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To solve this problem, we need to find the probability of \(A \cup B\), which is the probability that either event \(A\) or event \(B\) (or both) occur. The formula to calculate this is:

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

### Given Data:
- \( P(A) = 0.48 \)
- \( P(B) = 0.21 \)
- \( P(B^c | A^c) = 0.91 \)

### Finding \(P(A \cap B)\):
To determine \( P(A \cap B) \), we need to use the relationship with \( P(B^c | A^c) \). We can use the following identity:

\[
P(B^c | A^c) = \frac{P(B^c \cap A^c)}{P(A^c)}
\]

Since \( P(B^c \cap A^c) = 1 - P(A \cup B) \), and \( P(A^c) = 1 - P(A) \), we can find \( P(A \cup B) \) by rearranging the terms. Here’s the step-by-step calculation:

1. Calculate \( P(A^c) \):
   \[
   P(A^c) = 1 - P(A) = 1 - 0.48 = 0.52
   \]

2. Use \( P(B^c | A^c) \):
   \[
   P(B^c | A^c) = \frac{1 - P(A \cup B)}{0.52}
   \]
   Plug in \( P(B^c | A^c) = 0.91 \):
   \[
   0.91 = \frac{1 - P(A \cup B)}{0.52}
   \]

3. Solve for \( P(A \cup B) \):
   \[
   1 - P(A \cup B) = 0.91 \times 0.52
   \]
   \[
   1 - P(A \cup B) = 0.4732
   \]
   \[
   P(A \cup B) = 1 - 0.4732 = 0.5268
   \]

### Final Answer:
\[
\boxed{0.5268}
\]

Would you like a further breakdown or explanation of any part of this solution?

### Related Questions:
1. How do you calculate the probability of the intersection of two events?
2. What does conditional probability mean in this context?
3. How would the solution change if \( P(B^c | A^c) \) were different?
4. What is the complement rule in probability, and how is it used here?
5. How can you verify the accuracy of this probability calculation?

### Tip:
Always double-check the interpretation of conditional probabilities, especially when working with complements, as it can change the approach to the solution.

Let A and B be two events where P(A) = 0.48, P(B) = 0.21, and P(B^c | A^c) = 0.91. What is the probability of A or B?

Learn how to calculate the probability of A or B when given P(A) = 0.48, P(B) = 0.21, and P(B^c | A^c) = 0.91. This solution involves using conditional probability and the complement rule.

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