7. A digital clock displays hours, minutes, and seconds using two digits each (e.g., 10:09:18). Between 05:00:00 and 22:59:59 on the same day, the probability that all six digits on the clock are different is
7. $\frac{16}{135}$.

Note that 1 hour is 3600 seconds.
Thus, the total number of cases is $18 \times 3600$.
Let the time be $A B: C D: E F$. Then
$C, E \in\{0,1, \cdots, 5\}$.
(1) When $A=0, B=5$, $C, E$ have $\mathrm{A}_{4}^{2}$ possibilities, and once $C, E$ are chosen, $D, F$ have $\mathrm{A}_{6}^{2}$ possibilities. In this case, there are $\mathrm{A}_{4}^{2} \mathrm{~A}_{6}^{2}$ possible times;
(2) When $A=0, B \in\{6,7,8,9\}$, the selection of $B$ has $\mathrm{C}_{4}^{1}$ possibilities, $C, E$ have $\mathrm{A}_{5}^{2}$ possibilities, and the remaining $D, F$ have $\mathrm{A}_{6}^{2}$ possibilities. In this case, there are $\mathrm{C}_{4}^{1} \mathrm{~A}_{5}^{2} \mathrm{~A}_{6}^{2}$ possible times;
(3) When $A=1$, $C, E$ have $\mathrm{A}_{5}^{2}$ possibilities, and $B, D, F$ have $\mathrm{A}_{7}^{3}$ possibilities. In this case, there are $\mathrm{A}_{5}^{2} \mathrm{~A}_{7}^{3}$ possible times;
(4) When $A=2, B \in\{0,1\}$, $C, E$ have $\mathrm{A}_{4}^{2}$ possibilities, and $D, F$ have $\mathrm{A}_{6}^{2}$ possibilities. In this case, there are $\mathrm{C}_{2}^{1} \mathrm{~A}_{4}^{2} \mathrm{~A}_{6}^{2}$ possible times.
In summary, there are 7680 possible times.
Therefore, the required probability is $\frac{7680}{18 \times 3600}=\frac{16}{135}$.