Answer

Solution

1. 1

   For each white position $i\in\{2,\dots,n+1\}$, count its white neighbours among $\{i-1,i+1\}$:

2. 2

   \begin{itemize}[leftmargin=*,nosep]

3. 3
   • Position $2$: neighbours are $1$ (black) and $3$ (white if $n\ge 2$). White count = $1$.
4. 4
   • Position $n+1$: neighbours $n$ (white if $n\ge 2$), $n+2$ (black). White count = $1$.
5. 5
   • Positions $3,\dots,n$ (interior white): both neighbours white. White count = $2$.
6. 6

   \end{itemize}

7. 7

   Number of interior whites: $n-2$ (for $n\ge 2$; for $n=1$ trivially $0$ with 1 white at position $2$).

8. 8

   For $n\ge 2$:

9. 9
   $$P(\text{white neighbour})=\frac{1}{n}\sum_{i=2}^{n+1}\frac{\#\text{white neighbours of }i}{2}=\frac{1}{n}\cdot\frac{2\cdot 1+(n-2)\cdot 2}{2}=\frac{1}{n}\cdot\frac{2n-2}{2}=\frac{n-1}{n}=1-\frac{1}{n}.$$
10. 10

    For $n=1$: 1 white square (position 2), both neighbours black, so $P=0=1-1/1$. ✓

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