Quinn was on the internet researching prices of new and used cars. One website had 18 cars for sale, 3 of which were trucks.
If Quinn randomly chose to look at 13 of the cars on the website, what is the probability that exactly 2 of the chosen cars are trucks?
Write your answer as a decimal rounded to four decimal places.

To solve this problem, we need to calculate the probability that exactly 2 of the 13 cars chosen by Quinn are trucks. This is a hypergeometric probability problem because it involves selecting a subset of items from a larger set without replacement.

### Step 1: Define the variables for the hypergeometric distribution

- **Total cars (population size)**: \( N = 18 \)
- **Total trucks (successes in the population)**: \( K = 3 \)
- **Cars chosen (sample size)**: \( n = 13 \)
- **Trucks chosen (successes in the sample)**: \( k = 2 \)

### Step 2: Hypergeometric probability formula

The probability of getting exactly \( k \) successes (trucks) in the sample is given by the hypergeometric probability formula:

\[
P(X = k) = \frac{\binom{K}{k} \cdot \binom{N - K}{n - k}}{\binom{N}{n}}
\]

where:
- \(\binom{K}{k}\) is the number of ways to choose \( k \) successes from the \( K \) successes in the population.
- \(\binom{N - K}{n - k}\) is the number of ways to choose the remaining \( n - k \) items from the \( N - K \) items in the population that are not successes.
- \(\binom{N}{n}\) is the total number of ways to choose \( n \) items from the \( N \) items in the population.

### Step 3: Calculate each term

1. Calculate \(\binom{K}{k} = \binom{3}{2}\):

   \[
   \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3
   \]

2. Calculate \(\binom{N - K}{n - k} = \binom{18 - 3}{13 - 2} = \binom{15}{11}\):

   \[
   \binom{15}{11} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365
   \]

3. Calculate \(\binom{N}{n} = \binom{18}{13}\):

   \[
   \binom{18}{13} = \frac{18 \times 17 \times 16 \times 15 \times 14}{5 \times 4 \times 3 \times 2 \times 1} = 8568
   \]

### Step 4: Substitute the values into the formula

\[
P(X = 2) = \frac{\binom{3}{2} \cdot \binom{15}{11}}{\binom{18}{13}} = \frac{3 \cdot 1365}{8568} \approx 0.4776
\]

So, the probability that exactly 2 of the 13 cars chosen are trucks is approximately:

\[
\boxed{0.4776}
\]

Quinn was on the internet researching prices of new and used cars. One website had 18 cars for sale, 3 of which were trucks. If Quinn randomly chose to look at 13 of the cars on the website, what is the probability that exactly 2 of the chosen cars are trucks? Write your answer as a decimal rounded to four decimal places.

Learn how to calculate the probability of selecting exactly 2 trucks from 13 cars on a website with a total of 18 cars, including 3 trucks. This problem covers key probability concepts, including hypergeometric distribution and combinatorics, and is suitable for students in Grades 10-12.

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