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Five people, designated as \(A, B, C, D, E,\) are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that (a) there is exactly one person between \(A\) and \(B ?\) (b) there are exactly two people between \(A\) and \(B\) ? (c) there are three people between \(A\) and \(B ?\)

Short Answer

Expert verified
The probabilities of the given scenarios are: (a) \(\frac{2}{5}\), (b) \(\frac{1}{20}\), and (c) \(\frac{1}{60}\).

Step by step solution

01

Find Total Number of Arrangements

There are 5 people, and each person can occupy one of the 5 positions. Hence, the total number of arrangements is given by 5! (5 factorial). Total Arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120
02

a) Exactly one person between A and B

In this case, A and B can be thought of as a single entity, with one person in between them. This leaves us with a total of 4 positions (3 other people plus the A-B combination). We'll calculate the number of arrangements for this situation and then multiply it by two, considering A and B can swap their places. Number of arrangements = 4! = 4 × 3 × 2 × 1 = 24 Since A and B can swap their places, we have a total of 2 × 24 = 48 successful arrangements for this scenario. Probability = Successful arrangements / Total arrangements = 48 / 120 = \(\frac{2}{5}\)
03

b) Exactly two people between A and B

In this case, A and B can be thought of as "boundary" positions, with two people in between them. There are 3 possible ways to position people between A and B (leftmost, two internal positions, and rightmost). As before, we will multiply by two because A and B can swap places. Total arrangements = 3 × 2 = 6 Probability = Successful arrangements / Total arrangements = 6 / 120 = \(\frac{1}{20}\)
04

c) Three people between A and B

In this case, A and B must be at opposite ends of the line. Since there are two possible scenarios (A at the first position and B at the last position, or vice versa), we have a total of 2 successful arrangements. Probability = Successful arrangements / Total arrangements = 2 / 120 = \(\frac{1}{60}\) Summary: (a) Probability: \(\frac{2}{5}\) (b) Probability: \(\frac{1}{20}\) (c) Probability: \(\frac{1}{60}\)

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Key Concepts

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

Factorial
Factorial is a fundamental concept in mathematics, particularly in the field of probability and combinatorics. It refers to the product of all positive integers up to a given number. For example, the factorial of 5, denoted as 5!, is calculated as follows:
  • 5! = 5 × 4 × 3 × 2 × 1 = 120

This concept is crucial because it helps in determining the total number of ways to arrange a set of items, known as permutations. When dealing with probability problems, especially those involving order or arrangement, factorials make calculating possible outcomes more manageable.
Understanding factorials is key when examining problems that require counting permutations or combinations, as it provides the base for more complex calculations.
Permutations
Permutations refer to the different ways of arranging a group of objects. This is relevant when each object in a set is unique, and the order of arrangement matters.
If you have 5 distinct objects, like persons A, B, C, D, and E, the total number of permutations is the factorial of the number of objects, 5!.
  • Permutation of 5 objects: 5! = 120

In the original exercise, permutations are explored when determining the probability of specific arrangements, such as having one, two, or three people between A and B. By treating A and B as a unit and exploring their possible positions within the overall lineup, permutations are utilized to solve the arrangement puzzle.
These calculations guide us in understanding and solving the broader context of the problem, which often applies to various real-world situations.
Combinatorics
Combinatorics is a branch of mathematics concerning the counting, arrangement, and organization of a set of objects. It forms the backbone of probability theory.
In the context of the exercise, combinatorics helps us understand how to choose and arrange subsets of people to meet specific conditions, such as having exactly one or more individuals between two given persons.
This field encompasses techniques for calculating permutations and combinations, which are the building blocks of most probability-related problems. Understanding how to select or arrange items efficiently allows for a deeper comprehension of how probabilities are determined in more complex scenarios.
Combinatorics provides the strategies used to dissect problems into more manageable subproblems, making complex arrangements and probability limits easier to solve and understand.
Conditional Probability
Conditional probability refers to the likelihood of an event occurring, given that another event has already occurred. While the main exercise does not explicitly focus on conditional probability, understanding this concept is essential.
It allows us to refine probability calculations by focusing on specific scenarios that affect the likelihood of certain outcomes. In an arrangement problem, like arranging people in different configurations, conditional probability can become crucial when additional conditions are layered onto an event.
Suppose we expand this to a scenario where specific people cannot be next to each other, conditional probability helps determine the new set of outcomes that satisfy this constraint.
This concept is vital because it extends beyond basic probability to model real-world situations that rely on specific prior events or conditions influencing the probability of subsequent events.

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Most popular questions from this chapter

If it is assumed that all \(\left(\begin{array}{c}52 \\ 5\end{array}\right)\) poker hands are equally likely, what is the probability of being dealt (a) a flush? (A hand is said to be a flush if all 5 cards are of the same suit. (b) one pair? (This occurs when the cards have denominations \(a, a, b, c, d,\) where \(a, b, c,\) and \(d\) are all distinct. (c) two pairs? (This occurs when the cards have denominations \(a, a, b, b, c,\) where \(a, b,\) and \(c\) are all distinct.) (d) three of a kind? (This occurs when the cards have denominations \(a, a, a, b, c,\) where \(a, b\) and \(c\) are all distinct.) (e) four of a kind? (This occurs when the cards have denominations \(a, a, a, a, b .)\)

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