$14 \cdot 11$ First, $a$ is chosen at random from $\{1,2,3, \cdots, 99,100\}$, and then $b$ is chosen at random from the same set. The probability that the units digit of $3^{a}+7^{b}$ is 8 is 
(A) $\frac{1}{16}$.
(B) $\frac{1}{8}$.
(C) $\frac{3}{16}$.
(D) $\frac{1}{5}$.
(E) $\frac{1}{4}$.
(41st American High School Mathematics Examination, 1990)
[Solution] The last digit of $3^{n}$ changes periodically with the increase of $n$ (starting from 1) in the sequence $3,9,7,1$, and the last digit of $7^{m}$ changes periodically in the sequence $7,9,3,1$.

Since their periods are both 4, and among the 100 numbers from 1 to 100, the numbers of the form $4 k, 4 k+1, 4 k+2, 4 k+3$ are evenly distributed (each has 25 numbers).

In the above last digits, only the cases $7+1, 1+7, 9+9$ result in a last digit of 8. That is, $\square$
\begin{tabular}{c|c|c|c}
\hline$a$ & $4 k+2$ & $4 k+3$ & $4 k$ \\
\hline$b$ & $4 l+2$ & $4 l$ & $4 l+1$ \\
\hline The last digit of $3^{a}+7^{b}$ & 8 & 8 & 8 \\
\hline
\end{tabular}

In this way, from the four types of numbers, the number of pairs $(a, b)$ that can be chosen is $C_{4}^{1} C_{4}^{1}=16$, and only 3 of them meet the problem's requirements.
Thus, the required probability (possibility) is $\frac{3}{16}$.
Therefore, the answer is $(C)$.