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Four equally qualified runners, John, Bill, Ed, and Dave, run a 100 -meter sprint, and the order of finish is recorded. a. How many simple events are in the sample space? b. If the runners are equally qualified, what probability should you assign to each simple event? c. What is the probability that Dave wins the race? d. What is the probability that Dave wins and John places second? e. What is the probability that Ed finishes last?

Short Answer

Expert verified
Answer: The probabilities for the respective outcomes are as follows: Dave winning = 1/4, Dave winning and John placing second = 1/12, and Ed finishing last = 1/4.

Step by step solution

01

Determine the number of simple events in the sample space

For this, we count all the possible outcomes of finishing order among the four runners. Since there are four runners, and each position can be occupied by any one of the runners (assuming no tie), we have a total of 4! (4 factorial) possible permutations. The formula for the number of permutations is: n! = n(n-1)(n-2)...(n-(n-1)) Using this formula, we find the number of simple events in the sample space: 4! = 4 × 3 × 2 × 1 = 24
02

Assign probability to each simple event

Since the runners are equally qualified, each simple event has an equal probability of occurring. To find the probability of each simple event, we divide 1 (total probability) by the number of simple events (24). So, the probability of each simple event is: P(simple event) = 1/24
03

Find the probability that Dave wins the race

To find the probability that Dave wins, we need to know the total number of permutations where Dave finishes first. Since there are three other runners, and after Dave, there are 3! simple events for the rest of the runners. So, the total number of permutations with Dave winning is: 3! = 3 × 2 × 1 = 6 Now, we multiply the 6 simple events with the probability of each of those simple events: P(Dave wins) = 6 × (1/24) = 1/4
04

Find the probability that Dave wins and John places second

We find the total number of permutations where Dave finishes first and John finishes second. There are two other runners left (Bill and Ed), so there are 2! permutations for them: 2! = 2 × 1 = 2 Now, we multiply the 2 simple events with the probability of each of those simple events: P(Dave wins and John places second) = 2 × (1/24) = 1/12
05

Find the probability that Ed finishes last

To find the probability that Ed finishes last, we need to know the total number of permutations where Ed finishes fourth. Since there are three other runners, there are 3! simple events for the rest of the runners (excluding Ed). So, the total number of permutations with Ed finishing last is: 3! = 3 × 2 × 1 = 6 Now, we multiply the 6 simple events with the probability of each of those simple events: P(Ed finishes last) = 6 × (1/24) = 1/4

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Key Concepts

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

Sample Space
In probability theory, a sample space is the set of all possible outcomes of a particular experiment. When we're dealing with the outcomes of a race, such as the 100-meter sprint run by John, Bill, Ed, and Dave, each possible order of finish represents a simple event within the sample space.

For this race, we need to consider every possible combination of finishing positions for the four runners. Since each of the four runners can finish in any of the four positions, and no two runners can share a position, each unique permutation of finishers represents a different simple event. Thus, if each runner could potentially occupy each finishing position, the total number of simple events is given by the factorial of the number of runners, written as 4!.

  • The sample space for this scenario includes permutations because order matters: first, second, third, and fourth place.
  • Including all permutations ensures we've accounted for every possible outcome of the race.
Permutation
Permutation is a key concept in probability that involves arranging a set of elements in order. Permutation takes into account the arrangement of these elements, meaning the sequence is important.

In the context of our race, permutations allow us to list every possible order in which the four runners can finish. Since the position of each runner is significant (e.g., who comes first, who comes second), permutations are utilized rather than combinations.

  • To compute permutations, use the factorial operation, denoted as n!, where n is the number of distinct items to arrange.
  • For our four runners, the permutation calculation is 4!, meaning 4 × 3 × 2 × 1, which equals 24 total permutations.

This means there are 24 different ways that the runners could potentially finish the race, each representing a unique event in the sample space.
Equally Likely Events
Equally likely events in probability mean that each outcome has the same chance of occurring as any other outcome. This assumption simplifies calculations because probabilities are distributed uniformly across all possible outcomes.

For the race with our four runners, if we assume they are equally qualified, any order of finish is as probable as another. Hence, each permutation of the final results carries an equal probability of occurring.

  • Since each outcome is equally likely, the probability for any specific sequence of finishes is equal to 1 divided by the total number of permutations.
  • This is specifically calculated as 1/24 because there are 24 permutations.

This concept is powerful in situations where no particular outcome is favored, allowing for straightforward probability calculations using the concept of equally likely events.
Factorial Calculation
Factorial calculation plays a crucial role in solving permutation-related problems in probability. The factorial of a number is the product of all positive integers from 1 to that number. In mathematical terms, it's expressed as n!.

In our race scenario with four runners, the factorial calculation helps us determine the total number of permutations possible by arranging the finishing order of the runners. We compute this with 4!, which is 4 × 3 × 2 × 1, equaling 24.

  • Factorials are not only useful in permutations but also in other probability concepts requiring arrangement or order consideration.
  • Understanding how to calculate and use factorials is important for determining complex probabilities in real-world applications.

This concept is foundational for tackling problems that involve sorting, ranking, or ordering, which are common in probability exercises.

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Most popular questions from this chapter

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