\( 1 \leftarrow \begin{array}{l}\text { A probability experiment is conducted in which the sample space of the experiment is } S=\{1,2,3,4,5,6,7,8,9,10,11,12\} \text {, event } F=\{5,6,7,8,9,10\} \text {, and event } G=\{9,10,11,12\} \text {. Assume that each } \\ \text { outcome is equally likely. List the outcomes in } F \text { or } G \text {. Find } P(F \text { or } G) \text { by counting the number of outcomes in } F \text { or } G \text {. Determine } P(F \text { or } G) \text { using the general addition rule. } \\ \text { List the outcomes in } F \text { or } G \text {. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. } \\ \text { A. F or } G=\{\square \\ \text { (Use a comma to separate answers as needed.) } \\ \text { B. F or } G=\{ \}\end{array} \)

The outcomes in \( F \) or \( G \) can be found by taking the union of sets \( F \) and \( G \). This means we combine all unique elements from both sets: \( F = \{5, 6, 7, 8, 9, 10\} \) and \( G = \{9, 10, 11, 12\} \). Thus, \( F \) or \( G = \{5, 6, 7, 8, 9, 10, 11, 12\} \). To find \( P(F \text{ or } G) \), we count the outcomes. There are 8 outcomes in total (5, 6, 7, 8, 9, 10, 11, 12) out of 12 total possible outcomes in \( S \). Thus, \( P(F \text{ or } G) = \frac{8}{12} = \frac{2}{3} \). Using the general addition rule gives: \[ P(F) + P(G) - P(F \cap G) \] Calculating separately: - \( P(F) = \frac{6}{12} = \frac{1}{2} \) - \( P(G) = \frac{4}{12} = \frac{1}{3} \) - \( P(F \cap G) \) includes outcomes \( \{9, 10\} \), so \( P(F \cap G) = \frac{2}{12} = \frac{1}{6} \) Thus, \[ P(F \text{ or } G) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \] So the outcomes are: A. \( F \text{ or } G=\{5, 6, 7, 8, 9, 10, 11, 12\} \)