2000 AIME I Problems/Problem 5: Difference between revisions
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== Problem == | == Problem == | ||
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>? | 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>. | ||
'''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>. | *'''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>. | ||
'''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>. | *'''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>. | ||
<!-- Solution lost in edit conflict ~~~~ | |||
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 16:35, 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 
One marble is taken out of each box randomly. The probability that both marbles are black is 
and the probability that both marbles are white is 
where 
and 
are relatively prime positive integers. What is 
?
Solution
If we work with the problem for a little bit, we quickly see that their is no direct combinatorics way to calculate 
. The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.
Let 
represent the number of marbles in each box, and without loss of generality let 
. Then, 
, and since the 
may be reduced to form 
on the denominator of 
, 
. It follows that 
, so there are 2 pairs of 
and 
.
- Case 1: Then the product of the number of black marbles in each box is
, so the only combination that works is 
black in first box, and 
black in second. Then, 
so 
.
, so the only combination that works is 
black in first box, and 
black in second. Then, 
so 
.- Case 2: The only combination that works is 9 black in both. Thus,
. .
. Thus, 
.
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
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