https://file.aiquickdraw.com/mathbot/file/1726615526862ywics10j.jpg

Let's break down and solve the problems based on the image provided.

### Problem 1:
Let \( A \) and \( B \) be events with \( P(A) = 0.6 \), \( P(B) = 0.9 \), and \( P(A \cap B) = 0.5 \).

**(a) Are \( A \) and \( B \) independent? Explain.**  
Two events are independent if:
\[
P(A \cap B) = P(A) \cdot P(B)
\]
Let's check this condition:
\[
P(A) \cdot P(B) = 0.6 \cdot 0.9 = 0.54
\]
Since \( P(A \cap B) = 0.5 \) and \( 0.54 \neq 0.5 \), the events \( A \) and \( B \) are **not independent**.

**(b) Compute \( P(A \cup B) \).**  
We use the formula for the union of two events:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Substitute the given values:
\[
P(A \cup B) = 0.6 + 0.9 - 0.5 = 1.0
\]

**(c) Are \( A \) and \( B \) mutually exclusive? Explain.**  
Two events are mutually exclusive if:
\[
P(A \cap B) = 0
\]
Since \( P(A \cap B) = 0.5 \neq 0 \), the events \( A \) and \( B \) are **not mutually exclusive**.

---

### Problem 2:
Let \( A \) and \( B \) be events with \( P(A) = 0.4 \), \( P(B) = 0.5 \), and \( P(A \cup B) = 0.6 \).

**(a) Compute \( P(A \cap B) \).**  
We use the same formula for the union of two events:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Substitute the known values:
\[
0.6 = 0.4 + 0.5 - P(A \cap B)
\]
\[
P(A \cap B) = 0.9 - 0.6 = 0.3
\]

**(b) Are \( A \) and \( B \) mutually exclusive? Explain.**  
Since \( P(A \cap B) = 0.3 \neq 0 \), the events \( A \) and \( B \) are **not mutually exclusive**.

**(c) Are \( A \) and \( B \) independent? Explain.**  
Check if \( P(A \cap B) = P(A) \cdot P(B) \):
\[
P(A) \cdot P(B) = 0.4 \cdot 0.5 = 0.2
\]
Since \( P(A \cap B) = 0.3 \neq 0.2 \), the events \( A \) and \( B \) are **not independent**.

---

### Problem 3:
Let \( A \) and \( B \) be events with \( P(A) = 0.5 \), \( P(B) = 0.3 \), and \( P(A \cup B) = 0.8 \).

**(a) Compute \( P(A \cap B) \).**  
Using the formula for the union of two events:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Substitute the known values:
\[
0.8 = 0.5 + 0.3 - P(A \cap B)
\]
\[
P(A \cap B) = 0.8 - 0.8 = 0
\]

**(b) Are \( A \) and \( B \) mutually exclusive? Explain.**  
Since \( P(A \cap B) = 0 \), the events \( A \) and \( B \) are **mutually exclusive**.

**(c) Are \( A \) and \( B \) independent? Explain.**  
For independence, we check if \( P(A \cap B) = P(A) \cdot P(B) \). Since \( P(A \cap B) = 0 \) and \( P(A) \cdot P(B) = 0.5 \cdot 0.3 = 0.15 \), they are **not independent** because \( 0 \neq 0.15 \).

---

Do you want more details or have any questions?

### Related Questions:
1. What is the difference between mutually exclusive and independent events?
2. How do you compute \( P(A \cup B) \) if \( A \) and \( B \) are mutually exclusive?
3. Can two events be both mutually exclusive and independent? Why or why not?
4. How would the answers change if \( P(A \cap B) = P(A) \cdot P(B) \)?
5. What are some real-world examples of independent and mutually exclusive events?

### Tip:
Always start by checking if events are mutually exclusive or independent before calculating probabilities, as this affects the formulas you'll use.

Let A and B be events with specific probabilities (P(A), P(B), P(A ∩ B), P(A ∪ B)). Determine if A and B are independent or mutually exclusive, and compute the necessary probabilities.

Learn how to determine if two events A and B are independent or mutually exclusive. We calculate P(A), P(B), P(A ∩ B), and P(A ∪ B) using fundamental probability rules and check for independence and mutual exclusivity. Ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation