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The problem involves independent events \( A \) and \( B \), where you are given:

- \( P(A) = 0.70 \)
- \( P(B) = 0.50 \)

You are asked to find:

1. \( P(A \cap B) \) — the probability of the intersection of events \( A \) and \( B \).
2. \( P(A \cup B) \) — the probability of the union of events \( A \) and \( B \).

### Solution:

#### 1. \( P(A \cap B) \)

Since \( A \) and \( B \) are independent, the probability of the intersection is the product of the individual probabilities:

\[
P(A \cap B) = P(A) \times P(B)
\]

Substitute the values:

\[
P(A \cap B) = 0.70 \times 0.50 = 0.35
\]

#### 2. \( P(A \cup B) \)

The probability of the union of two events is given by:

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

Substitute the values:

\[
P(A \cup B) = 0.70 + 0.50 - 0.35 = 0.85
\]

### Final Answers:
- \( P(A \cap B) = 0.35 \)
- \( P(A \cup B) = 0.85 \)

Would you like further details or explanations? Here are 5 related questions to expand your understanding:

1. How would the result change if the events \( A \) and \( B \) were not independent?
2. What is the general formula for \( P(A \cap B) \) when events are dependent?
3. How do you interpret \( P(A \cup B) \) in real-world scenarios?
4. Can you derive \( P(A^c \cup B^c) \) using De Morgan’s laws?
5. How does this concept of independent events extend to three or more events?

**Tip:** When dealing with independent events, the key property is that \( P(A \cap B) = P(A) \times P(B) \). This simplifies many probability calculations.

Suppose A and B are independent events such that P(A) = 0.70 and P(B) = 0.50. Find P(A ∩ B) and P(A ∪ B).

Learn how to find the intersection and union probabilities of two independent events A and B, with P(A) = 0.70 and P(B) = 0.50. This detailed step-by-step solution covers key probability concepts for high school students.

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