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Let's break down and solve each part of the problem based on the provided probability table:

|                | Miles/Year < 10,000 | 10,000 ≤ Miles/Year < 20,000 | 20,000 ≤ Miles/Year |
|----------------|---------------------|-----------------------------|---------------------|
| **Accident**   | 0.04                | 0.10                        | 0.28                |
| **No accident**| 0.21                | 0.16                        | 0.21                |

### (a) Find the probability that the client drives less than 10,000 miles/year and has an accident.

From the table:
- The probability that a client drives less than 10,000 miles per year and has an accident is given directly as **0.04**.

\[
\text{P(< 10,000 miles and Accident)} = 0.04
\]

---

### (b) Find the probability that a client drives less than 10,000 miles/year or has an accident.

Here we need to find \( P(\text{< 10,000 miles} \cup \text{Accident}) \), the union of two events:

\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

Where:
- \( A \) = the event of driving less than 10,000 miles per year.
- \( B \) = the event of having an accident.
- \( P(A \cap B) \) = the intersection (already found in part (a)).

From the table:
- \( P(A) \) = probability of driving less than 10,000 miles/year = \( 0.04 + 0.21 = 0.25 \).
- \( P(B) \) = probability of having an accident (for all mileage ranges) = \( 0.04 + 0.10 + 0.28 = 0.42 \).
- \( P(A \cap B) \) = 0.04.

So:

\[
P(A \cup B) = 0.25 + 0.42 - 0.04 = 0.63
\]

---

### (c) Find the probability that a client does not drive less than 10,000 miles per year and does not have an accident.

This is the complement of the event "drives less than 10,000 miles/year or has an accident" (part b). So we use:

\[
P(\text{Not (<10,000 miles or Accident)}) = 1 - P(\text{<10,000 miles} \cup \text{Accident})
\]

From part (b), \( P(\text{< 10,000 miles} \cup \text{Accident}) = 0.63 \), so:

\[
P(\text{Not (<10,000 miles or Accident)}) = 1 - 0.63 = 0.37
\]

---

Would you like more details on any part of the process? Here are some related questions for practice:

1. What is the probability of a client driving between 10,000 and 20,000 miles/year and not having an accident?
2. Find the probability that a client drives more than 20,000 miles/year and has no accident.
3. What is the total probability of not having an accident, regardless of mileage?
4. Calculate the probability that a client drives between 10,000 and 20,000 miles/year or does not have an accident.
5. What is the complement of driving between 10,000 and 20,000 miles/year and having an accident?

**Tip:** When working with "or" probabilities, always remember to subtract the overlap (intersection) between events.

After examining their clients' records, Crashorama Auto Insurance calculated the probabilities related to mileage and accidents. The table shows the joint probabilities for clients' mileage categories and accident occurrences. Questions:
(a) Find the probability that the client drives less than 10,000 miles/year and has an accident.
(b) Find the probability that a client drives less than 10,000 miles/year or has an accident.
(c) Find the probability that a client does not drive less than 10,000 miles per year and does not have an accident.

Explore how to calculate probabilities related to accidents and mileage with Crashorama Insurance's data. This problem includes concepts of intersection, union, and complements of events. Suitable for students in Grades 10-12.

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