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The question on the screen involves determining probabilities related to car repairs for a person who owns two cars. Here's a breakdown of how to approach each part:

### Given Information:
- 16% of cars will need to be repaired once.
- 5% will need repairs twice.
- 3% will require three or more repairs.

Thus, the total probability that a car will need any repair is:
- \( P(\text{Repair}) = 16\% + 5\% + 3\% = 24\% \).

This means the probability that a car will **not** need a repair is:
- \( P(\text{No Repair}) = 1 - 0.24 = 0.76 \).

### (a) The probability that neither car will need repair:
For both cars not to need repair, the events are independent. Hence, the probability is:
\[
P(\text{Neither}) = P(\text{No Repair for Car 1}) \times P(\text{No Repair for Car 2}) = 0.76 \times 0.76 = 0.5776.
\]
So, the probability is **0.5776**.

### (b) The probability that both cars will need repair:
For both cars to need a repair, the probability is:
\[
P(\text{Both}) = P(\text{Repair for Car 1}) \times P(\text{Repair for Car 2}) = 0.24 \times 0.24 = 0.0576.
\]
Thus, the probability is **0.0576**.

### (c) The probability that at least one car will need repair:
The probability of at least one car needing repair is the complement of neither car needing repair:
\[
P(\text{At least one}) = 1 - P(\text{Neither}) = 1 - 0.5776 = 0.4224.
\]
Thus, the probability is **0.4224**.

### Summary of Results:
- (a) Probability that neither car will need repair: **0.5776**.
- (b) Probability that both cars will need repair: **0.0576**.
- (c) Probability that at least one car will need repair: **0.4224**.

Would you like further details on how these probabilities were calculated?

Here are 5 related questions to practice:
1. What is the probability that exactly one car will need repair?
2. If you own three cars, what is the probability that none will need repair?
3. What is the probability that all three cars will need repair?
4. What is the probability that at least one car will need repair if you own three cars?
5. How does the probability of needing repair change if the repair rate increases to 30%?

**Tip:** When calculating probabilities for multiple independent events, multiply the individual probabilities for each event.

A consumer organization estimates that over a 1-year period 16% of cars will need to be repaired once, 5% will need repairs twice, and 3% will require three or more repairs. If you own two cars, what is the probability that (a) neither will need repair? (b) both will need repair? (c) at least one car will need repair?

Solve a car repair probability problem involving two cars using independent events. Learn how to calculate the probability that neither, both, or at least one car will need repair. The problem utilizes key probability concepts like the multiplication rule and the complement rule. Suitable for high school students in Grades 10-12.

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