Difference between revisions of "2004 AIME II Problems/Problem 2"

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(I might've done it wrong, but solution, box)
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== Solution ==
 
== Solution ==
{{solution}}
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The [[probability]] that Terry picks two red candies is <math>\frac{10 * 9}{20 * 19} = \frac{9}{38}</math><!--<math>\frac{{10}\choose {2}}{{20}\choose {2}} = \frac{9}{38}</math>-->, and the probability that Mary picks two red candies is <math>\frac{7*8}{18*17} = \frac{28}{153}</math> (the same goes for the blue candies). The probability that Terry picks two different candies is <math>\frac{2*10*10}{20*19} = \frac{10}{19}</math><!--<math>\frac{{10}\choose {1} {10}\choose{1}}{{20}\choose {2}} = \frac{10}{19}</math>-->, and the probability that Mary picks two red candies is <math>\frac{2*9*9}{18*17} = \frac{18}{34}</math><!--<math>\frac{{9}\choose {1}{9}\choose {1}}{{18}\choose {2}} = \frac{18}{34}</math>-->. Thus, the probability becomes:
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:<math>2 * \frac{9}{38} * \frac{28}{153} + \frac{10}{19} * \frac{9}{17}</math>
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:<math>= \frac{118}{323}</math>
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Therefore, the answer is <math>441</math>.
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== See also ==
 
== See also ==
* [[2004 AIME II Problems/Problem 1 | Previous problem]]
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{{AIME box|year=2004|num-b=1|num-a=3|n=II}}
* [[2004 AIME II Problems/Problem 3 | Next problem]]
 
* [[2004 AIME II Problems]]
 

Revision as of 20:14, 12 February 2007

Problem

A jar has 10 red candies and 10 blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

The probability that Terry picks two red candies is $\frac{10 * 9}{20 * 19} = \frac{9}{38}$, and the probability that Mary picks two red candies is $\frac{7*8}{18*17} = \frac{28}{153}$ (the same goes for the blue candies). The probability that Terry picks two different candies is $\frac{2*10*10}{20*19} = \frac{10}{19}$, and the probability that Mary picks two red candies is $\frac{2*9*9}{18*17} = \frac{18}{34}$. Thus, the probability becomes:

$2 * \frac{9}{38} * \frac{28}{153} + \frac{10}{19} * \frac{9}{17}$
$= \frac{118}{323}$

Therefore, the answer is $441$.

See also

2004 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions