Question

Refer to Exercise 36 (Section 4.2 ), 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

- To calculate the probability that Dave finishes first and Ed finishes last, we already have found the probabilities of those two events individually. Now, we need to find the probability of both events occurring together. To find this probability, we need to calculate the number of ways Dave finishes first and Ed finishes last, and then divide it by the total number of possible orders of finish. There are 2 remaining spots for John and Bill to finish, which means there are 2! ways for them to finish (2 ways). So, the probability of Dave finishing first and Ed finishing last is (Number of ways Dave finishes first and Ed finishes last) / (total number of possible orders of finish) = 2 / 24 = 1/12. Answer: The probability that Dave finishes first and Ed finishes last is 1/12.

Step-by-step solution

Calculate the total number of possible orders of finish

To find the total number of possible orders of finish for all the 4 runners, we will use the permutation formula where n = 4 and r = 4: nPr = n! / (n - r)! In this case, both n and r are equal to 4, so nPr = 4! / 0! which is equal to 24. So, there are 24 different possible orders of finish.

Calculate the probability that Dave wins the sprint

If Dave wins the sprint, the remaining 3 runners, John, Bill, and Ed can finish the race in any order. So, for calculating this probability, we will break it up into two parts: Probability that Dave wins: (number of ways Dave wins) / (total number of possible orders of finish) There are 3! ways the other runners can finish or 6 ways, so the probability that Dave wins is 6/24 = 1/4.

Calculate the probability that Dave wins and John places second

Now, we know that Dave has to be first and John has to be second. So, there are 2 remaining spots for Bill and Ed to finish. Probability that Dave wins and John places second: (Number of ways that Dave wins and John places second) / (total number of possible orders of finish) There are 2! ways for Bill and Ed to finish or 2 ways, so the probability that Dave wins and John places second is 2/24 = 1/12.

Calculate the probability that Ed finishes last

In this case, Ed has to finish last. So, there are 3 remaining spots for Dave, John, and Bill to finish. We can calculate this probability as: Probability that Ed finishes last: (Number of ways Ed finishes last) / (total number of possible orders of finish) There are 3! ways for Dave, John, and Bill to finish or 6 ways, so the probability that Ed finishes last is 6/24 = 1/4. So, to answer the original questions: a) There are 24 possible orders of finish. b) The probability that Dave wins the sprint is 1/4. c) The probability that Dave wins and John places second is 1/12. d) The probability that Ed finishes last is 1/4.

Key concepts

Permutation Formula

When calculating the different ways in which a set of objects can be arranged, permutation is a fundamental concept in probability and combinatorics. The permutations of a set depend on the number of objects to be arranged and how many of them are being positioned.

In mathematical terms, the number of permutations of 'n' objects taken 'r' at a time is given by the permutation formula, which is expressed as
\[ nPr = \frac{n!}{(n - r)!} \] Here, the exclamation point denotes a factorial, which is the product of all positive integers from 1 up to that number (more on this below). The permutation formula is helpful in various scenarios, such as determining the number of possible orders of finish in a race. To illustrate, with four runners and four places, every single permutation represents a unique order in which the runners can finish. This is why in the solution, for 4 runners racing to finish at 4 different spots, there are \( 4! / (4 - 4)! = 24 \) different possible orders.

In the given problem, since we are looking for all possible arrangements with no restrictions, the permutation formula simplifies because 'n' is equal to 'r'. This simplicity allows for easy computation, particularly when we use factorial notation.

Probability Calculations

Probability calculations are critical for determining how likely an event is to occur. The heart of probability lies in the ratio of the number of favorable outcomes to the total number of possible outcomes. This is represented by the formula:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]The concept becomes clearer through the exercises like the race scenario. For example, if one wanted to determine the likelihood of Dave winning the race, we would first identify the favorable outcomes — which in this case is the number of ways Dave can win (6). We then divide that by the total number of arrangements permissible (24). This calculation, \( 6/24 = 1/4 \), tells us that Dave has a one in four chance of winning, assuming all runners have equal chances.

Understanding and manipulating these ratios is essential in analyzing outcomes and probabilities in a wide array of situations, from games of chance to forecasting events in the natural and social sciences. Remember that probabilities are always between 0 and 1, where 0 means something cannot happen and 1 means it will certainly happen.

Factorial Notation

Factorial notation is a mathematical concept used to describe the product of all positive integers up to a certain number. Represented by an exclamation point (!), the factorial of a number 'n' is the product of all whole numbers from 1 to 'n' and is symbolized by \( n! \). Factorials grow at an incredibly rapid rate, making them significant in combinatorics and probability calculations.

For example:
\[ 1! = 1 \]\[ 2! = 2 \times 1 = 2 \]\[ 3! = 3 \times 2 \times 1 = 6 \]In the race problem, factorial notation simplifies the computation of possible outcomes. When the problem asks for the total number of orders four runners can finish, it refers to \( 4! \), which calculates to \( 4 \times 3 \times 2 \times 1 = 24 \). The factorial of zero, \( 0! \), is defined to be 1, which comes into play when computing a permutation where n equals r; this is why in the permutation formula given in the original solution, 4! is divided by 0!, yielding 24.(no division by zero issues occur).

Factorial notation is a powerful tool to understand because it eases the handling of large calculations in probability and permutations, enabling students to approach these problems with greater confidence.