For the given condition, the sample space $$=6^4$$

For favourable condition

Case I : For $$D_4$$ there are $${ }^6 C_1$$ way. Now, it appears on any one of $$D_1, D_2, D_3$$ i.e. $${ }^3 C_1 \cdot 1$$, for other two there are $$5 \times 5$$ ways.

$$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1 \cdot 1 \cdot 5^2=450 \text { Ways }$$

Case II : For $$\mathrm{D}_4$$ there are $${ }^6 \mathrm{C}_1$$ ways. Now, it appears on any two of $$D_1, D_2, D_3$$ i.e. $${ }^3 \mathrm{C}_2 .1 .1$$, for other one there are 5 ways.

$$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_2 \cdot 1^2 \cdot 5=90 \text { Ways }$$

Case III : For $$\mathrm{D}_4$$ there are $${ }^6 \mathrm{C}_1$$ ways. Now, it appears on all i.e. $$1^3$$.

$$\Rightarrow{ }^6 \mathrm{C}_1 \cdot 1^3=6$$

Total number of favourable cases

$$\begin{aligned} & =450+90+6 \\ & =546 \text { ways } \end{aligned}$$

Thus, Probability $$=\frac{546}{6^4}=\frac{91}{216}$$