1993 AHSME Problem 24

A box contains 3 shiny pennies and 4 dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is a/b that it will take more than four draws until the third shiny penny appears and a/b is in lowest terms, then a+b=

\text{(A) } 11\quad \text{(B) } 20\quad \text{(C) } 35\quad \text{(D) } 58\quad \text{(E) } 66

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.