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| − | ==Problem==
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| − | At FenZhu High School, <math>7</math>th graders have a 60\% of chance of having a dog and <math>8</math>th graders have a 40\% chance of having a dog. Suppose there is a classroom of <math>30</math> <math>7</math>th grader and <math>10</math> <math>8</math>th graders. If exactly one person owns a dog, then the probability that a <math>7</math>th grader owns the dog is <math>\frac{m}{n},</math> for relatively prime positive integers <math>m</math> and <math>n.</math> Find <math>m+n.</math>
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| − | ==Solution==
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| − | The probability that a <math>7</math>th grader has the only dog is
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| − | <cmath>
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| − | A = \left(30 \cdot \left(\frac{2}{5}\right)^{29} \cdot \frac{3}{5}\right) \cdot \left(\frac{3}{5}\right)^{10}
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| − | </cmath>
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| − | and the probability for an <math>8</math>th grader is
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| − | <cmath>
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| − | B = \left(\frac{2}{5}\right)^{30} \cdot \left(10 \cdot \left(\frac{3}{5}\right)^{9} \cdot \frac{2}{5}\right).
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| − | </cmath>
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| − | Then,
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| − | <cmath>
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| − | \frac{A}{B} = \frac{30 \cdot 2^{29} \cdot 3 \cdot 3^{10}}{2^{30} \cdot 10 \cdot 3^9 \cdot 2} = \frac{30 \cdot 3 \cdot 3}{2 \cdot 10 \cdot 2} = \frac{27}{4}.
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| − | </cmath>
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| − | The probability is thus <math>\frac{A}{A + B} = \frac{27}{31}</math> so the
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| − | sum is <math>27 + 31 = \boxed{58}</math>
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