24. Suppose non-zero digits are selected in the following way: the probability of digit $d$ being selected is $\log g_{10}(d+1)-\log _{10} d$. What is the probability that the digit 2 is selected from the following sets of numbers that can be selected with a probability of $-\frac{1}{2}$?
$$
\begin{array}{l}
\text { (A) }\{2,3\} \text {, } \\
\text { (B) }\{3,4\} \text {; (C) } \\
\{4,5,6,7,8\} \text {; } \\
\text { (D) }\{5,6,7,8,9\} \\
\text { (E) }\{4,5,6,7,8,9\} \text {. } \\
\end{array}
$$
24. Let $\operatorname{Pr}\left\{d_{1}, d_{2}, \cdots, d_{n}\right\}$ denote the probability that one of the numbers $d_{1}, d_{2}, \cdots, d_{n}$ is selected.

Note that $P_{r}\{d\}=\log \frac{d+1}{d}$ and $P_{r}\{d, d+1\}$
$$
=\log \frac{d+1}{d}+\log \frac{d+2}{d+1}=\log \frac{d+2}{d} .
$$

We look for a set whose probability is $2 P r\{Z\}$. For this, note that $2 P \cdot\{2\}=2 \log \frac{3}{2}$
$$
\begin{array}{l}
=\log _{4}^{9}=\log \frac{5}{4}+\log _{5}^{6}+\cdots+\log _{8}^{9} \\
=P_{r}\{4,5,6,7,8\} .
\end{array}
$$
(C) True.