AHSME 1993 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 $a+b=$ $\text{(A) } 11 \text{(B) } 20 \text{(C) } 35 \text{(D) } 58 \text{(E) } 66$