5. A square 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 tossed upwards. The probability that the disc, upon landing, does not intersect with the gaps between the tiles is approximately $\qquad$
5. $\frac{4}{9}$.

To ensure the circular disc does not overlap the gaps between the tiles, the center of the disc must fall within a smaller regular hexagon that is centered with the tile and has sides parallel to the tile edges, with a distance of $6 \sqrt{3} \mathrm{~cm}$ (as shown in Figure 5). Construct $O C_{1} \perp A_{1} A_{2}$, 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 $C_{2} O=C_{1} O-C_{1} C_{2}=12 \sqrt{3}$.
Also, since $C_{2} O=\frac{\sqrt{3}}{2} B_{2} O$, therefore,
$$
B_{2} O=\frac{2}{\sqrt{3}} C_{2} O=\frac{2}{\sqrt{3}} \times 12 \sqrt{3}=24 \text {. }
$$

And $B_{1} B_{2}=B_{2} O$, so the side length of the smaller regular hexagon is $24 \mathrm{~cm}$. Therefore, the required probability is
$$
P=\frac{\text { area of the smaller regular hexagon }}{\text { area of the regular hexagon tile }}=\frac{B_{1} B_{2}^{2}}{A_{1} A_{2}^{2}}=\frac{24^{2}}{36^{2}}=\frac{4}{9} \text {. }
$$