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The uploaded image contains a math quiz focusing on the "Probability of Events," specifically analyzing the relationship between two events, \( A \) and \( B \), with probabilities \( P(A) = \frac{1}{2} \), \( P(A \cup B) = \frac{2}{3} \), and \( P(A \cap B) = \frac{1}{6} \). The task is to find \( P(B) \). Here’s how we can solve it step by step:

### Solution:

The formula for the union of two events is:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

### Substitute the given values:
\[
\frac{2}{3} = \frac{1}{2} + P(B) - \frac{1}{6}
\]

### Simplify:
First, find a common denominator for the fractions:
- \( \frac{1}{2} = \frac{3}{6} \),
- \( \frac{2}{3} = \frac{4}{6} \).

Thus, the equation becomes:
\[
\frac{4}{6} = \frac{3}{6} + P(B) - \frac{1}{6}.
\]

Combine terms:
\[
\frac{4}{6} = \frac{2}{6} + P(B).
\]

Solve for \( P(B) \):
\[
P(B) = \frac{4}{6} - \frac{2}{6} = \frac{2}{6} = \frac{1}{3}.
\]

### Final Answer:
\[
P(B) = \frac{1}{3}.
\]

Would you like a detailed explanation of the derivation or have other related questions? Here are five related questions you might explore:

1. How do you calculate \( P(A \cap B^c) \) using these probabilities?
2. How does the independence of events \( A \) and \( B \) affect \( P(A \cup B) \)?
3. Can you derive \( P(B^c) \) from \( P(B) \)?
4. How do you check if \( A \) and \( B \) are mutually exclusive using the provided values?
5. What is the probability of \( A^c \cap B \), and how do you calculate it?

### Tip:
Always remember that \( P(A \cup B) \) can be rearranged to find any missing component if the values for \( P(A) \), \( P(B) \), and \( P(A \cap B) \) are provided.

Events A and B are given such that P(A) = 1/2, P(A ∪ B) = 2/3, and P(A ∩ B) = 1/6. Find P(B).

Learn how to calculate P(B) using the given probabilities P(A), P(A ∪ B), and P(A ∩ B). This step-by-step guide covers key probability concepts, including the addition rule of probability, and is suitable for high school students studying IB Mathematics.

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