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The combination formula is a fundamental concept used in probability and combinatorics. It allows us to determine the number of ways to choose a subset of items from a larger set, where the arrangement does not matter. In this exercise, we use the combination formula to find out how many possible samples of two officers can be selected from a set of five. To calculate combinations, the formula used is \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). Here, \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. - **Example calculation:** For this problem, \( n = 5 \) (the total number of officers) and \( r = 2 \) (the number of officers to select). - Apply the values to the formula: \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \] This calculation tells us there are 10 different possible pairs of officers we could select, showing how combinations help simplify scenarios where the order of selection is not important.

Probability calculation is the process of determining the likelihood of a specific event occurring. It's a crucial part of predicting outcomes in random scenarios, like this one about officer selection. Given that each pair is equally likely to be chosen, probability offers a way to quantify fairness and randomness. To determine the probability of selecting a particular pair, we use the formula: - **Probability of a single event:** \[ \text{Probability (P)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \] - In this problem, each pair is equally probable: - There are 10 possible pairs, so the probability of picking any specific pair (say, (P, V)) is: \[ P(\text{specific pair}) = \frac{1}{10} = 0.1 \] This simply means that any pair of officers has a 10% chance of being the chosen team to travel to the conference. Additionally, to find the probability that a specific officer, such as the activity coordinator A, is selected, count the number of pairs including that officer: - We have four pairs: (P, A), (V, A), (S, A), and (T, A). - Thus, the probability that A is included is: \[ \frac{4}{10} = \frac{2}{5} = 0.4 \] This tells us there's a 40% probability that the activity coordinator is chosen in any random draw of two officers.

Listing possible outcomes is a helpful approach in probability to ensure every scenario is considered. It provides a clear structure, guaranteeing nothing is missed and is especially useful for visual learners who benefit from seeing all possibilities laid out. For this problem, listing helps identify the complete set of officer pairs, confirming calculations are based on full and accurate information. - **Steps to list outcomes:** - Start with a clear set of all available items—in this case, the five officers P, V, S, T, and A. - Systematically pair each officer with every other officer without repeating pairs. - Avoid counting the same pair in reverse, for example, (P, V) is the same as (V, P). - **Outcome list created:** - (P, V), (P, S), (P, T), (P, A) - (V, S), (V, T), (V, A) - (S, T), (S, A) - (T, A) This careful listing reveals there are exactly 10 unique outcomes, matching the total number derived from the combination formula. By seeing all outcomes laid out, one can easily understand how many total options are available and confirm that all calculations, such as probability assessments, are accurate.