If 5 out of 20 tires in storage are defective and 5 of them randomly chosen for inspection (that is each tire has the same chance of being selected), what is the probability that two of the defective tires will be included?

To solve this problem, we will use the **hypergeometric probability formula**. This formula is suitable for problems where we are dealing with selections without replacement from a finite population.

### Step-by-Step Solution:

1. **Understand the Parameters:**
   - Total number of tires: \( N = 20 \)
   - Number of defective tires: \( D = 5 \)
   - Number of non-defective tires: \( N - D = 15 \)
   - Number of tires chosen: \( n = 5 \)
   - Number of defective tires we want in the selection: \( k = 2 \)

2. **Apply the Hypergeometric Distribution Formula:**
   The hypergeometric probability formula is:

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

   Where:
   - \( \binom{D}{k} \) is the number of ways to choose \( k \) defective tires from \( D \) defective tires.
   - \( \binom{N - D}{n - k} \) is the number of ways to choose \( n - k \) non-defective tires from \( N - D \) non-defective tires.
   - \( \binom{N}{n} \) is the total number of ways to choose \( n \) tires from \( N \) tires.

3. **Calculate the Individual Components:**

   - \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)
   - \( \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \)
   - \( \binom{20}{5} = \frac{20!}{5!(20-5)!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 15504 \)

4. **Compute the Probability:**

   \[
   P(X = 2) = \frac{\binom{5}{2} \cdot \binom{15}{3}}{\binom{20}{5}} = \frac{10 \times 455}{15504}
   \]

   \[
   P(X = 2) = \frac{4550}{15504} \approx 0.2934
   \]

### Conclusion:

The probability that exactly two of the defective tires will be included in the selection of 5 tires is **approximately 0.2934**, or **29.34%**.

Would you like more details on this calculation, or do you have any questions?

Here are 5 related questions that may interest you:
1. What is the probability of selecting exactly three defective tires?
2. How does the hypergeometric distribution differ from the binomial distribution?
3. How would the probability change if there were 6 defective tires instead of 5?
4. What is the expected number of defective tires in a random selection of 5 tires?
5. Can the hypergeometric distribution be approximated by the binomial distribution in some cases?

**Tip:** When dealing with probabilities in scenarios without replacement, consider using the hypergeometric distribution formula.

Explore the hypergeometric probability formula to calculate the likelihood of selecting exactly two defective tires from a random inspection of 5 tires out of 20, where 5 tires are defective and 15 are non-defective. Get a detailed step-by-step solution and understand the math behind it.

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