1. **State the problem:** We have a sample space $S = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\}$, event $F = \{2, 3, 4, 5, 6\}$, and event $G = \{6, 7, 8, 9\}$. We want to list the outcomes in $F \text{ or } G$ (i.e., $F \cup G$), find $P(F \cup G)$ by counting outcomes, and then find $P(F \cup G)$ using the general addition rule. 2. **List the outcomes in $F \cup G$:** The union of two sets includes all elements in either set without repetition. $$F \cup G = \{2, 3, 4, 5, 6\} \cup \{6, 7, 8, 9\} = \{2, 3, 4, 5, 6, 7, 8, 9\}$$ 3. **Count the number of outcomes in $F \cup G$:** There are 8 outcomes. 4. **Calculate $P(F \cup G)$ by counting:** Since all outcomes in $S$ are equally likely and $|S|=12$, $$P(F \cup G) = \frac{|F \cup G|}{|S|} = \frac{8}{12} = \frac{2}{3}$$ 5. **Use the general addition rule:** $$P(F \cup G) = P(F) + P(G) - P(F \cap G)$$ Calculate each probability: $$P(F) = \frac{|F|}{|S|} = \frac{5}{12}$$ $$P(G) = \frac{|G|}{|S|} = \frac{4}{12} = \frac{1}{3}$$ $$F \cap G = \{6\} \Rightarrow |F \cap G| = 1$$ $$P(F \cap G) = \frac{1}{12}$$ Apply the formula: $$P(F \cup G) = \frac{5}{12} + \frac{4}{12} - \frac{1}{12} = \frac{8}{12} = \frac{2}{3}$$ **Final answer:** - Outcomes in $F \cup G$ are $\{2, 3, 4, 5, 6, 7, 8, 9\}$. - $P(F \cup G) = \frac{2}{3}$ by both counting and the general addition rule.