$7 \cdot 6$ Assuming a random number selector can only choose from the numbers $1,2, \cdots, 9$, 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.
[Solution] 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 5 is never chosen, and event $B$ represent the event that no even number is ever chosen. Let $p(E)$ denote the probability of event $E$. 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 $1-\frac{8^{n}+5^{n}-4^{n}}{9^{n}}$.