14・13 Suppose non-zero digits are selected in such a way that the probability of selecting a digit $d$ is $\lg (d+1)-\lg d$. What is the set of digits from which the probability of selecting a 2 is $-\frac{1}{2}$ of the probability of selecting a digit from the following sets?
(A) $\{2,3\}$.
(B) $\{3,4\}$.
(C) $\{4,5,6,7,8\}$.
(D) $\{5,6,7,8,9\}$.
(E) $\{4,5,6,7,8,9\}$.
(36th American High School Mathematics Examination, 1985)
[Solution] The probability that the number 2 is selected according to the problem is
$$
\lg (2+1)-\lg 2=\lg \frac{3}{2},
$$

Notice that the probability of the numbers in set $(C)$ being selected is
$$
\begin{aligned}
& \lg \frac{5}{4}+\lg \frac{6}{5}+\lg \frac{7}{6}+\lg \frac{8}{7}+\lg \frac{9}{8} \\
= & \lg \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6} \cdot \frac{8}{7} \cdot \frac{9}{8} \\
= & \lg \frac{9}{4}=\lg \left(\frac{3}{2}\right)^{2}=2 \lg \frac{3}{2},
\end{aligned}
$$

Thus, the probability of the numbers in set $(C)$ being selected is twice the probability of 2 being selected. Therefore, the answer is $(C)$.