Given: \[\] \( P(A) = \frac{2}{3} \) \[\] \( P(A \cup B) = \frac{3}{4} \) \[\] We use the formula for the union of two sets: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute the known values: \[ \frac{3}{4} = \frac{2}{3} + P(B) - P(A \cap B) \] Rearrange the equation: \[ P(A \cap B) = P(B) - \frac{1}{12} \] Now, considering the condition: \[ 0 \leq P(A \cap B) \leq P(A) \] Substitute \( P(A \cap B) \): \[ 0 \leq P(B) - \frac{1}{12} \leq \frac{2}{3} \] Solving this inequality: 1. Add \( \frac{1}{12} \) to all parts of the inequality: \[ \frac{1}{12} \leq P(B) \leq \frac{2}{3} + \frac{1}{12} \] 2. Simplify the right-hand side: \[ P(B) \leq \frac{3}{4} \] Thus, the final range for \( P(B) \) is: \[ \frac{1}{12} \leq P(B) \leq \frac{3}{4} \]