# 9. Solution.

1st method. An elementary outcome in a random experiment is a triplet of positions where children in red caps stand. Consider the event $A$ "all three red caps are next to each other." This event is favorable in 10 elementary outcomes. The event $B$ "two red caps are next to each other, and the third is separate" is favorable in 60 elementary outcomes (10 ways to choose two adjacent places, and the third place must be one of the 6 places not adjacent to the already chosen ones). The total number of ways to choose a triplet is $C_{10}^{3}=120$. Therefore, the required probability is $\mathrm{P}(A)+\mathrm{P}(B)=\frac{70}{120}=\frac{7}{12}$.

2nd method. Number the children in red caps. Consider the events $A_{1}$ "the second and third caps are next to each other," $A_{2}$ "the first and third caps are next to each other," and $A_{3}$ "the first and second caps are next to each other." We need to find the probability of the union:

$$
\begin{aligned}
\mathrm{P}\left(A_{1} \cup A_{2} \cup A_{3}\right) & =\mathrm{P}\left(A_{1}\right)+\mathrm{P}\left(A_{2}\right)+\mathrm{P}\left(A_{3}\right)- \\
- & \mathrm{P}\left(A_{1} \cap A_{2}\right)-\mathrm{P}\left(A_{1} \cap A_{3}\right)-\mathrm{P}\left(A_{2} \cap A_{3}\right)+\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3}\right) .
\end{aligned}
$$

The probability of event $A_{1}$ is 2/9 (if the second cap occupies some position, then for the third cap, nine positions remain, but only two of them are adjacent to the second). The probabilities of events $A_{2}$ and $A_{3}$ are the same:

$$
\mathrm{P}\left(A_{1}\right)=\mathrm{P}\left(A_{2}\right)=\mathrm{P}\left(A_{3}\right)=\frac{2}{9}
$$

The event $A_{1} \cap A_{2}$ consists in the third cap being between the first and second. The probability of this is $\frac{2}{9} \cdot \frac{1}{8}=\frac{1}{36}$. The same probability applies to the other two pairwise intersections:

$$
\mathrm{P}\left(A_{1} \cap A_{2}\right)=\mathrm{P}\left(A_{1} \cap A_{3}\right)=\mathrm{P}\left(A_{2} \cap A_{3}\right)=\frac{1}{36} .
$$

The event $A_{1} \cap A_{2} \cap A_{3}$ is impossible: $\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)=0$. Therefore,

$$
\mathrm{P}\left(A_{1} \cup A_{2} \cup A_{3}\right)=3 \cdot \frac{2}{9}-3 \cdot \frac{1}{36}+0=\frac{2}{3}-\frac{1}{12}=\frac{7}{12}
$$

The 3rd method is similar to the 2nd method from the second variant. Other methods of solution are also possible.
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 probability of the event "all three together" is correctly found | 1 |
| Solution is missing or incorrect | 0 |

X Internet Olympiad in Probability Theory and Statistics. Invitational Round

Answers and Solutions

Variant 2

## Short Answer Questions

| Question | Answer | Question | Answer |
| :---: | :---: | :---: | :---: |
| 1 | 1345 | 4 | 9.5 hours (570 minutes) |
| 2 | $7 / 15$ | 5 | $2 / 7$ |
| 3 | 56 | 6 | 315 |

Note. Questions $1-6$ are considered completed if the correct answer is given.

## Extended Answer Questions