1. The problem involves calculating probabilities of combined events using the addition rule. 2. The addition rule for probabilities states: $$P(A \text{ or } B) = P(A) + P(B) - P(A \cap B)$$ This rule accounts for the overlap between events A and B to avoid double counting. 3. For the first problem, calculating $P(\text{gray or mixed})$: Given: $$P(\text{gray}) = \frac{3}{8}, \quad P(\text{mixed}) = \frac{3}{8}$$ Assuming gray and mixed are mutually exclusive (no overlap), then: $$P(\text{gray or mixed}) = P(\text{gray}) + P(\text{mixed}) = \frac{3}{8} + \frac{3}{8} = \frac{6}{8}$$ 4. For the second problem, calculating $P(\text{blue or too big})$: Given: $$P(\text{blue}) = \frac{7}{14}, \quad P(\text{too big}) = \frac{11}{14}$$ We need to subtract the intersection $P(\text{blue} \cap \text{too big})$ to avoid double counting. 5. Since $P(\text{blue} \cap \text{too big})$ is not provided, the expression remains: $$P(\text{blue or too big}) = \frac{7}{14} + \frac{11}{14} - P(\text{blue} \cap \text{too big})$$ 6. Without the value of $P(\text{blue} \cap \text{too big})$, the exact probability cannot be computed. Final answers: - $P(\text{gray or mixed}) = \frac{6}{8} = \frac{3}{4}$ - $P(\text{blue or too big})$ cannot be determined without $P(\text{blue} \cap \text{too big})$.