Example 6 A square ground is paved with regular hexagonal tiles with a side length of $36 \mathrm{~cm}$. Now, a circular disc with a radius of $6 \sqrt{3} \mathrm{~cm}$ is thrown upwards. The probability that the disc, after landing, does not intersect with the gaps between the tiles is approximately $\qquad$ .
Explanation: As shown in Figure 5, to ensure that the circular disc does not overlap with the gaps between the tiles, the center of the circular disc must fall within a smaller regular hexagon that is concentric with the tiles, with sides parallel to the tile edges and a distance of \(6 \sqrt{3} \mathrm{~cm}\) from the tile edges.

Draw \(O C_{1} \perp A_{1} A_{2}\),

and \(C_{1} C_{2}=6 \sqrt{3} \mathrm{~cm}\).
Since \(A_{1} A_{2}=A_{2} O=36, A_{2} C_{1}=18\), therefore,
\(C_{1} O=\frac{\sqrt{3}}{2} A_{2} O=18 \sqrt{3}\).
Then \(C_{2} O=C_{1} O-C_{1} C_{2}\)
\(=18 \sqrt{3}-6 \sqrt{3}=12 \sqrt{3}\).
Also, \(C_{2} O=\frac{\sqrt{3}}{2} B_{2} O\), so,
\(B_{2} O=\frac{2}{\sqrt{3}} \times 12 \sqrt{3}=24\).
Since \(B_{1} B_{2}=B_{2} O\), the side length of the smaller regular hexagon is \(24 \mathrm{~cm}\).
Thus, the required probability is
\[
\begin{array}{l}
P=\text { area of the smaller regular hexagon } \\
\text { area of the larger regular hexagon } \\
=\left(\frac{B_{1} B_{2}}{A_{1} A_{2}}\right)^{2}=\left(\frac{24}{36}\right)^{2}=\frac{4}{9} .
\end{array}
\]

Note: The key to solving this problem lies in accurately understanding "the circular disc does not overlap with the gaps between the tiles," and then converting the problem into the ratio of the areas of two regular hexagons.