11. It is necessary to clearly follow the requirements for the volume of work if it is specified in the assignment.

## Task 1 (Maximum 12 points)

Let there be 5 banks in the country Alpha. The probability that a bank will close is the same for all banks and is equal to 0.05. Banks close independently of each other. At some point, a crisis occurred in the country Alpha, and the probability that a bank will close became 0.25 (it remains the same for all banks, and banks still close independently of each other). Find how much, in absolute terms, the probability that at least one bank will close has changed. Round your answer to the nearest hundredth.
# Solution:

Let's introduce the following events: event A - at least one bank in country Alpha has closed, event $\mathrm{A}_{\mathrm{i}}$ - the $\mathrm{i}$-th bank has closed, event $\mathrm{B}_{\mathrm{i}}$ - the $\mathrm{i}$-th bank has not closed. Then the opposite of event A (denoted as event B) states that no bank has closed. The sum of the probabilities of these events is 1. Therefore, $\mathrm{P}(\mathrm{A})=1-\mathrm{P}(\mathrm{B})$. Event B implies that the first bank did not close, the second bank did not close, ..., the fifth bank did not close. Clearly, event B is the intersection of events $\mathrm{B}_{\mathrm{i}}$. According to the multiplication theorem of probabilities for independent events, $\mathrm{P}(\mathrm{B})$ is the product of $\mathrm{P}\left(\mathrm{B}_{\mathrm{i}}\right)$. For each $i$, $\mathrm{P}\left(\mathrm{B}_{\mathrm{i}}\right)=1-\mathrm{P}\left(\mathrm{A}_{\mathrm{i}}\right)$. Therefore:

$$
\mathrm{P}(\mathrm{A})=1-\mathrm{P}(\mathrm{B})=1-\left(1-\mathrm{P}\left(\mathrm{A}_{\mathrm{i}}\right)\right)^{5}=1-(1-\mathrm{p})^{5}
$$

Solving this problem for two scenarios (before and after the crisis) we get:

$$
\begin{gathered}
\mathrm{P}\left(\mathrm{A}_{\text {before crisis }}\right)=1-(1-0.05)^{5}=0.2262191 ; \\
\mathrm{P}\left(\mathrm{A}_{\text {after crisis }}\right)=1-(1-0.25)^{5}=0.7626953 ; \\
|\Delta \mathrm{P}(\mathrm{A})|=\left|\mathrm{P}\left(\mathrm{A}_{\text {before crisis }}\right)-\mathrm{P}\left(\mathrm{A}_{\text {after crisis }}\right)\right|=0.5364762 \approx 0.54 .
\end{gathered}
$$

Answer: 0.54.

| Points | Evaluation Criteria for Task Completion |
| :---: | :--- |
| **12** | A correct and justified sequence of all solution steps.  All transformations and calculations are performed correctly.  The correct answer is obtained. |
| **8** | A correct sequence of all solution steps.  Gaps in the justification of the choice of probability theory formulas or  computational errors or typos that do not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
|  | The task is not solved, but its solution is significantly advanced, i.e.:  a significant part of the solution is performed correctly, possibly inaccurately (for example, the probability that no bank will close is correctly obtained); |
|  another part is either not performed or performed incorrectly, possibly even with |  |
| logical errors (for example, the change in the probability that at least one bank will close is not found). |  |
| In this case, the solution may be incomplete. |  |

## Task 2 (Maximum 15 points)

Economist Igor received a salary in 49 banknotes (denominations 50, 100, 500, and 1000 units). Find the minimum possible value of the maximum number of banknotes of one denomination that Igor has.

## Solution:

To find the minimum possible value of the maximum number of banknotes of one denomination, it is necessary to use as many banknotes of other denominations as possible. Since there are 4 types of banknotes, this means that by the pigeonhole principle, there must be at least 13 $(48 / 4 + 1 = 13)$ banknotes of one denomination. This will be the answer, as deviating from the distribution of 12 banknotes of three denominations and 13 banknotes of one denomination will either increase or keep the maximum number of banknotes of one denomination the same.

Answer: 13 banknotes.

| Points | Evaluation Criteria for Task Completion |
| :---: | :--- |
| **15** | A correct and justified sequence of all solution steps.  All transformations and calculations are performed correctly.  The correct answer is obtained. |
| **10** | A correct sequence of all solution steps.  Gaps in the justification of the minimum possible value of the maximum number of banknotes of one denomination or computational errors or typos that do not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
|  | The task is not solved, but its solution is significantly advanced, i.e.: |
| a significant part of the solution is performed correctly, possibly inaccurately (for example, |  |
| the minimum possible value of the maximum number of banknotes of one denomination is found); |  |
| another part is either not performed or performed incorrectly, possibly even with |  |
| logical errors (for example, the minimum is incorrectly justified). |  |
| In this case, the solution may be incomplete. |  |
| **0** | All cases of the solution that do not meet the criteria for scoring 5 and 10 points. |

## Task 3 (Maximum 18 points)

First-year student Masha took 10 economics books from the library. Two of them are different books by the same author (Adam Smith), and one book is by John Keynes. The authors of the other books do not interest Masha (they are not Keynes or Smith). It is important to her that the books by Smith stand together on her bookshelf, and the book by Keynes does not stand next to any of the books by Smith. In how many ways can Masha arrange all 10 books on her bookshelf?

## Solution:

Since the books by Smith must stand together on the bookshelf, we will consider them as one book. Thus, we now have 9 books and 9