<s>[INST] How to find probability of having at least 2 out of 3 students selected to sit together? I have the problem. 25 students are seated in a circle at the campfire night. 3 students are selected (the probability of being selected for each student is the same) and asked to join a game. Find the probability of having at least 2 out of 3 students selected to sit together.
I tried. 


*

*There are $ \binom{25}{3} $ ways to select.

*If 3 students selected to sit together, then we have $ 25 $ ways.

*If there are exact 2 students selected to sit together, then we have $ 25 \times 21 $ ways.

*Probability we need to find is $ \dfrac{25 +25 \times 21 }{ \binom{25}{3}} =\dfrac{11}{46}.$
Is my solution true?    
 [/INST] Yes, your solution is correct. 
Refer to the diagram:
$\hspace{4cm}$
You considered two cases:
Case 1: triple students (order does not matter):
$$(\color{red}1,2,3),(\color{red}2,3,4),(\color{red}3,4,5),...,(\color{red}{23},24,25),(\color{red}{24},25,1),(\color{red}{25},1,2) \Rightarrow 25$$
Case 2: double students (order does not matter):
$$(\color{red}1,2,4),(1,2,5),...,(1,2,24) \Rightarrow 24-4+1=21\\
(\color{red}2,3,5),(2,3,6),...,(2,3,25)\Rightarrow 25-5+1=21\\
\vdots\\
(\color{red}{25},1,3),(25,1,4),...,(25,1,23)\Rightarrow 23-3+1=21$$
Hence the result.
Alternatively, select any student, say $1$. Now we will consider the probability of selecting neighboring one or two students. Refer to the diagram (order matters):
$\hspace{5cm}$
Hence:
$$\mathbb P(1,(2 \text{ or } 25))+P(1,3,(2 \text{ or } 4 \text{ or } 25))+P(1,24,(2 \text{ or } 23 \text{ or } 25))+\\
P(1,\color{red}4,(2 \text{ or } 3 \text{ or } 5 \text{ or } 25))+\cdots+P(1,\color{red}{23},(2 \text{ or } 22 \text{ or } 24 \text{ or } 25))=\\
\frac2{24}+\frac{3}{24\cdot 23}+\frac{3}{24\cdot 23}+\frac{4}{24\cdot 23}+\cdots+\frac{4}{24\cdot 23}=\\
\frac{2\cdot 23+3+3+4\cdot (23-4+1)}{23\cdot 24}=\frac{132}{23\cdot 24}=\frac{11}{46}.$$
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