Solution
= Solution
To find the <probability> of exactly one pair, choose its repeated face in six ways, the distinct face in five ways, and which of the three dice has that distinct face in three ways. Hence
$$
\boxed{P(T)=\frac{6\cdot5\cdot3}{216}=\frac5{12}.}
$$
As a check, the disjoint events $S,D,T$ exhaust the sample space, and $1/36+5/9+5/12=1$.