1. (1970 American High School Mathematics Exam Question) If a number is randomly selected from a set of five-digit numbers, where the sum of its digits is 43, the probability that this number is divisible by 11 is ( ).
A. $\frac{2}{5}$
B. $\frac{1}{5}$
C. $\frac{1}{6}$
D. $\frac{1}{11}$
E. $\frac{1}{15}$
1. B. Reason: In decimal, the maximum value for each digit is 9, so the sum of the five digits $d_{1}+d_{2}+d_{3}+d_{4}+d_{5}$ can be at most 45. Since the sum of the digits is 43, there are the following cases:
(1) One of the digits is 7, and the rest are 9, giving 5 possibilities:
79999, 97999, 99799, 99979, 99997.
(2) Two of the digits are 8, and the rest are 9, giving 10 possibilities:
$$
\begin{array}{l}
88999, 89899, 89989, 89998, 98899, \\
98989, 98998, 99889, 99898, 99988 .
\end{array}
$$

Among the above numbers, only $97999, 99979, 98989$ are divisible by 11.
In summary, the required probability $P=\frac{3}{15}=\frac{1}{5}$.