2023 SSMO Team Round Problems/Problem 10

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 $11 + 243 = \boxed{254}.$

~SMO_Team

Page

Toolbox

Search

2023 SSMO Team Round Problems/Problem 10

Problem

Solution