4. From 6 pairs of different shoes, the probability of picking 4 shoes such that at least 2 of them form a pair is ( ).
(A) $\frac{6}{11}$
(B) $\frac{17}{33}$
(C) $\frac{19}{33}$
(D) $\frac{7}{165}$
4.B.

From 6 pairs of different shoes, if 4 are chosen at random, the probability that no 2 shoes form a pair is $\frac{\mathrm{C}_{6}^{4} \cdot 2^{4}}{\mathrm{C}_{12}^{4}}=\frac{16}{33}$. Therefore, the probability that at least 2 shoes form a pair is $1-\frac{16}{33}=\frac{17}{33}$.