Of all the new vehicles of a certain model that are sold, 23% require repairs to be done under warranty during the first year of service. A particular dealership sells 15 such vehicles. What is the probability that fewer than three of them require warranty repairs?

To solve this problem, we use the **binomial distribution** because we are dealing with a fixed number of trials (selling 15 vehicles), each trial has two possible outcomes (a vehicle either requires warranty repair or it does not), and each vehicle's outcome is independent of the others.

### Binomial Distribution

The binomial distribution formula for the probability of exactly \( k \) successes in \( n \) trials is:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

where:
- \( n = 15 \) is the number of trials (vehicles sold),
- \( k \) is the number of successes (vehicles requiring warranty repairs),
- \( p = 0.23 \) is the probability of a vehicle requiring repairs,
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient.

### Calculating the Probability of Fewer Than Three Vehicles Needing Repairs

To find the probability that **fewer than three** vehicles require warranty repairs, we need to calculate:

\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\]

#### 1. Calculating \(P(X = 0)\):

\[
P(X = 0) = \binom{15}{0} (0.23)^0 (1 - 0.23)^{15}
\]

\[
P(X = 0) = 1 \cdot 1 \cdot (0.77)^{15}
\]

\[
P(X = 0) = (0.77)^{15} \approx 0.0418
\]

#### 2. Calculating \(P(X = 1)\):

\[
P(X = 1) = \binom{15}{1} (0.23)^1 (1 - 0.23)^{14}
\]

\[
P(X = 1) = 15 \cdot 0.23 \cdot (0.77)^{14}
\]

\[
P(X = 1) = 15 \cdot 0.23 \cdot 0.0543 \approx 0.1874
\]

#### 3. Calculating \(P(X = 2)\):

\[
P(X = 2) = \binom{15}{2} (0.23)^2 (1 - 0.23)^{13}
\]

\[
P(X = 2) = \frac{15 \cdot 14}{2} \cdot (0.23)^2 \cdot (0.77)^{13}
\]

\[
P(X = 2) = 105 \cdot 0.0529 \cdot 0.0705 \approx 0.3905
\]

### Total Probability

Now, sum up these probabilities:

\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\]

\[
P(X < 3) = 0.0418 + 0.1874 + 0.2701
\]

\[
P(X < 3) \approx 0.4993
\]

### Conclusion

The probability that fewer than three vehicles out of the 15 sold require warranty repairs during the first year of service is **approximately 0.4993** or **49.93%**.

---

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the probability change if the dealership sold 20 vehicles instead of 15?
2. What is the probability that exactly three vehicles require warranty repairs?
3. How would you calculate the probability of more than five vehicles requiring repairs?
4. If the probability of a vehicle requiring repairs increased to 30%, how would the probability calculations change?
5. Can you derive the expected number of vehicles requiring repairs in the first year?

**Tip:** For probabilities involving "at least" or "at most" conditions, consider using the cumulative distribution function (CDF) of the binomial distribution.

Learn how to calculate the probability of fewer than three vehicles requiring warranty repairs out of 15 sold, using the binomial distribution formula. This problem involves probability theory and is suitable for advanced high school or college students.

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