Mock AIME 2 Pre 2005 Problems/Problem 3
Problem
In a box, there are 
green balls, blue balls, 
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 
balls be removed. The probability that balls are drawn can be expressed as 
, where 
and 
are relatively prime positive integers. Compute 
.
Solution
Note that the desired probability is equivalent to the probability for 
randomly drawn balls to be of different colors. There is a total of 
to choose the balls of different colors, and 
total ways. Thus, the answer is ![\[\dfrac{4 \cdot 4 \cdot 2}{\binom{13}{6}} = \dfrac{8}{429} \implies \boxed{437}.\]](http://latex.artofproblemsolving.com/5/f/8/5f812b5d00a2e62d9abce1049e538a5f9e4c0b2e.png)
-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 | ||
