32. On a twelve-sided die, the numbers $1,2,3,4$ are each marked on two faces, and the numbers 5,6,7,8 are each marked on one face. Observing, it is found that the probability of each of the 12 faces appearing is the same. The probability that the sum of the results of two rolls of this twelve-sided die is 6 is ().
(A) $\frac{1}{9}$
(B) $\frac{5}{144}$
(C) $\frac{1}{6}$
(D) $\frac{1}{12}$
(E) *
32. (A).
$$
6=5+1=4+2=3+3=2+4=1+5 \text {. }
$$

Due to the probability of the first roll showing $5,4,3,2,1$ being $\frac{1}{12}, \frac{1}{6}, \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}$; the probability of the second roll showing $1,2,3,4,5$ being $\frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{12}$, the required probability is
$$
\frac{1}{12} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6} \times 3+\frac{1}{6} \times \frac{1}{12}=\frac{1}{9} .
$$