3. In the range $1 \sim 2000$, if a number is randomly selected, the probability that the integer can be divided by 6 but not by 4 is ).
(A) $\frac{1}{4}$
(B) $\frac{83}{1000}$
(C) $\frac{167}{2000}$
(D) $\frac{3}{4}$
3. C.

Let event $A$ be "the number is divisible by 6", and event $B$ be "the number is divisible by 4".
Since $333<\frac{2000}{6}<334$, then $P(A)=\frac{333}{2000}$.
The least common multiple of 6 and 4 is $12, 166<\frac{2000}{12}<167$, so there are exactly 166 numbers that are divisible by both 6 and 4, i.e., $P(A B)=\frac{166}{2000}$.
Therefore, the required probability is
$$
P(A)-P(A B)=\frac{333}{2000}-\frac{166}{2000}=\frac{167}{2000} .
$$