Example 6 (1972 US Mathematical Olympiad) Suppose a random number selector can only choose from the nine digits $1,2, \cdots, 9$, and makes these choices with equal probability. Determine the probability that the product of the $n$ numbers selected after $n$ choices $(n>1)$ is divisible by 10.
To ensure that the product of the $n$ selected numbers is divisible by 10, at least one of the selections must be 5, and at least one of the selections must be an even number $2, 4, 6, 8$.

Let event $A$ represent the event that no 5 is selected, and event $B$ represent the event that no even number is selected. Let $P(X)$ denote the probability of event $X$. The probability we seek is $1-P(A \cup B)$, thus we have
$$
\begin{aligned}
1-P(A \cup B) & =1-P(A)-P(B)+P(A \cap B) \\
& =1-\left(\frac{8}{9}\right)^{n}-\left(\frac{5}{9}\right)^{n}+\left(\frac{4}{9}\right)^{n}=1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}} .
\end{aligned}
$$

Therefore, the probability that the product of the $n$ selected numbers is divisible by 10 is $1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}}$.