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.