This problem is asking for the probability of $A$ or $B$, which can be calculated using the formula for the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

We are given:

• $P(A) = 0.52$
• $P(B) = 0.18$
• $P(B^c | A^c) = 0.9$

To find $P(A \cup B)$, we need to determine $P(A \cap B)$. Here's how we can proceed:

Step 1: Calculate $P(A^c)$

Since $A^c$ represents the complement of $A$, we know: $P(A^c) = 1 - P(A) = 1 - 0.52 = 0.48$

Step 2: Use the conditional probability $P(B^c | A^c)$

We are given that: $P(B^c | A^c) = 0.9$

This is the probability that $B$ does not occur given that $A$ does not occur. Using the definition of conditional probability: $P(B^c | A^c) = \frac{P(B^c \cap A^c)}{P(A^c)}$ Substituting values: $0.9 = \frac{P(B^c \cap A^c)}{0.48}$

Solving for $P(B^c \cap A^c)$: $P(B^c \cap A^c) = 0.9 \times 0.48 = 0.432$

Step 3: Calculate $P(A^c \cap B)$

We know: $P(A^c) = P(A^c \cap B) + P(A^c \cap B^c)$ Thus: $0.48 = P(A^c \cap B) + 0.432$ Solving for $P(A^c \cap B)$: $P(A^c \cap B) = 0.48 - 0.432 = 0.048$

Step 4: Calculate $P(A \cap B)$

Using the total probability rule: $P(B) = P(A \cap B) + P(A^c \cap B)$ Substituting values: $0.18 = P(A \cap B) + 0.048$ Solving for $P(A \cap B)$: $P(A \cap B) = 0.18 - 0.048 = 0.132$

Step 5: Calculate $P(A \cup B)$

Finally, we can compute: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.52 + 0.18 - 0.132 = 0.568$

So, the probability of $A$ or $B$ is 0.568.

Would you like further details or have any questions?

Here are 5 follow-up questions to enhance your understanding:

1. How do you interpret conditional probability in real-world scenarios?
2. What does the complement of an event represent in probability theory?
3. How would the formula for $P(A \cup B)$ change if $A$ and $B$ were mutually exclusive?
4. How does knowing $P(A \cap B)$ affect our understanding of the relationship between events $A$ and $B$?
5. Can you think of an example where events are dependent, and how would that affect $P(A \cup B)$?

Tip: Always verify that the sum of probabilities for all possible outcomes equals 1, which ensures consistency in probability calculations.