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Difference between revisions of "2024 SSMO Speed Round Problems/Problem 6"

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==Problem==
 
==Problem==
  
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There are <math>4</math> people and <math>4</math> houses. Each person independently randomly chooses a house to live in. The expected number of inhabited houses can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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There are <math>4</math> people and <math>4</math> houses. Each person independently randomly chooses a house to live in. The expected number of inhabited houses can be expressed as <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
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==Solution==
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==Solution 1==
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Let <math>a_n</math> equal <math>1</math> if the nth house is occupied and <math>0</math> if the nth house is unoccupied. Thus, we are trying to find <math>\mathbb{E}[a_1+a_2+a_3+a_4] = 4\mathbb{E}[a_1]</math> by the Linearity of Expectation. Instead of calculating the probability that a single house is occupied, we instead calculate the probability that it is unoccupied. Each person has a <math>\frac{1}{4}</math> chance of residing in that house, and thus a <math>\frac{3}{4}</math> chance of not residing in that house. The probability that none of the people stay in that house is <math>\left (\frac{3}{4} \right)^4 = \frac{81}{256}</math>. Thus, the probability that any given house is occupied, or <math>\mathbb{E}[a_1]</math>, is <math>1 - \frac{81}{256} = \frac{175}{256}</math>. Thus, our answer is <math>4\cdot \frac{175}{256} = \frac{175}{64}</math>, <math>175+64=\boxed{239}</math>.  
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For each house, the expected probability that it is uninhabited is <math>\left(\frac{3}{4}\right)^4.</math> So, the expected probability that it is inhabited is <math>1-\left(\frac{3}{4}\right)^4 = \frac{175}{256}.</math> Thus, the expected number of inhabited houses is <cmath>4\cdot\left(\frac{175}{256}\right) = \frac{175}{64}\implies175+64 = \boxed{239}.</cmath>
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~SMO_Team
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==Solution 2==
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Let <math>a_n</math> equal <math>1</math> if the nth house is occupied and <math>0</math> if the nth house is unoccupied. Thus, we are trying to find <math>\mathbb{E}[a_1+a_2+a_3+a_4] = 4\mathbb{E}[a_1]</math> by the Linearity of Expectation. Instead of calculating the probability that a single house is occupied, we instead calculate the probability that it is unoccupied. Each person has a <math>\tfrac{1}{4}</math> chance of residing in that house, and thus a <math>\tfrac{3}{4}</math> chance of not residing in that house. The probability that none of the people stay in that house is <math>\left (\tfrac{3}{4} \right)^4 = \tfrac{81}{256}</math>. Thus, the probability that any given house is occupied, or <math>\mathbb{E}[a_1]</math>, is <math>1 - \tfrac{81}{256} = \tfrac{175}{256}</math>. Thus, our answer is <math>4\cdot \tfrac{175}{256} = \tfrac{175}{64}</math>, <math>175+64=\boxed{239}</math>.  
  
 
-Vivdax
 
-Vivdax

Latest revision as of 14:27, 10 September 2025

Problem

There are $4$ people and $4$ houses. Each person independently randomly chooses a house to live in. The expected number of inhabited houses can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

For each house, the expected probability that it is uninhabited is $\left(\frac{3}{4}\right)^4.$ So, the expected probability that it is inhabited is $1-\left(\frac{3}{4}\right)^4 = \frac{175}{256}.$ Thus, the expected number of inhabited houses is \[4\cdot\left(\frac{175}{256}\right) = \frac{175}{64}\implies175+64 = \boxed{239}.\]

~SMO_Team

Solution 2

Let $a_n$ equal $1$ if the nth house is occupied and $0$ if the nth house is unoccupied. Thus, we are trying to find $\mathbb{E}[a_1+a_2+a_3+a_4] = 4\mathbb{E}[a_1]$ by the Linearity of Expectation. Instead of calculating the probability that a single house is occupied, we instead calculate the probability that it is unoccupied. Each person has a $\tfrac{1}{4}$ chance of residing in that house, and thus a $\tfrac{3}{4}$ chance of not residing in that house. The probability that none of the people stay in that house is $\left (\tfrac{3}{4} \right)^4 = \tfrac{81}{256}$. Thus, the probability that any given house is occupied, or $\mathbb{E}[a_1]$, is $1 - \tfrac{81}{256} = \tfrac{175}{256}$. Thus, our answer is $4\cdot \tfrac{175}{256} = \tfrac{175}{64}$, $175+64=\boxed{239}$.

-Vivdax

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