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Difference between revisions of "2000 AIME I Problems/Problem 5"

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== Problem ==
 
== Problem ==
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Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is <math>25.</math> One marble is taken out of each box randomly. The probability that both marbles are black is <math>27/50,</math> and the probability that both marbles are white is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?
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Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is <math>25.</math> One marble is taken out of each box randomly. The [[probability]] that both marbles are black is <math>27/50,</math> and the probability that both marbles are white is <math>m/n,</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. What is <math>m + n</math>?
  
 
== Solution ==
 
== Solution ==
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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>.
 
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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'''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''': 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 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}\cdot \frac{6}{15} = \frac{1}{25}</math>. <math>m + n = 26</math>.
  
 
Thus, <math>m + n = \boxed{026}</math>.
 
Thus, <math>m + n = \boxed{026}</math>.
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<!-- Solution lost in edit conflict ~~~~
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Let <math>b_1, s_1, b_2, s_2</math> respectively reprsent the number of black, and total marbles in each box. Then <math>\frac{b_1}{s_1} \cdot \frac{b_2}{s_2} = \frac{27}{50}</math>, so <math>50 | s_1s_2</math> and <math>s_1+s_2 = 25</math>. It follows that <math>5|s_1, s_2</math> and the possible pairs are <math>5,20</math> and <math>10,15</math>. For the first case, we find that <math>b_1b_2 = 54</math>, and since <math>b_1 < s_1, b_2 < s_2</math>, the only possibilities for <math>b_1,b_2</math> are <math>3,18</math>. It then follows that <math>\frac{2}{5}</math> and <math>\frac{2}{20}</math> of the marbles are white, and the answer is <math>\frac{m}{n} = \frac{2}{5} \cdot \frac{2}{20} = \frac{1}{25} \Longrightarrow m+n=\boxed{026}</math>. If we check the other case we get the same answer.-->
 
== 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}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 17:35, 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
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All AIME Problems and Solutions
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