2023 SSMO Speed Round Problems/Problem 7

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 $27 + 31 = \boxed{58}$