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Difference between revisions of "2000 AIME I Problems/Problem 5"
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== Solution ==
 
== Solution ==
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Let <math>A</math> be the first box. Let <math>B</math> be the second box. Let <math>a</math> be the number of marbles total in <math>A</math>, and <math>b</math> in <math>B</math>.
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If we work with the problem for a little bit, we quickly see that their is no direct combinatorics way to calculate <math>m/n</math>. The [[Principle of Inclusion-Exclusion]] still requires us to find the individual probability of each box.
  
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Then, <math>a + b = 25</math>, and since the <math>ab</math> may be reduced to form <math>50</math> on the denominator of <math>\frac{27}{50}</math>, 50 is the lowest, and <math>50|ab</math>.
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Let <math>a, b</math> represent the number of marbles in each box, and [[without loss of generality]] let <math>a>b</math>. Then, <math>a + b = 25</math>, and since the <math>ab</math> may be reduced to form <math>50</math> on the denominator of <math>\frac{27}{50}</math>, <math>50|ab</math>. It follows that <math>5|a,b</math>, so there are 2 pairs of <math>a</math> and <math>b: (20,5),(15,10)</math>.
  
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Then there are 2 pairs of <math>a</math> and <math>b: (20,5),(15,10)</math>, since <math>a</math> and <math>b</math> are identical in computing probability.
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'''Case 1''': Then the product of the number of black marbles in each box is <math>54</math>, so the only combination that works is <math>18</math> black in first box, and <math>3</math> black in second. Then, <math>P(\text{both white}) = \frac{2}{20} \cdot \frac{2}{5} = \frac{1}{25},</math> so <math>m + n = 26</math>.
  
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'''Case 1''': The only combination that works is 18 black in first box, 3 black in second. Then, <math>P(\text{both white}) = \frac{2}{20} * \frac{2}{5} = \frac{1}{25},</math> so <math>m + n = 26</math>.
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'''Case 2''': The only combination that works is 9 black in both. Thus, <math>P(\text{both white}) = \frac{1}{10}\cdot \frac{6}{15} = \frac{1}{25}</math>. <math>m + n = 26</math>.
  
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'''Case 2''': The only combination that works is 9 black in both. Thus, <math>P(\text{both white}) = \frac{1}{10}*\frac{6}{15} = \frac{1}{25}</math>. <math>m + n = 26</math>.
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Thus, <math>m + n = \boxed{026}</math>.
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Thus, <math>m + n = 26</math>.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2000|n=I|num-b=4|num-a=6}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 16:34, 31 December 2007

Problem

Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$?

Solution

If we work with the problem for a little bit, we quickly see that their is no direct combinatorics way to calculate $m/n$. The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.

Let $a, b$ represent the number of marbles in each box, and without loss of generality let $a>b$. Then, $a + b = 25$, and since the $ab$ may be reduced to form $50$ on the denominator of $\frac{27}{50}$, $50|ab$. It follows that $5|a,b$, so there are 2 pairs of $a$ and $b: (20,5),(15,10)$.

Case 1: Then the product of the number of black marbles in each box is $54$, so the only combination that works is $18$ black in first box, and $3$ black in second. Then, $P(\text{both white}) = \frac{2}{20} \cdot \frac{2}{5} = \frac{1}{25},$ so $m + n = 26$.

Case 2: The only combination that works is 9 black in both. Thus, $P(\text{both white}) = \frac{1}{10}\cdot \frac{6}{15} = \frac{1}{25}$. $m + n = 26$.

Thus, $m + n = \boxed{026}$.

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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