$14 \cdot 27$ A box contains three shiny new coins and four dull old coins. Coins are drawn one by one without replacement from the box. If the probability that more than 4 draws are required to remove all the new coins is $\frac{a}{b}$, and $\frac{a}{b}$ is in lowest terms, then $a+b$ is
(A) 11 .
(B) 20 .
(C) 35 .
(D) 58 .
(E) 66 .
(44th American High School Mathematics Examination, 1993)
[Solution] Let $p_{3}, p_{4}$ be the probabilities of having all three shiny coins drawn when drawing three and four coins, respectively. Then $\frac{a}{b}=1-\left(p_{3}+p_{4}\right)$.
$$
p_{3}=\frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5}=\frac{1}{35},
$$
(i.e., the probability of drawing a shiny coin on the first, second, and third draws)
$$
\begin{aligned}
p_{4} & =\frac{4}{7} \cdot \frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4}+\frac{3}{7} \cdot \frac{4}{6} \cdot \frac{2}{5} \cdot \frac{1}{4}+\frac{3}{7} \cdot \frac{2}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} \\
& =\frac{3}{35} .
\end{aligned}
$$
(i.e., the sum of the probabilities of drawing an old coin on the first, second, or third draw, and a shiny coin on the fourth draw)

Thus,
$$
\frac{a}{b}=1-\left(\frac{1}{35}+\frac{3}{35}\right)=\frac{31}{35} .
$$

Therefore,
$$
a+b=66 \text {, }
$$

So the answer is $(E)$.