7. What? Where? When? The game wheel in the TV quiz is 

![](https://cdn.mathpix.com/cropped/2024_05_06_e03b48fad60e6548aec7g-04.jpg?height=457&width=665&top_left_y=1693&top_left_x=296)
divided into 13 identical sectors. The sectors are numbered from 1 to 13. At the beginning of the game, each sector contains an envelope with a question. Players choose a random sector using a spinner with an arrow. If this sector has already been chosen before, there is no envelope in it, and then the next sector clockwise is played. If it is also empty, the next one, and so on, until a non-empty sector is found. Before the break, the players played 6 sectors. a) (from 6th grade, 1 point) What is more likely: that among the played sectors there is sector №1 or that among the played sectors there is sector №8?

b) (from 7th grade, 5 points). Find the probability that as a result, the six sectors played in a row are numbered from №1 to №6.
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 such sequence. For example, the sequence $1,1,2,3,3,6$ and its various permutations $3,2,3,1,1,6,6,1,3,2,1,3$, etc., are suitable.

The number of sequences favorable to $A$ does not depend on the number of unselected sectors, and is therefore the same for any circle with more than six sectors. It is convenient to count these sequences on a circle with seven sectors. Then, regardless of the sequence of fallen sectors, there will be only one unplayed sector, and the played sectors will be arranged consecutively. There are a total of $7^{6}$ sequences. The number of sequences in which sector №7 is not played is exactly 7 times less: $7^{2}$.

Returning to the larger circle. There are a total of $13^{6}$ equally possible sequences on it. Therefore, the probability of event $A$ is $\frac{7^{5}}{13^{6}} \approx 0.00348$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e03b48fad60e6548aec7g-05.jpg?height=400&width=509&top_left_y=828&top_left_x=291)

The chairman is B.

The first place gives the candidate 3 points, the second place 2 points, the third place 1 point, and the fourth place 0 points. After collecting all the ballots, the electoral commission sums the points for each candidate. The winner is the one with the highest total points.

After the vote, it turned out that V (who scored the fewest points) withdrew his candidacy due to a transfer to another school. They did not vote again, but simply crossed out V from all the ballots. In each ballot, three candidates remained. Therefore, the first place was worth 2 points, the second place 1 point, and the third place 0 points. The points were summed up again.

Could it happen that the candidate who previously had the most points ended up with the fewest after V's withdrawal? If it could, provide an example of how the votes could have been distributed. If it could not, explain why.

| Council Member | 3 | 2 | 1 | 0 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | A | B | $\mathbf{5}$ | $\Gamma$ |
| 2 | A | B | $\mathbf{5}$ | $\Gamma$ |
| 3 | A | B | Б | $\Gamma$ |
| 4 | $\Gamma$ | Б | A | $\mathbf{B}$ |
| 5 | $\Gamma$ | Б | A | B |
| 6 | $\Gamma$ | Б | A | B |
| 7 | $\Gamma$ | Б | $\mathbf{A}$ | B |

| Council Member | 2 | 1 | 0 |
| :---: | :---: | :---: | :---: |
| 1 | A | Б | Г |
| 2 | A | Б | Г |
| 3 | A | Б | Г |
| 4 | Г | Б | A |
| 5 | Г | Б | A |
| 6 | Г | Б | A |
| 7 | Г | Б | A |

Solution. This is possible. Suppose that seven members voted, and the votes were distributed as shown in the table. The points were distributed as follows:

$$
\text { A }-13, \text { Б }-11, \text { В }-6, \Gamma-12
$$

Candidate A had the most points, and candidate V had the fewest.

When V withdrew his candidacy, the preference table changed, as the rows shortened due to the removal of V. The point totals also changed:

$$
\text { A }-6, \text { Б }-7, \Gamma-8
$$

Strangely, the candidate with the lowest rating changed the election results simply by no longer participating, and the leader became the underdog.

Answer: Yes, it could. $9^{3}$. Grandma's pies. In one bag there are two cabbage pies, in another two cherry pies, and in the third one cabbage and one cherry pie. The pies look and weigh the same, so it is unknown which is which. The grandson needs one pie for school. Grandma wants to give him a cherry pie, but she herself is confused by her pies and can determine the filling only by breaking a pie. The grandson does not want a broken pie; he wants a whole one.

a) (from 7th grade, 1 point). Show that grandma can act in such a way that the probability of giving the grandson a whole cherry pie is $2 / 3$.

b) (from 7th grade, 2 points). Is there a strategy that increases the probability of giving the grandson a whole cherry pie to more than $2 / 3$? If yes, find this strategy. If no, prove its absence.

Solution. a) Let grandma break a random pie. If it is a cabbage pie, then she should give the grandson any pie from any other bag. If the broken pie is a cherry pie, then she should give the grandson the second pie from the same bag.

We will show that in this case, the probability of giving the grandson a cherry pie is $2 / 3$. Let's denote the bags as КК, КВ, and ВВ according to their contents.

The broken pie can be a cabbage pie from the КК bag (probability $1 / 3$), or it can be a cabbage pie from the КВ bag (probability $1 / 6$). In the first case, any other pie will be a cherry pie with a probability of $3 / 4$, and in the second case, with a probability of $1 / 2$.

The broken pie can be a cherry pie from the ВВ bag (probability $1 / 3$), or from the КВ bag (probability $1 / 6$). In the first case, the second pie from the same bag will be a cherry pie with a probability of 1, and in the second case, with a probability of 0. The total probability that the grandson will get a cherry pie is

$$
\mathrm{P}(B)=\frac{1}{3} \cdot \frac{3}{4}+\frac{1}{6} \cdot \frac{1}{2}+\frac{1}{3} \cdot 1+\frac{1}{6} \cdot 0=\frac{2}{3}
$$

Understanding all the reasoning can be helped by the experiment tree, which shows all the probabilities (see fig.).

![](https://cdn.mathpix.com/cropped/2024_05_06_e03b48fad60e6548aec7g-06.jpg?height=412&width=1125&top_left_y=1573&top_left_x=543)

In the figure, the large letters indicate the selected pie, and the ovals outline the sets from which a random pie is chosen for the grandson. The event "the grandson gets a cherry pie" is shaded.

b) Let grandma break a random pie from each bag in turn. As soon as a cherry pie is broken from any bag, she should give the grandson the second pie from the same bag.

We will show that the probability of giving the grandson a cherry pie in this case is $3 / 4$. The grandson will get a cabbage pie only in one of two cases. Either grandma opened the ВК bag and broke the В pie (the probability of this is $1 / 3 \cdot 1 / 2=1 / 6$), or grandma opened the КК bag, broke one of the pies, then took the ВК bag and broke the В pie (the probability of this is $1 / 3 \cdot 1 / 2 \cdot 1 / 2=1 / 12$). Therefore, the grandson will get a cabbage pie with a probability of $1 / 6+1 / 12=1 / 4$, and a cherry pie with a probability of $3 / 4>2 / 3$.

The downside of this method is the risk of breaking half the pies. We were unable to find a strategy that gives a probability higher than 3/4. We would be happy if someone could.