20. (1996 American High School Mathematics Examination) 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. $\frac{7}{12}$
20. D. Reason; Let $(a, b, c)$ be the numbers obtained from rolling 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, among which 8 contain the number 2. The required probability is $\frac{8}{15}$.