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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.

Let A and B be two events where P(A) = 0.52, P(B) = 0.18, and P(B^c | A^c) = 0.9. What is the probability of A or B?

Learn how to calculate the probability of events A or B with given conditional probabilities, using key concepts like the complement rule and conditional probability. This example explores real-world applications of probability and is suitable for high school students.

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