1993 AHSME Problems/Problem 24

Problem

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 $a+b=$

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

Solution

$\fbox{E}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1993 AHSME Problems/Problem 24

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