1. **Problem Statement:** We are given a Minor Baseball team with 9 starting players, and we want to analyze probabilities related to the order in which players bat. 2. **Part i:** The probability that the coach selects the same order as last game is $\frac{1}{362880}$. - The denominator $362880$ comes from the total number of ways to arrange 9 players, which is $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880$. - The numerator is 1 because there is exactly one specific order that matches the last game. 3. **Part ii:** Probability that the best hitter is in the 3rd position and the fastest runner is in the 1st position (assuming they are different people). - Fix the best hitter in position 3 and the fastest runner in position 1. - The remaining 7 players can be arranged in any order in the remaining 7 positions. - Number of favorable orders = $7! = 5040$. - Total possible orders = $9! = 362880$. - Probability = $\frac{7!}{9!} = \frac{5040}{362880} = \frac{1}{72}$. 4. **Part iii:** Create a question using the indirect method $n(A')$. - Example question: "What is the probability that the coach does NOT put the best hitter in the third position?" - To solve, use the indirect method: - Total possible orders = $9!$. - Number of orders with best hitter in 3rd position = $8!$ (fix best hitter in 3rd, arrange others). - So, $n(A') = 9! - 8!$. - Probability = $\frac{n(A')}{9!} = 1 - \frac{8!}{9!} = 1 - \frac{1}{9} = \frac{8}{9}$. **Final answers:** - i. Probability = $\frac{1}{362880}$. - ii. Probability = $\frac{1}{72}$. - iii. Probability that best hitter is NOT in 3rd position = $\frac{8}{9}$.