Difference between revisions of "2000 AIME I Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | {{ | + | 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>. |
| + | |||
| + | 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>. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | '''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}*\frac{6}{15} = \frac{1}{25}</math>. <math>m + n = 26</math>. | ||
| + | |||
| + | 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}} | ||
Revision as of 17:26, 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
Let 
be the first box. Let 
be the second box. Let 
be the number of marbles total in , and 
in .
Then, 
, and since the 
may be reduced to form 
on the denominator of 
, 50 is the lowest, and 
.
Then there are 2 pairs of and 
, since and are identical in computing probability.
Case 1: The only combination that works is 18 black in first box, 3 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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
