8、A digital clock uses two digits to display hours, minutes, and seconds (such as $10: 09: 18$). Between $05: 00: 00 \sim 22: 59: 59$ on the same day, the probability that all six digits on the clock are different is $\qquad$ -
8. $\frac{16}{135}$.

Analysis: One hour is 3600 seconds, making a total of $18 \times 3600$ scenarios. Let the time be $A B: C D: E F$, then $C, E \in\{0,1,2,3,4,5\}$.
(1) When $A=0, B=5$, $C, E$ have $A_{4}^{2}$ possibilities. Once $C, E$ are chosen, the remaining $D, F$ have $A_{6}^{2}$ possibilities. In this case, there are $A_{4}^{2} \cdot A_{6}^{2}$ possible times; (2) When $A=0, B \in\{6,7,8,9\}$, the selection of $B$ has $C_{4}^{1}$ possibilities, $C, E$ have $A_{5}^{2}$ possibilities, and the remaining $D, F$ have $A_{6}^{2}$ possibilities. In this case, there are $C_{4}^{1} \cdot A_{5}^{2} \cdot A_{6}^{2}$ possible times; (3) When $A=1$, $C, E$ have $A_{5}^{2}$ possibilities, and $B, D, F$ have $A_{7}^{3}$ possibilities. In this case, there are $A_{5}^{2} \cdot A_{7}^{3}$ possible times; (4) When $A=2, B \in\{0,1\}$, $C, E$ have $A_{4}^{2}$ possibilities, and $D, F$ have $A_{6}^{2}$ possibilities. In this case, there are $C_{2}^{1} \cdot A_{4}^{2} \cdot A_{6}^{2}$ possible times. Therefore, (1)(2)(3)(4) together have 7680 possibilities; hence the probability is $\frac{7680}{18 \times 3600}=\frac{16}{135}$