Let's analyze and solve the problem from the image.

We are given:

• $P(A) = 0.2$
• $P(B) = 0.3$
• $P(A \cup B) = 0.4$

We need to find the following:

Part (a) $P(A \cap B)$

We can use the formula for the union of two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substitute the known values: $0.4 = 0.2 + 0.3 - P(A \cap B)$ $0.4 = 0.5 - P(A \cap B)$ $P(A \cap B) = 0.5 - 0.4 = 0.1$

Part (b) $P(\overline{A} \cup \overline{B})$

We use De Morgan's Law, which states: $P(\overline{A} \cup \overline{B}) = 1 - P(A \cap B)$ We already know that $P(A \cap B) = 0.1$, so: $P(\overline{A} \cup \overline{B}) = 1 - 0.1 = 0.9$

Part (c) $P(\overline{A} \cap B)$

The probability of $\overline{A} \cap B$ can be found as: $P(\overline{A} \cap B) = P(B) - P(A \cap B)$ Substitute the known values: $P(\overline{A} \cap B) = 0.3 - 0.1 = 0.2$

Part (d) $P(\overline{A} | B)$

We use the conditional probability formula: $P(\overline{A} | B) = \frac{P(\overline{A} \cap B)}{P(B)}$ Substitute the known values: $P(\overline{A} | B) = \frac{0.2}{0.3} = \frac{2}{3}$

Summary of Results:

• (a) $P(A \cap B) = 0.1$
• (b) $P(\overline{A} \cup \overline{B}) = 0.9$
• (c) $P(\overline{A} \cap B) = 0.2$
• (d) $P(\overline{A} | B) = \frac{2}{3}$

Would you like more details on any part, or do you have any other questions?

Here are 5 related questions you might find helpful:

1. How do you calculate $P(\overline{A})$ from $P(A)$?
2. Can you explain why we use De Morgan’s Law in part (b)?
3. What is the formula for conditional probability?
4. How can you verify the result for $P(A \cap B)$ using a Venn diagram?
5. What happens to these probabilities if events A and B are independent?

Tip: When solving probability problems, always check if the events are mutually exclusive or independent, as this influences the formulas used!