4. From the numbers $1,2, \cdots, 20$, if 3 different numbers are chosen, then the probability that these 3 numbers form an arithmetic sequence is
A. $\frac{3}{19}$
B. $\frac{1}{19}$
C. $\frac{3}{38}$
D. $\frac{1}{38}$
Let the three numbers be $x, y, z$, then $x, z$ have the same parity, i.e., $x, z$ are chosen from 10 odd numbers or 10 even numbers. Let $A=$ “these 3 numbers form an arithmetic sequence”, thus $P(A)=\frac{2 A_{10}^{2}}{A_{20}^{3}}=\frac{1}{38}$, so the answer is $D$.