Zeke was on the internet researching prices of new and used cars. One website had 12 cars for sale, 5 of which were vans.
If Zeke randomly chose to look at 9 of the cars on the website, what is the probability that exactly 4 of the chosen cars are vans?
Write your answer as a decimal rounded to four decimal places.

To solve this problem, we need to calculate the probability that exactly 4 of the 9 cars chosen by Zeke are vans. This is a problem involving combinations and probabilities.

Here’s how we can solve it step by step:

### Step 1: Identify the Variables
- Total cars: \( 12 \)
- Vans: \( 5 \)
- Non-vans (cars that are not vans): \( 12 - 5 = 7 \)
- Cars chosen: \( 9 \)
- We want exactly 4 vans in the selection.

### Step 2: Use the Hypergeometric Distribution
The probability of selecting exactly \( k \) vans out of \( n \) selections from a group containing \( K \) vans out of a total of \( N \) items can be modeled using the hypergeometric distribution formula:

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

Where:
- \( N = 12 \) (total cars),
- \( K = 5 \) (vans),
- \( n = 9 \) (cars chosen),
- \( k = 4 \) (vans wanted).

### Step 3: Plug in the Values
1. Calculate the number of ways to choose 4 vans out of 5: \( \binom{5}{4} \).
2. Calculate the number of ways to choose the remaining 5 cars from the 7 non-vans: \( \binom{7}{5} \).
3. Calculate the total number of ways to choose 9 cars out of 12: \( \binom{12}{9} \).

\[
\binom{5}{4} = 5, \quad \binom{7}{5} = 21, \quad \text{and} \quad \binom{12}{9} = 220
\]

### Step 4: Compute the Probability
Substituting into the formula:

\[
P(X = 4) = \frac{\binom{5}{4} \binom{7}{5}}{\binom{12}{9}} = \frac{5 \times 21}{220}
\]

Calculating:

\[
P(X = 4) = \frac{105}{220} = 0.4773
\]

So, the probability that exactly 4 of the chosen cars are vans is **0.4773**.

Would you like more details, or have any questions?

Here are some related questions:
1. What is the probability that all chosen cars are vans?
2. What is the probability of choosing no vans among the cars?
3. How would the probability change if there were 6 vans instead of 5?
4. How do you calculate the probability for a different number of vans (e.g., 3 vans)?
5. What is the expected number of vans when choosing 9 cars?

**Tip:** The hypergeometric distribution is useful for scenarios without replacement, where the probability of selecting an item changes with each selection.

Learn how to calculate the probability that exactly 4 out of 9 randomly chosen cars are vans from a selection of 12 cars, where 5 are vans. This problem uses the hypergeometric distribution formula to determine the likelihood of specific outcomes. Ideal for advanced high school students.

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