Difference between revisions of "2000 AIME I Problems/Problem 5"
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For completeness, the fraction <math>\frac{81}{150}</math> may be tested. <math>150</math> is the highest necessary denominator that needs to be tested, since the maximum the denominator <math>(w_1+ b_1)(w_2 + b_2)</math> can be when the sum of all integer variables is <math>25</math> is when the variables are <math>6, 6, 6, </math> and <math>7</math>, in some permutation, which gives <math>154</math>. If <math>b_1 \cdot b_2 = 81</math>, this forces <math>b_1 = b_2 = 9</math>, since all variables must be integers under <math>25</math>. The denominator becomes <math>(9 + w_1)(9 + w_2) = 150</math>, and since there are now <math>25 - 18 = 7</math> white marbles total, the denominator becomes <math>(9 + w_1)(16 - w_1) = 150</math>. Testing <math>w_1 = 1</math> gives a solution, and thus <math>w_2 = 6</math>. The complete solution for this case is <math>(w_1, w_2, b_1, b_2) = (1, 6, 9, 9)</math>. Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is <math>\frac{6 \cdot 1}{ 150}</math>, which still simplifies to <math>\frac {1}{25}</math>. | For completeness, the fraction <math>\frac{81}{150}</math> may be tested. <math>150</math> is the highest necessary denominator that needs to be tested, since the maximum the denominator <math>(w_1+ b_1)(w_2 + b_2)</math> can be when the sum of all integer variables is <math>25</math> is when the variables are <math>6, 6, 6, </math> and <math>7</math>, in some permutation, which gives <math>154</math>. If <math>b_1 \cdot b_2 = 81</math>, this forces <math>b_1 = b_2 = 9</math>, since all variables must be integers under <math>25</math>. The denominator becomes <math>(9 + w_1)(9 + w_2) = 150</math>, and since there are now <math>25 - 18 = 7</math> white marbles total, the denominator becomes <math>(9 + w_1)(16 - w_1) = 150</math>. Testing <math>w_1 = 1</math> gives a solution, and thus <math>w_2 = 6</math>. The complete solution for this case is <math>(w_1, w_2, b_1, b_2) = (1, 6, 9, 9)</math>. Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is <math>\frac{6 \cdot 1}{ 150}</math>, which still simplifies to <math>\frac {1}{25}</math>. | ||
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+ | ==Solution 3== | ||
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+ | We know that <math>\frac{27}{50} = \frac{b_1}{t_1} \cdot \frac{b_2}{t_2}</math>, where <math>b_1</math> and <math>b_2</math> are the number of black marbles in the first and the second box respectively, and <math>t_1</math> and <math>t_2</math> is the total number of marbles in the first and the second boxes respectively. So, <math>t_1 + t_2 = 25</math>. Then, we can realize that <math>\frac{27}{50} = \frac{9}{10} \cdot \frac{3}{5} = \frac{9}{10} \cdot \frac{9}{15}</math>, which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes <math>\frac{1}{10} \cdot \frac{6}{15} = \frac{1}{25}</math>. The answer is <math>1 + 25 = \boxed{026}</math> | ||
== See also == | == See also == |
Latest revision as of 12:18, 22 February 2019
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 1
If we work with the problem for a little bit, we quickly see that there 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 .
- Case 2: The only combination that works is 9 black in both. Thus, . .
Thus, .
Solution 2
Let and represent the white and black marbles in boxes 1 and 2.
Since there are marbles in the box:
From the fact that there is a chance of drawing one black marble from each box:
Thinking of the numerator and denominator separately, if was not a reduced fraction when calculating out the probability, then . Since , this forces the variables to be and in some permutation. Without loss of generality, let and .
The denominator becomes:
Since there have been black marbles used, there must be white marbles. Substituting that in:
Since the factors of that are greater than are and , the quantity must equal one of those. However, since , testing and for does not give a correct product. Thus, must be a reduced form of the actual fraction.
First assume that the fraction was reduced from , yielding the equations and . Factoring and saying WLOG that gives or . Trying the first pair and setting the denominator equal to 100 gives:
Since , the pairs and can be tried, since each box must contain at least one white marble. Plugging in gives the true equation , so the number of marbles are
Thus, the chance of drawing 2 white marbles is in lowest terms, and the answer to the problem is
For completeness, the fraction may be tested. is the highest necessary denominator that needs to be tested, since the maximum the denominator can be when the sum of all integer variables is is when the variables are and , in some permutation, which gives . If , this forces , since all variables must be integers under . The denominator becomes , and since there are now white marbles total, the denominator becomes . Testing gives a solution, and thus . The complete solution for this case is . Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is , which still simplifies to .
Solution 3
We know that , where and are the number of black marbles in the first and the second box respectively, and and is the total number of marbles in the first and the second boxes respectively. So, . Then, we can realize that , which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes . The answer is
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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