Let \( A \) be the event that the first throw shows a number less than 4 (i.e., 1, 2, or 3). \[ P(A) = \frac{3}{6} = \frac{1}{2} \] Let \( B \) be the event that the second throw shows at least 4 (i.e., 4, 5, or 6). \[ P(B) = \frac{3}{6} = \frac{1}{2} \] Since the two throws are independent, the probability of both events happening simultaneously is: \[ P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \] We need to find the probability of \( A \) OR \( B \) occurring: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] \[ P(A \cup B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4} \]