11. (1993 American High School Mathematics Examination) 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 the number of draws required to remove all the new coins is more than 4 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
11. E. Reason: 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)
$$
\begin{aligned}
p_{4} & =\frac{4}{7} \cdot \frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} \cdot \frac{3}{7} \cdot \frac{4}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} \\
& =\frac{3}{35} .
\end{aligned}
$$
(i.e., the probability of drawing an old coin on the first, second, or third draw, and a shiny coin on the fourth draw)
Thus, $\quad \frac{a}{b}=1-\left(\frac{1}{35}+\frac{3}{35}\right)=\frac{31}{35}$,
so $a+b=66$.