. Therefore, the two grades to the right of it can only be $(4,4),(4,5)$, or $(5,5)$. The first two options are not suitable, as then the number of fours would be greater than the number of fives, which co... | (2,3,4,5,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,938 |
13. There are fewer than 30 people in the class. The probability that a randomly chosen girl is an excellent student is $\frac{3}{13}$, and the probability that a randomly chosen boy is an excellent student is $\frac{4}{11} \cdot$ How many excellent students are there in the class? | # Solution.
According to classical probability theory, the probability that a randomly chosen girl is an excellent student is equal to the ratio of the number of excellent girl students to the total number of girls in the class. Accordingly,
$$
\frac{3}{13}=\frac{\text { number of girls-excellent students }}{\text { ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,939 |
14. Anya, Borya, and Vasya decided to go to the "Fir Tree". They agreed to meet at the bus stop, but they don't know when each of them will arrive. Each of them can arrive at a random time between 15:00 and 16:00. Vasya is the most patient of all: if he arrives and there is neither Anya nor Borya at the stop, he will w... | # Solution.
Since Anya will not wait at all, the boys will only go to the "Fir Tree" together if Anya arrives last. Clearly, these events are independent, so
$$
\begin{gathered}
\mathrm{P}(\{\text { all three will go to the "Fir Tree" together }\})= \\
=\mathrm{P}(\{\text { Anya will arrive last }\}) \mathrm{P}(\{\te... | 0.124 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,940 |
15. Three tired cowboys entered a saloon and hung their hats on a buffalo horn at the entrance. When the cowboys left deep into the night, they were unable to distinguish one hat from another, and therefore picked three hats at random. Find the probability that none of them took their own hat. | # Solution.
Note that the cowboys have 6 different ways to take the hats. Indeed, the first cowboy can take any of the three hats, the second one can take one of the remaining two, and the third will take the last hat.
At the same time, in only two out of the six scenarios, none of the cowboys will take their own hat... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,941 |
17. On a roulette wheel, any number from 0 to 2007 can come up with equal probability. The roulette wheel is spun repeatedly. Let $P_{k}$ be the probability that at some point the sum of the numbers that have come up in all the spins equals $k$. Which number is greater: $P_{2007}$ or $P_{2008} ?$ | # Solution.
Reasoning in the same way as in the previous problem (the only difference is that instead of the number 6 - the number 2008), we find:
$$
P_{n}=\frac{1}{2008} \sum_{k=0}^{2007} P_{n-k} ; \frac{2007}{2008} P_{n}=\frac{1}{2008} \sum_{k=1}^{2007} P_{n-k} ; P_{n}=\frac{1}{2007} \sum_{k=1}^{2007} P_{n-k}
$$
S... | P_{2007} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,942 |
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 | 10,943 |
19. On the occasion of the start of winter holidays, all the boys from 8th grade class "V" went to the shooting range. It is known that there are $n$ boys in 8th grade class "V". At the shooting range where the boys went, there are $n$ targets. Each boy randomly selects a target for himself, and some boys could have ch... | # Solution.
a) Let $p$ be the probability that a given target is hit. Then, by symmetry, the expected number of hit targets is $np$. Thus, it remains to find $p$. Note that the probability that a given boy will hit a given target is $\frac{1}{n}$. Therefore, the probability that a given boy will not hit a given target... | cannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,944 |
8. Solution. Let's find the probability of the opposite event “the groups will not be able to contact each other.” Consider pairs of tourists where the first tourist is from the first group and the second is from the second group. There are a total of $5 \cdot 8=40$ such pairs. Therefore, the probability that none of t... | Answer: $1-(1-p)^{40}$.
| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| Probability of the opposite event is found, last step is missing | 2 |
| Principle of solution is correct, but the number of pairs is calculated incorrectly | 1 |
| Solution is missing or completely inc... | 1-(1-p)^{40} | Other | math-word-problem | Yes | Yes | olympiads | false | 10,945 |
# 9. Solution.
1st method. An elementary outcome in a random experiment is a triplet of places 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 se... | 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 | 10,946 |
# 8. Solution.
Let's find the probability of the opposite event "the groups will not be able to contact each other." Consider pairs of tourists where the first tourist is from the first group and the second is from the second group. There are a total of $6 \cdot 7=42$ such pairs. Therefore, the probability that none o... | Answer: $1-(1-p)^{42}$
| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| Probability of the opposite event is found, last step is missing | 2 |
| Principle of solution is correct, but the number of pairs is calculated incorrectly | 1 |
| Solution is missing or completely inco... | 1-(1-p)^{42} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,947 |
# 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 | 10,948 |
1. Heating season. The municipal government of Zatonsk has adopted a rule: heating in houses should be turned on no earlier than October 26, but only if the average temperature over the previous three days is below $8^{\circ} \mathrm{C}$. The city has two districts: Coastal and Riverside.
In the Coastal district, the ... | # Solution.
a) In Zarechny District, the average temperature over a three-day period is calculated. For the period of October 23-25, the average temperature is $8.33^{\circ} \mathrm{C}$, so heating should not be turned on on October 26. However, for the period from October 24 to 26, the average temperature is below $8... | ) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,950 |
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 | 10,951 |
3. Translator (from 6th grade. 1 point). Vasily Petrov is completing an English language assignment. In this assignment, there are 10 English expressions and their Russian translations in random order. The task is to establish the correct correspondences between the expressions and their translations. 1 point is given ... | Solution. It cannot happen that exactly nine correct answers are guessed, because in this case the tenth answer would also be correct. Therefore, the desired probability is zero.
Answer: 0 | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,952 |
4. Defective Coins (from 6th grade, 1 point). For the anniversary of the St. Petersburg Mathematical Olympiads, the mint struck three commemorative coins. One coin turned out to be correct, the second coin had two eagles on both sides, and the third coin had tails on both sides. The director of the mint randomly select... | Solution. It is known that heads came up. Therefore, either the first or the second coin was chosen. In total, heads are depicted three times on these coins. In one of these three equally likely cases, tails is on the opposite side of the coin, and in the other two cases, heads are on the opposite side. Answer: $\frac{... | \frac{2}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,953 |
5. Intersecting Diagonals (from 7th grade. 2 points). In a convex hexagon, two random diagonals are chosen independently of each other. Find the probability that these diagonals intersect inside the hexagon (inside means not at a vertex). | Solution. A hexagon has a total of 9 diagonals, and thus $\frac{9 \cdot 8}{2}=36$ pairs of diagonals. It remains to determine how many pairs of diagonals intersect inside the hexagon. Diagonals in a hexagon can be of two types: main (connecting opposite vertices) and non-main. For example, diagonal $A C$ is non-main. T... | \frac{5}{12} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,954 |
6. Three targets. A shooter shoots at three targets until all are hit. The probability of hitting a target with one shot is $p$.
a) (from 7th grade. 2 points). Find the probability that exactly 5 shots will be required.
b) (from 8th grade. 2 points). Find the expected number of shots. | # Solution.
a) The desired event occurs only if the first four shots hit two targets, and the last shot was successful. The probability of this is
$$
C_{4}^{2} p^{2}(1-p)^{2} \cdot p=6 p^{3}(1-p)^{2}
$$
b) Note that with a hit probability of $p$, the average number of shots needed to hit the first target is $1 / p$.... | )6p^{3}(1-p)^{2};b)\frac{3}{p} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,955 |
7. Pairings (from 8th grade. 2 points). At the tournament, 10 tennis players arrived, 4 of whom are from Russia. According to the rules, for the first round, the players are randomly paired. Find the probability that in the first round all Russian players will be paired only with other Russian players. | Solution. Let's take one Russian woman. In pa-
the probability that another Russian woman will be paired with her is $\frac{3}{9}=\frac{1}{3}$.
Given this condition, let's consider the two... | \frac{1}{21} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,956 |
8. (for 9th grade. 2 points). In triangle $A B C$, angle $A$ is equal to $40^{\circ}$. The triangle is randomly thrown onto a table. Find the probability that vertex $A$ will be to the east of the other two vertices. | Solution. Draw a line through vertex $A$ running strictly from south to north. The event “Vertex $A$ is east of the other two” occurs if and only if vertices $B$ and $C$ are located in the western half-plane from the drawn line. The diagram shows the two extreme positions of the triangle where vertices $B$ and $C$ are ... | \frac{7}{18} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,957 |
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 | 10,958 |
10. Disks (from 9th grade. 3 points). At a familiar factory, metal disks with a diameter of 1 m are cut out. It is known that a disk with a diameter of exactly 1 m weighs exactly 100 kg. During manufacturing, there is a measurement error, and therefore the standard deviation of the radius is 10 mm. Engineer Sidorov bel... | Solution. Given $\mathrm{E} R=0.5 \mathrm{m}, \mathrm{D} R=10^{-4}$ (sq.m). Let's find the expected value of the area of one disk:
$$
\mathrm{ES}=\mathrm{E}\left(\pi R^{2}\right)=\pi \mathrm{E} R^{2}=\pi\left(D R+\mathrm{E}^{2} R\right)=\pi\left(10^{-4}+0.25\right)=0.2501 \pi
$$
Thus, the expected value of the mass o... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,959 |
11. Intersecting Diagonals 2 (from 9th grade. 3 points). In a convex polygon with an odd number of vertices, equal to $2 n+1$, two random diagonals are chosen independently of each other. Find the probability that these diagonals intersect inside the polygon. | Solution. In a $(2n+1)$-gon,
$$
C_{2 n+1}^{2}-(2 n+1)=\frac{(2 n+1) \cdot 2 n}{2}-(2 n+1)=2 n^{2}-n-1
$$
diagonals, and, consequently, the number of pairs of diagonals is
$$
C_{2 n^{2}-n-1}^{2}=\frac{\left(2 n^{2}-n-1\right)\left(2 n^{2}-n-2\right)}{2}
$$
Two intersecting diagonals are uniquely determined by the ch... | \frac{n(2n-1)}{3(2n^{2}-n-2)} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,960 |
12. Stunning news. 18 scientists have arrived at the conference, and exactly 10 of them know the stunning news. During the break (coffee break), all scientists randomly pair up, and in each pair, each scientist who knows the news tells it to the other if they haven't heard it yet.
a) (from 6th grade. 1 point). Find th... | Solution. a) After the coffee break, the news will be known by everyone who ended up in a pair where there is at least one "knower" (someone who knew the news beforehand). Since these scientists form an even number of pairs, their number is even. Therefore, the probability that there will be 13 of them is zero.
b) The... | )0;b)\frac{1120}{2431};)14\frac{12}{17} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,961 |
13. Small Tetris. A tall rectangle of width 2 is open at the top, and $\Gamma$-trimino (see figure) falls into it in a random orientation.
a) (from 9th grade. 3 points). $k$ trimino have fallen. Find the expected height of the resulting polygon.
b) (from 10th grade. 6 points). 7 trimino have fallen. Find the probabil... | Solution. a) Let the random variable "height of the resulting polygon" be denoted by $X$. Clearly, $X=2 k-\left(I_{2}+I_{3}+\ldots+I_{k}\right)$, where $I_{j}$ is the indicator of the event "trimino with numbers $j$ and $j-1$ form a block of height 3". This can only happen in two cases, as shown in the left figure. The... | \frac{15k+1}{8};0.133 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,962 |
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 there are no two hats of the same size. The gentlemen leave one by one in a random order.
When seeing off each gentleman, the doorman of the club tr... | 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 | 10,964 |
16. Draws (from 9th grade. 6 points). Two hockey teams of equal strength agreed to play until the total score reaches 10. Find the mathematical expectation of the number of times a draw occurred. | Solution. If $2n$ is the maximum total score, then the game can be viewed as a random walk of length $2n$: at each step, the score difference either increases or decreases by one.
Let $I_{... | 1.707 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,965 |
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 | 10,967 |
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 uniformly and randomly distributed among the chocolate eggs, meaning that in a randomly selected egg, each cro... | 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 | 10,968 |
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