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Chapter 2: Problem 43

(a) If \(N\) people, including \(A\) and \(B,\) are randomly arranged in a line, what is the probability that \(A\) and \(B\) are next to each other? (b) What would the probability be if the people were randomly arranged in a circle?

Short Answer

Expert verified
(a) The probability that A and B are next to each other when N people are randomly arranged in a line is \(\frac{2*(N-1)!}{N!}\). (b) The probability that A and B are next to each other when N people are randomly arranged in a circle is \(\frac{2*(N-2)!}{(N-1)!}\).

Step by step solution

01

Calculate the total number of arrangements

There are N people to be arranged in a line, and each person can be arranged in any position. Using the permutation formula, the total number of possible arrangements for N people is N!.
02

Calculate the number of arrangements with A and B next to each other

Now, consider A and B as a single unit (AB or BA). This new unit and the remaining N-2 people need to be arranged in a line. There are (N-1) units now, so the number of arrangements for this scenario is (N-1)!. Since A and B can be arranged as AB or BA within their unit, we need to multiply by 2 to account for this. So the number of arrangements where A and B are next to each other is 2*(N-1)!.
03

Calculate the probability

Now we will find the probability that A and B are next to each other by dividing the number of favorable arrangements by the total number of arrangements. Probability = (Number of arrangements with A and B next to each other) / (Total number of arrangements) Probability = \(\frac{2*(N-1)!}{N!}\) #Part (b) - Arranging people in a circle#
04

Calculate the total number of arrangements

For arranging people in a circle, we can fix one person's position to account for the rotation and arrange the remaining (N-1) people. Using the permutation formula, the total number of possible arrangements for N people in a circle is (N-1)!.
05

Calculate the number of arrangements with A and B next to each other

Similar to part (a), consider A and B as a single unit (AB or BA). This new unit and the remaining N-2 people need to be arranged in a circle. There are now (N-2) units, so the number of arrangements for this scenario is (N-2)!. Since A and B can be arranged as AB or BA within their unit, we need to multiply by 2 to account for this. So the number of arrangements where A and B are next to each other is 2*(N-2)!.
06

Calculate the probability

Now we will find the probability that A and B are next to each other when people are arranged in a circle, by dividing the number of favorable arrangements by the total number of arrangements. Probability = (Number of arrangements with A and B next to each other) / (Total number of arrangements) Probability = \(\frac{2*(N-2)!}{(N-1)!}\)

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Key Concepts

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

Permutations
Permutations involve the arrangement of objects in a specific order. When talking about permutations in probability, we often consider how many ways we can arrange a set of unique items. For example, if we have three different books, there are several ways to arrange them on a shelf. Each unique arrangement represents a different permutation.

Permutations are important in calculating probabilities, especially when the order of arrangement matters. In exercises involving permutations, we often use the formula for permutations of a set size:
  • The permutation formula for arranging \(n\) distinct objects is given by \(n!\), where \(!\) denotes factorial.
  • This means you multiply all positive integers up to \(n\) together. For instance, for 3 objects, \(3! = 3 \times 2 \times 1 = 6\).
Factorials
Factorials might sound complex, but they are simply a shorthand for multiplying a series of descending natural numbers. The term "factorial" applies to non-negative integers, and it is symbolized by the exclamation mark, \(!\).

Here's how it works:
  • The factorial of 0, represented as \(0!\), is defined as 1.
  • For a positive integer \(n\), \(n!\) is calculated by multiplying all integers from 1 up to \(n\).
  • For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials play a crucial role in permutations, combinations, and various probability calculations. They help determine the number of possible arrangements or selections in a given scenario.
Circular Arrangements
Circular arrangements refer to arranging objects in a circle, where rotations of a configuration are considered identical. This type of arrangement is different from linear arrangements because each position is related to another in a circular manner.

In a circular permutation, we fix one position and arrange the remaining objects relative to that fixed point:
  • For \(n\) objects, the number of circular permutations is \((n-1)!\), because we can arrange the remaining \(n-1\) objects in relation to a fixed point.
  • When two specific objects need to be next to each other in a circle, they can be treated as a single unit, reducing the problem to arranging \(n-1\) units.
Understanding circular arrangements is essential when studying problems where the positioning of objects in a loop affects the outcome.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, arranging, and selecting objects. It's a powerful tool used in probability and statistics to solve problems related to arrangement and combination.

Key concepts in combinatorics include:
  • Permutations: As discussed, permutations focus on the arrangement of objects where the order is important.
  • Combinations: When the order does not matter, we study combinations, which count the number of ways to choose items from a larger set.
  • Binomial Coefficient: Denoted as \(\binom{n}{k}\), it represents the number of combinations possible by selecting \(k\) items from \(n\) options.
Combinatorics underpins many probability scenarios by providing methods to calculate how likely particular arrangements or selections are under defined conditions.

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Most popular questions from this chapter

The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired? (b) Rebecca and Elise will be chosen to represent their schools but will not play each other? (c) either Rebecca or Elise will be chosen to represent her school?

A system is composed of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector \(\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right),\) where \(x_{i}\) is equal to 1 if component \(i\) is working and is equal to 0 if component \(i\) is failed. (a) How many outcomes are in the sample space of this experiment? (b) Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components \(1,3,\) and 5 are all working. Let \(W\) be the event that the system will work. Specify all the outcomes in \(W\). (c) Let \(A\) be the event that components 4 and 5 are both failed. How many outcomes are contained in the event \(A ?\) (d) Write out all the outcomes in the event \(A W\).

There are 5 hotels in a certain town. If 3 people check into hotels in a day, what is the probability that they each check into a different hotel? What assumptions are you making?

Two symmetric dice have had two of their sides painted red, two painted black, one painted yellow, and the other painted white. When this pair of dice is rolled, what is the probability that both dice land with the same color face up?

In a hand of bridge, find the probability that you have 5 spades and your partner has the remaining \(8 .\)
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