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Chapter 4: Problem 14

Choosing officers A campus club consists of five officers: president \((\mathrm{P}),\) vice president \((\mathrm{V}),\) secretary \((\mathrm{S}),\) treasurer (T), and activity coordinator (A). The club can select two officers to travel to New Orleans for a conference; for fairness, they decide to make the selection at random. In essence, they are choosing a simple random sample of size \(n=2\) a. What are the possible samples of two officers? b. What is the chance that a particular sample of size 2 will be drawn? c. What is the chance that the activity coordinator will be chosen?

Short Answer

Expert verified
a. 10 samples: (P, V), (P, S), (P, T), (P, A), (V, S), (V, T), (V, A), (S, T), (S, A), (T, A). b. Each sample has a probability of 0.1. c. Probability of choosing A is 0.4.

Step by step solution

01

Identify Combinatorial Elements

To find the possible samples, recognize that we're dealing with combinations as the order doesn't matter. We have a total of 5 officers, and we want to choose 2 for the conference.
02

Calculate Total Combinations

The total number of ways to choose 2 officers from 5 is given by the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]In this scenario, \( n = 5 \) and \( k = 2 \). So, we calculate:\[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]
03

List Possible Samples

From the total combinations calculated, list all possible pairs of officers: 1. (P, V) 2. (P, S) 3. (P, T) 4. (P, A) 5. (V, S) 6. (V, T) 7. (V, A) 8. (S, T) 9. (S, A) 10. (T, A) These are the 10 possible samples of size 2.
04

Calculate Probability of a Sample

Every sample has an equal chance of being chosen. Since there are 10 possible samples:\[ \text{Probability of one specific sample (e.g., (P, V))} = \frac{1}{10} = 0.1 \]
05

Calculate Probability of Including Activity Coordinator

To determine the chance that the activity coordinator (A) will be chosen, count the number of pairs that include A:1. (P, A)2. (V, A)3. (S, A)4. (T, A)There are 4 such pairs, so the probability is:\[ \frac{4}{10} = 0.4 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simple Random Sampling
Simple random sampling is a statistical method where each member of a population has an equal chance of being chosen. In the context of the club officers attending a conference, this method ensures fairness. Each officer, whether the president, vice president, secretary, treasurer, or activity coordinator, has an identical probability of being selected.
This method is powerful because it prevents bias. It's crucial in experiments and surveys to ensure that the selection doesn't favor one group over another.
In practical terms, imagine writing each officer's name on a slip of paper, placing them in a hat, and drawing two at random. By this means, every possible pair of officers has the same chance of being selected, thereby achieving a simple random sample.
Combination Formula
The combination formula is fundamental in combinatorics when the order of selection does not matter. This is perfect for scenarios like the one with the club officers, where we're interested in simply selecting pairs, not in the order they are chosen.
The formula is given as \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), which computes how many combinations are possible for choosing \( k \) items from \( n \) items.
For the five officers, where two need to be selected, the formula becomes:
  • \( n = 5 \)
  • \( k = 2 \)
Plug these values into the formula: \[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]This calculation tells us there are 10 different combinations or groups of pairs that can be made from five individuals.
Probability Calculation
Probability is a way to quantify how likely an event is to occur. In this exercise, we're calculating two probabilities: the likelihood of picking a particular pair of officers and the likelihood of picking the activity coordinator.
Since each of the 10 possible pairs of officers is equally likely, the probability of any one specific pair being chosen, such as (P, V), is \( \frac{1}{10} = 0.1 \). This reflects the concept that each outcome has an equal chance.

To find the probability of selecting the activity coordinator, we first count the number of combinations that include them. There are four such pairs: (P, A), (V, A), (S, A), and (T, A). Hence, the probability that the activity coordinator is selected is: \[ \frac{4}{10} = 0.4 \]This shows a straightforward and fair chance calculation, guiding informed decisions using probability concepts.

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