Difference between revisions of "Mock AIME 2 Pre 2005 Problems/Problem 3"

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== Solution ==
 
== Solution ==
Note that the desired probability is equivalent to the probability for <math>6</math> randomly drawn balls to be of different colors. There is a total of <math>4 \cdot 4 \cdot 2 \cdot 1 \cdot 1 \cdot 1</math> to choose the balls of different colors, and <math>\binom{13}{6}</math> total ways. Thus, the answer is <cmath>\dfrac{4 \cdot 4 \cdot 2}{\binom{13}{6}} = \dfrac{8}{429} \implies \boxed{437}.</cmath>
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Note that the desired probability is equivalent to the probability for <math>6</math> randomly drawn balls to be of different colors. There is a total of <math>4 \cdot 4 \cdot 2 \cdot 1 \cdot 1 \cdot 1</math> to choose the balls of different colors, and <math>\binom{13}{6}</math> total ways. Thus, the answer is <cmath>\dfrac{4 \cdot 4 \cdot 2}{\binom{13}{6}} = \dfrac{8}{429} \implies \boxed{437}.</cmath>  
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-MP8148
  
 
== See also ==
 
== See also ==
 
{{Mock AIME box|year=Pre 2005|n=2|num-b=2|num-a=4|source=14769}}
 
{{Mock AIME box|year=Pre 2005|n=2|num-b=2|num-a=4|source=14769}}

Latest revision as of 17:15, 4 August 2019

Problem

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 $m+n$.

Solution

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 \[\dfrac{4 \cdot 4 \cdot 2}{\binom{13}{6}} = \dfrac{8}{429} \implies \boxed{437}.\]

-MP8148

See also

Mock AIME 2 Pre 2005 (Problems, Source)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15