10. (14 points) Given the set $A$ consisting of the positive integers $1,2, \cdots, 1000$, first randomly select a number $a$ from set $A$, and after selecting, return $a$ to set $A$; then randomly select another number $b$ from set $A$. Find the probability that $\frac{a}{b}>\frac{1}{3}$.
10. Solution 1 Note that
$$
P\left(\frac{a}{b}>\frac{1}{3}\right)=1-P\left(\frac{a}{b} \leqslant \frac{1}{3}\right) \text {. }
$$

By $\frac{a}{b} \leqslant \frac{1}{3} \Leftrightarrow a \leqslant \frac{1}{3} b \Leftrightarrow a \leqslant\left[\frac{1}{3} b\right]$.
Then $P\left(\frac{a}{b} \leqslant \frac{1}{3}\right)=\frac{\sum_{b=1}^{1000}\left[\frac{1}{3} b\right]}{1000^{2}}$
$$
=\frac{\sum_{k=1}^{332} 3 k+333+333}{1000^{2}}=\frac{333}{2000} \text {. }
$$

Therefore, $P\left(\frac{a}{b}>\frac{1}{3}\right)=1-\frac{333}{2000}=\frac{1667}{2000}$.
Solution 2 When $a=1$, $b=1$ or 2, there are 2 ways;
When $a=2$, $b$ increases to $3, 4, 5$, there are $2+3$
ways;
When $a=3$, $b$ increases to $6, 7, 8$, there are $2+2 \times 3$ ways; 
When $a=333$, $b$ has $2+332 \times 3$ ways; when $334 \leqslant a \leqslant 1000$, $b$ has 1000 ways.
Thus, $P\left(\frac{a}{b}>\frac{1}{3}\right)$
$$
\begin{array}{l}
=\frac{2+(2+3)+(2+2 \times 3)+\cdots+(2+332 \times 3)+667 \times 1000}{1000^{2}} \\
=\frac{333(2+166 \times 3)+667 \times 1000}{1000^{2}} \\
=\frac{1667}{2000} .
\end{array}
$$