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Difference between revisions of "2023 SSMO Speed Round Problems/Problem 7"

(Created page with "==Problem== 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 ther...")
 
 
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==Problem==
 
==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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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>
  
 
==Solution==
 
==Solution==

Latest revision as of 13:21, 3 July 2023

Problem

At FenZhu High School, $7$th graders have a 60% of chance of having a dog and $8$th graders have a 40% chance of having a dog. Suppose there is a classroom of $30$ $7$th grader and $10$ $8$th graders. If exactly one person owns a dog, then the probability that a $7$th grader owns the dog is $\frac{m}{n},$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

The probability that a $7$th grader has the only dog is \[A = \left(30 \cdot \left(\frac{2}{5}\right)^{29} \cdot \frac{3}{5}\right) \cdot \left(\frac{3}{5}\right)^{10}\] and the probability for an $8$th grader is \[B = \left(\frac{2}{5}\right)^{30} \cdot \left(10 \cdot \left(\frac{3}{5}\right)^{9} \cdot \frac{2}{5}\right).\]

Then, \[\frac{A}{B} = \frac{30 \cdot 2^{29} \cdot 3 \cdot 3^{10}}{2^{30} \cdot 10 \cdot 3^9 \cdot 2} = \frac{30 \cdot 3 \cdot 3}{2 \cdot 10 \cdot 2} = \frac{27}{4}.\]

The probability is thus $\frac{A}{A + B} = \frac{27}{31}$ so the sum is $27 + 31 = \boxed{58}$

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