[Solution] Let $(a, b, c)$ be the numbers rolled on a die 3 times. According to the problem, the scenarios where the sum of the first two rolls equals the third roll are:
$$
\begin{array}{l}
(1,1,2),(1,2,3),(1,3,4),(1,4,5),(1,5,6),(2,1,3),(2,2,4), \\
(2,3,5),(2,4,6),(3,1,4),(3,2,5),(3,3,6),(4,1,5),(4,2,6), \\
(5,1,6)
\end{array}
$$

In total, there are 15 such scenarios, and 8 of them include the number 2. The required probability is $\frac{8}{15}$. Therefore, the answer is $(D)$.
14.24 A regular die is rolled 3 times. If the sum of the first two rolls equals the number on the third roll, the probability that at least one of the rolls is a 2 is
(A) $\frac{1}{6}$.
(B) $\frac{91}{216}$.
(C) $\frac{1}{2}$.
(D) $\frac{8}{15}$.
(E) $\frac{7}{12}$.
(47th American High School Mathematics Examination, 1996)