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
$$\frac{1}{12} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6} \times 3+\frac{1}{6} \times \frac{1}{12}=\frac{1}{9} .$$