7. Let an "operation" be the act of randomly changing a known positive integer $n$ to a smaller non-negative integer (each number has the same probability). Then, the probability that the numbers $10$, $100$, and $1000$ all appear during the process of performing several operations to reduce 2019 to 0 is $\qquad$
7. $\frac{1}{1112111}$.

Let the operation process produce $a_{1}, a_{2}, \cdots, a_{n}$ in addition to $2019, 10, 100, 1000, 0$.
$$
\begin{array}{l}
\text { Then the probability is } \frac{1}{2019} \times \frac{1}{10} \times \frac{1}{100} \times \frac{1}{1000} \prod_{k=1}^{n} \frac{1}{a_{k}} . \\
\text { Therefore, } p=\sum \frac{1}{2019} \times \frac{1}{10} \times \frac{1}{100} \times \frac{1}{1000} \prod_{k=1}^{n} \frac{1}{a_{k}} \\
=\frac{1}{2019} \times \frac{1}{1000} \times \frac{1}{100} \times \frac{1}{10} \prod_{k=10,100,1000}^{2018}\left(1+\frac{1}{k}\right) \\
=\frac{1}{1001 \times 101 \times 11}=\frac{1}{1112111} .
\end{array}
$$