1993 AMC 12 Problem 24

View the three shiny pennies as occupying three of seven draw positions

Subtract the cases in which all three shiny pennies occur among the first four positions

The shiny pennies occupy a uniformly chosen $3$-element subset of the $7$ draw positions. There are $\binom73=35$ possibilities. The third shiny penny appears by draw $4$ exactly when all three shiny positions lie among the first four, which occurs in $\binom43=4$ cases. The requested probability is $$1-\frac4{35}=\frac{31}{35}.$$ Therefore $a+b=31+35=66.$ Thus the correct answer is E.