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The problem provides the following probabilities:

- \( P(A) = 0.4 \)
- \( P(B) = 0.7 \)
- \( P(A \cup B) = 0.8 \)

We are tasked to find:
a) \( P(A \cup B') \)  
b) \( P(A' \cap B) \)

### Solution:

#### a) \( P(A \cup B') \)
We know the complement rule \( P(B') = 1 - P(B) \). Thus:
\[
P(B') = 1 - 0.7 = 0.3
\]

The union of \( A \) and \( B' \) can be calculated as:
\[
P(A \cup B') = P(A) + P(B') - P(A \cap B')
\]

From the given, \( P(A \cap B') = P(A) - P(A \cap B) \), so we need \( P(A \cap B) \). Using the formula:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Substitute the known values:
\[
0.8 = 0.4 + 0.7 - P(A \cap B)
\]
\[
P(A \cap B) = 0.3
\]

Thus:
\[
P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.3 = 0.1
\]

Finally:
\[
P(A \cup B') = P(A) + P(B') - P(A \cap B') = 0.4 + 0.3 - 0.1 = 0.6
\]

#### b) \( P(A' \cap B) \)
The intersection of \( A' \) and \( B \) is:
\[
P(A' \cap B) = P(B) - P(A \cap B)
\]
Substitute the values:
\[
P(A' \cap B) = 0.7 - 0.3 = 0.4
\]

### Final Answers:
a) \( P(A \cup B') = 0.6 \)  
b) \( P(A' \cap B) = 0.4 \)

Let me know if you'd like further clarification! Here are 5 related questions to deepen your understanding:

1. How is the formula for \( P(A \cup B) \) derived, and why is subtraction of \( P(A \cap B) \) necessary?
2. What is the interpretation of \( P(A \cup B') \) in practical terms?
3. How would \( P(A' \cap B') \) be calculated with the same data?
4. What changes if \( P(A) \) and \( P(B) \) are not mutually exclusive?
5. Can \( P(A \cup B) \) ever exceed \( 1 \)? Why or why not?

**Tip:** Always verify if events are mutually exclusive or independent to simplify calculations in probability problems.

14. Given that P(A) = 0.4, P(B) = 0.7 and that P(A ∪ B) = 0.8, find:
a) P(A ∪ B')
b) P(A' ∩ B)

Learn how to find the probabilities P(A ∪ B') and P(A' ∩ B) given P(A) = 0.4, P(B) = 0.7, and P(A ∪ B) = 0.8. This problem covers essential probability concepts such as the complement rule and the inclusion-exclusion principle. Suitable for students in Grades 10-12.

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