4. From the 20 numbers $1,2, \cdots, 20$, if 3 different numbers are chosen, then the probability that these 3 numbers form an arithmetic sequence is $($ )
A. $\frac{1}{5}$
B. $\frac{1}{10}$
C. $\frac{3}{19}$
D. $\frac{1}{38}$
Solution: D.
From these 20 numbers, any 3 numbers can be chosen in $C_{20}^{3}=1140$ ways. If the 3 numbers $a, b, c$ form an arithmetic sequence, then $a+c=2b$, so $a$ and $c$ must both be odd or both be even, and once $a$ and $c$ are determined, $b$ is also determined.
Therefore, the required probability is $p=\frac{C_{10}^{2}+C_{10}^{2}}{C_{20}^{3}}=\frac{3}{38}$.