4. Six different cards are marked with the numbers 2, $2$, 4, $4$, 6, $6$, respectively. Three cards are drawn from them. The probability that the numbers on these three cards can serve as the lengths of the sides of a triangle is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{5}$
(C) $\frac{2}{3}$
(D) $\frac{3}{4}$
4. B.

If the numbers on the three cards drawn are all different, there are $2 \times 2 \times 2=8$ ways to draw them; if the numbers on the three cards drawn are the same, there are $3 \times 4=12$ ways to draw them. Therefore, from six different cards, there are $8+12=20$ ways to draw three cards.

To form the sides of a triangle with the three numbers, the only possibilities are:
$$
(2,4,4),(4,4,6),(2,6,6),(4,6,6) \text {. }
$$

Since the numbers on different cards can be repeated, the number of ways to draw three cards such that the numbers can form the sides of a triangle is $4 \times 2=8$.
Therefore, the required probability is $\frac{8}{20}=\frac{2}{5}$.