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Chapter 7: Problem 69

Use counting arguments from the preceding chapter. Horse Races The seven contenders in the fifth horse race at Aqueduct on February 18,2002, were: Pipe Bomb, Expect a Ship, All That Magic, Electoral College, Celera, Cliff Glider, and Inca Halo. \({ }^{5}\) You are interested in the first three places (winner, second place, and third place) for the race. a. Find the cardinality \(n(S)\) of the sample space \(S\) of all possible finishes of the race. (A finish for the race consists of a first, second, and third place winner.) b. Let \(E\) be the event that Electoral College is in second or third place, and let \(F\) be the event that Celera is the winner. Express the event \(E \cap F\) in words, and find its cardinality.

Short Answer

Expert verified
a. The cardinality of the sample space \(n(S)\) is 210 possible finishes of the race. b. The event \(E \cap F\) is when Electoral College is in second or third place and Celera is the winner. The cardinality of this event is 10 possible finishes.

Step by step solution

01

a. Cardinality of the Sample Space

To find the cardinality of the sample space, we will use the counting principle. There are 7 horses in the race and we are interested in the first three places. So, let's break down the number of possible finishes based on each position: 1. Number of choices for the winner (1st place): There are 7 contenders, so we have 7 possibilities for the winner. 2. Number of choices for the 2nd place: Since the winner has already been chosen, there are 6 contenders left to compete for the 2nd place. 3. Number of choices for the 3rd place: After the first two places have been filled, there are 5 contenders left to compete for the 3rd place. Apply the counting principle, the cardinality of the sample space \(n(S)\) is the product of the number of choices for each position: $$n(S) = 7 \times 6 \times 5 = 210$$ So, there are 210 possible finishes of the race.
02

b. Cardinality of the Event \(E \cap F\)

The event \(E \cap F\) implies that Electoral College is in second or third place and Celera is the winner. We need to find the cardinality of this event. Since Celera is the winner (1st place) in this event, we are left with 6 horses competing for the remaining 2nd and 3rd places. And, we know that Electoral College has to be in one of these places. Let's break down the number of choices for each position: 1. Number of choices for the 2nd place: As Electoral College can be in either 2nd or 3rd place, suppose it finishes in 2nd place. Therefore, there is only 1 choice for the 2nd place. 2. Number of choices for the 3rd place: Since Electoral College and Celera are already in the 1st and 2nd places, only 5 horses are left to contend for the 3rd place position. This gives us 1 choice for 2nd place and 5 choices for 3rd place, resulting in 5 possible finishes with Electoral College in 2nd place. Now, let's examine the case where Electoral College is in third place: 1. Number of choices for the 2nd place: There are 5 remaining horses to compete for 2nd place as Electoral College will be in 3rd place. 2. Number of choices for the 3rd place: Since Electoral College is finishing in 3rd place, there is only 1 choice for the 3rd place. This gives us 5 choices for 2nd place and 1 choice for 3rd place, resulting in 5 possible finishes with Electoral College in 3rd place. Adding up the possibilities for each case, we find the cardinality of the event \(E \cap F\) is: $$n(E \cap F) = 5 + 5 = 10$$ So, there are 10 possible finishes where Electoral College is in second or third place and Celera is the winner.

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Key Concepts

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

Understanding Sample Space
In probability, the concept of a sample space is foundational. A sample space, often denoted by the letter S, is the set of all possible outcomes of a particular experiment or random trial. For instance, when flipping a coin, the sample space consists of two outcomes: heads or tails. Similarly, in the context of the given horse race example, the sample space includes all the different ways the horses can finish in the top three positions.

To visualize the sample space, imagine writing down every single combination or sequence that could arise, where each sequence represents a unique finish of the first, second, and third place winners. The size of the sample space is crucial because it helps determine probabilities over the entire set of outcomes. The larger the sample space, the more potential outcomes there are to consider when calculating probabilities.
The Cardinality of a Set
The term cardinality refers to the number of elements in a set. In probability and statistics, understanding the cardinality is essential because it represents the total number of possible outcomes, which is pivotal for calculating probabilities. Cardinality is usually denoted by n(S) when referring to a sample space S.

In the horse race example provided, to find the cardinality of the sample space, the counting principle is applied. Since the race involves selecting a sequence of different horses to finish in the first, second, and third place out of seven contenders, the cardinality is represented as the product of the number of choices for each position, which results in 210 different possible outcomes. This is an integral part of the solution that leads to understanding the likelihood of various events within the context of the race.
Calculating Permutations
The concept of permutations comes into play when we want to count the number of ways we can arrange a set of items in order. This is particularly relevant in scenarios where the order is important, such as in race finishes, seating arrangements, or in ordering a sequence of numbers. A permutation is an ordered combination of items that are distinct from one another.

Applying permutations to the horse race example, the first, second, and third place finishes are calculated by considering the specific order in which horses can cross the finish line. The formula often used is P(n, r) = n! / (n-r)!, where n is the total number of items to choose from, and r is the number of items to be arranged. However, in simplified cases like this horse race, we can use the counting principle directly by multiplying the choices available as we progress through each finish position. The permutation concept helps us understand that there are 10 specific ways that the event E 鈭� F鈥攖he intersection of Electoral College being second or third, and Celera being the winner鈥攃an occur. By understanding permutations, students not only solve the problem at hand but also gain a valuable tool for analyzing similar problems across various contexts.

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Most popular questions from this chapter

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