Mock AIME 2 Pre 2005 Problems/Problem 3


In a box, there are $4$ green balls, $4$ blue balls, $2$ red balls, a brown ball, a white ball, and a black ball. These balls are randomly drawn out of the box one at a time (without replacement) until two of the same color have been removed. This process requires that at most $7$ balls be removed. The probability that $7$ balls are drawn can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.


Note that the desired probability is equivalent to the probability for $6$ randomly drawn balls to be of different colors. There is a total of $4 \cdot 4 \cdot 2 \cdot 1 \cdot 1 \cdot 1$ to choose the balls of different colors, and $\binom{13}{6}$ total ways. Thus, the answer is \[\dfrac{4 \cdot 4 \cdot 2}{\binom{13}{6}} = \dfrac{8}{429} \implies \boxed{437}.\]


See also

Mock AIME 2 Pre 2005 (Problems, Source)
Preceded by
Problem 2
Followed by
Problem 4
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