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Refer to Exercise 4.14, in which a 100 -meter sprint is run by John, Bill, Ed, and Dave. Assume that all of the runners are equally qualified, so that any order of finish is equally likely. Use the \(m n\) Rule or permutations to answer these questions: a. How many orders of finish are possible? b. What is the probability that Dave wins the sprint? c. What is the probability that Dave wins and John places second? d. What is the probability that Ed finishes last?

Short Answer

Expert verified
Answer: The probability of Dave winning the sprint and John placing second is 1/12.

Step by step solution

01

Find all the possible orders of finish using permutations

First, we need to find the total number of possible orders of finish for the four runners. Since all four runners are equally qualified, any order of finish is equally likely. We can use the permutation formula to solve this: Permutations = \(n!\), where n is the number of runners In this case, n = 4. Permutations = \(4! = 4 \times 3 \times 2 \times 1 = 24\) So, there are 24 possible orders of finish.
02

Calculate the probability of Dave winning the sprint

To find the probability of Dave winning the sprint, we need to count the number of permutations in which Dave finishes first. Since there are 3 other runners, these permutations would look like: 1D__ (3 different permutations) D____ Permutations with Dave winning = 1 × \(\begin{pmatrix}3\\1\end{pmatrix}\) Since there are 3! = 6 permutations for the other three runners, there are a total of 6 permutations where Dave finishes first. Therefore, the probability of Dave winning the sprint = \(\frac{\text{number of outcomes with Dave winning}}{\text{total possible outcomes}} = \frac{6}{24} = \frac{1}{4}\)
03

Calculate the probability of Dave winning and John placing second

To find the probability of Dave winning and John placing second, we need to count the number of permutations where Dave finishes first and John finishes second. For the remaining two runners, there are 2! = 2 permutations. Therefore, the probability of Dave winning and John placing second = \(\frac{\text{number of outcomes with Dave winning and John placing second}}{\text{total possible outcomes}} = \frac{2}{24} = \frac{1}{12}\)
04

Calculate the probability of Ed finishing last

To find the probability of Ed finishing last, we need to count the number of permutations where Ed finishes in the last place. Since there are 3! = 6 permutations for the other three runners (without Ed) to finish first, second, and third, there are a total of 6 permutations where Ed finishes last. Therefore, the probability of Ed finishing last = \(\frac{\text{number of outcomes with Ed finishing last}}{\text{total possible outcomes}} = \frac{6}{24} = \frac{1}{4}\) To summarize: a. There are 24 possible orders of finish. b. The probability of Dave winning the sprint is \(\frac{1}{4}\). c. The probability of Dave winning and John placing second is \(\frac{1}{12}\). d. The probability of Ed finishing last is \(\frac{1}{4}\).

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Key Concepts

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

Permutations
Permutations are a fundamental concept in probability, especially when the order of outcomes matters. When you're dealing with permutations, you're looking at the arrangements of a set of objects where order is important. For instance, in the context of a race with four runners - John, Bill, Ed, and Dave - the order they finish matters. So, if Dave finishes first, and Ed finishes last, that's a distinct permutation from when Ed finishes first and Dave finishes last.

To calculate permutations, we use the factorial of the number of items to determine how many different orders are possible. The formula is given by \(n!\) where \(n\) is the number of items. For our four runners, the calculation is \(4!\), which results in:
  • 4 × 3 × 2 × 1 = 24
This tells us there are 24 unique ways the runners can finish the race, assuming all possible sequences of finishes are considered.
Probability Calculation
The process of calculating probability involves finding how often a particular event occurs relative to the possible total outcomes. Probability ranges from 0 to 1, where 0 means the event does not occur, and 1 means it unequivocally does. In the race example, to find Dave's probability of winning, we need to find how many permutations have him finishing first.

When calculating probabilities, you divide the number of favorable outcomes by the total number of possible outcomes. Since there are 6 ways for the remaining 3 runners to finish after Dave (3! = 6), his probability of winning is given by:
  • Number of favorable outcomes for Dave winning: 6
  • Total possible outcomes: 24
  • Dave’s winning probability: \(\frac{6}{24} = \frac{1}{4}\)
Order of Finish
The order of finish is crucial in understanding permutations and probability. This concept deals with the different sequences in which participants complete an action. Each distinct sequence is a unique permutation. For instance, in a sprint race of four individuals, the sequence in which they cross the finish line is evaluated.

Visualize a scenario where we want Dave first, and John second. There's a specific set order (i.e., Dave finishes first, followed by John). With the remaining two runners, the order doesn't matter as much, but it still affects the overall permutations. Thus, when order is specified, you end up calculating permutations for each subsequent position, guiding us to solve specific probability questions on position rankings.
Equally Likely Outcomes
The assumption of equally likely outcomes is often a grounding principle in probability. It indicates that each outcome of an event has the same chance of occurring. When applied to our sprint race, it suggests that each order of runners finishing is as probable as any other, assuming all are equally qualified.

This principle allows for simpler probability calculations because each permutation of the race's outcome holds the same likelihood. For practical purposes, this assumption levels the playing field: no participant is seen as more likely to win solely based on this likelihood model. Thus, all permutations, whether involving Dave or Ed finishing first, are treated with equal initial probability unless further qualifiers are added.

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