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Difference between revisions of "2023 SSMO Team Round Problems/Problem 10"

(Created page with "==Problem== There exists a lane of infinite cars. Each car has a <math>\frac{1}{3}</math> chance of being high quality and a <math>\frac{2}{3}</math> chance of being low quali...")
 
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==Solution==
 
==Solution==
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We will find a general formula for the probability that John buys <math>k</math> cars before giving up. Clearly, John must have seen the third low-quality car after buying <math>k</math> high-quality cars. The number of ways to order this is <cmath>\binom{k + 2}{2}.</cmath>
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The probability of getting <math>k</math> high-quality cars and <math>3</math> low-quality cars is <cmath>\left(\frac{1}{3} \right)^k \cdot \left(\frac{2}{3} \right)^3.</cmath>
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Thus, the probability that John buys <math>k</math> cars before giving up is <cmath>\frac{\binom{k + 2}{2} \cdot 2^3}{3^{k + 3}}.</cmath>
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Now, the complement of the question is the probability that John gives up after buying fewer than <math>5</math> cars. Therefore, we evaluate the above expression for <math>k = 0, 1, 2, 3, 4</math> and sum the results to get 
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<cmath>\frac{232}{243}.</cmath>
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This means the desired probability is  <cmath>1 - \frac{232}{243} = \frac{11}{243},</cmath>  so the final answer is  <cmath>11 + 243 = \boxed{254}.</cmath>

Revision as of 21:25, 9 September 2025

Problem

There exists a lane of infinite cars. Each car has a $\frac{1}{3}$ chance of being high quality and a $\frac{2}{3}$ chance of being low quality. John goes down the row of cars buying high-quality cars. However, after John sees 3 low-quality cars, he gives up on buying additional cars. Let the probability that he buys at least $5$ cars before giving up as $\frac{m}{n}$. Find $m+n$.

Solution

We will find a general formula for the probability that John buys $k$ cars before giving up. Clearly, John must have seen the third low-quality car after buying $k$ high-quality cars. The number of ways to order this is \[\binom{k + 2}{2}.\]

The probability of getting $k$ high-quality cars and $3$ low-quality cars is \[\left(\frac{1}{3} \right)^k \cdot \left(\frac{2}{3} \right)^3.\]

Thus, the probability that John buys $k$ cars before giving up is \[\frac{\binom{k + 2}{2} \cdot 2^3}{3^{k + 3}}.\]

Now, the complement of the question is the probability that John gives up after buying fewer than $5$ cars. Therefore, we evaluate the above expression for $k = 0, 1, 2, 3, 4$ and sum the results to get \[\frac{232}{243}.\]

This means the desired probability is \[1 - \frac{232}{243} = \frac{11}{243},\] so the final answer is \[11 + 243 = \boxed{254}.\]

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