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The problem involves determining the probability that at least one of the four cars sold will require repairs within the first five months. 

We are given:
- The probability that a car will need repairs within the first five months, \( P(\text{repairs}) = 0.6 \).
- The dealer sells four cars, so we have \( n = 4 \) cars.
- We need to calculate the probability that **at least one** of these cars will require repairs.

### Step-by-step solution:

1. **Complementary Probability Approach**:
   - The complement of "at least one car needing repairs" is "no cars need repairs." 
   - First, we calculate the probability that **none** of the four cars will need repairs.
   - If the probability that a car will need repairs is \( 0.6 \), then the probability that a car **will not** need repairs is:
     \[
     P(\text{no repairs}) = 1 - P(\text{repairs}) = 1 - 0.6 = 0.4
     \]

2. **Calculating the Probability for No Repairs in All 4 Cars**:
   Since each car's repair need is independent, the probability that none of the four cars will need repairs is:
   \[
   P(\text{none of 4 need repairs}) = (0.4)^4 = 0.4 \times 0.4 \times 0.4 \times 0.4 = 0.0256
   \]

3. **Finding the Probability for At Least One Car Needing Repairs**:
   The probability that at least one of the cars will need repairs is the complement of the probability that none of them will need repairs:
   \[
   P(\text{at least one needs repairs}) = 1 - P(\text{none need repairs}) = 1 - 0.0256 = 0.9744
   \]

So, the probability that at least one of the four cars will require repairs in the first five months is **0.9744**.

Would you like further details or have any questions?

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Here are five related questions:
1. How would the result change if the probability of repair per car was 0.8 instead of 0.6?
2. What is the probability that exactly two cars out of four will require repairs?
3. How does this probability change if the number of cars sold is five instead of four?
4. Can we generalize the formula to handle different numbers of cars and repair probabilities?
5. How would you calculate the probability if repairs for each car were not independent?

**Tip**: Using complementary probabilities often simplifies problems involving "at least one" scenarios!

The probability that a certain make of car will need repairs in the first five months is 0.6. A dealer sells four such cars. What is the probability that at least one of them will require repairs in the first five months? Round your final answer to four decimal places.

Determine the probability that at least one of four cars will require repairs in the first five months when the probability of repair is 0.6. Explore key probability concepts including the complement rule and independent events.

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