Let \( A \) and \( B \) be independent events with \( P(A)=\frac{1}{3} \) and \( P(B)=\frac{1}{8} \). Find \( P(A \cap B) \) and \( P(A \cup B) \).

To find \( P(A \cap B) \) and \( P(A \cup B) \) for the independent events \( A \) and \( B \), we can use the following formulas: 1. For independent events, the probability of both events occurring (the intersection) is given by: \[ P(A \cap B) = P(A) \cdot P(B) \] 2. The probability of either event occurring (the union) is given by: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Given: - \( P(A) = \frac{1}{3} \) - \( P(B) = \frac{1}{8} \) First, we calculate \( P(A \cap B) \): \[ P(A \cap B) = P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{1}{8} = \frac{1}{24} \] Next, we calculate \( P(A \cup B) \): \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the values we have: \[ P(A \cup B) = \frac{1}{3} + \frac{1}{8} - \frac{1}{24} \] To perform this calculation, we need a common denominator. The least common multiple of 3, 8, and 24 is 24. We convert each fraction: \[ \frac{1}{3} = \frac{8}{24}, \quad \frac{1}{8} = \frac{3}{24}, \quad \frac{1}{24} = \frac{1}{24} \] Now substituting these values into the equation: \[ P(A \cup B) = \frac{8}{24} + \frac{3}{24} - \frac{1}{24} = \frac{8 + 3 - 1}{24} = \frac{10}{24} = \frac{5}{12} \] Thus, the final results are: \[ P(A \cap B) = \frac{1}{24} \] \[ P(A \cup B) = \frac{5}{12} \]