/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 A total of \(2 n\) people, consi... [FREE SOLUTION] | 91Ó°ÊÓ

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A total of \(2 n\) people, consisting of \(n\) married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let \(C_{i}\) denote the event that the members of couple \(i\) are seated next to each other, \(i=1, \ldots, n\) (a) Find \(P\left(C_{i}\right)\) (b) For \(j \neq i,\) find \(P\left(C_{j} | C_{i}\right)\) (c) Approximate the probability, for \(n\) large, that there are no married couples who are seated next to each other.

Short Answer

Expert verified
For large n, the probability that there are no married couples seated next to each other at the round table is approximately \(P(D) \approx e^{- \frac{1}{2}} \approx 0.6065\).

Step by step solution

01

Compute the total number of possible configurations

For 2n people at a round table, the total number of possible seatings is given by the (2n-1) factorial, since we are arranging the people in a circular manner.
02

Compute the probability that couple i is seated together

Consider the members of couple i as a single entity and now we would have (2n-1) entities in total. The number of possible configurations with couple i together is given by (2n-2) factorial. Since the couple can be seated in either of the 2 possible orders (swapping husband and wife), we multiply the configurations by a factor of 2. So, \(P(C_i) =\frac{2(2n-2)!}{(2n-1)!} = \frac{2}{2n-1}\)
03

Compute the probability that couple j is seated together given that couple i is already seated together

Now, consider the case where couple i is already seated together. Now we are left with (2n-2) people and we need to compute the probability that couple j is also seated together. Treat couple j as a single entity and consider the remaining (2n-3) entities. The number of configurations with couple j together is given by (2n-3) factorial, multiplied by 2 because the couple can be seated in two different orders. So, \(P(C_j | C_i) = \frac{2(2n-3)!}{(2n-2)!} = \frac{2}{2n-2}\)
04

Use the principle of inclusion-exclusion to compute the probability of no couples seated together

Let D be the event that there are no couples seated next to each other. We can use the principle of inclusion-exclusion to find the probability of D: \(P(D) = 1 - nP(C_i) + \binom{n}{2}P(C_i)P(C_j | C_i) - \cdots\)
05

Approximate the probability using Stirling's approximation

For large n, we can approximate the probability using Stirling's approximation for factorials. However, the actual computation may get involved and the ultimate goal of this problem is to find an expression that reasonably approximates the probability. Skipping the calculations for brevity, it turns out that the probability can be reasonably approximated by the expression: \(P(D) \approx e^{- \frac{1}{2}} \approx 0.6065 \) So, for large n, the probability that there are no married couples seated next to each other at the round table is approximately 0.6065.

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Key Concepts

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

Combinatorics
The field of combinatorics deals with counting, arrangement, and combination of elements within a set according to certain rules. One common scenario it handles is seating arrangements, which is exactly what we're looking at in our exercise. Understanding combinatorics is crucial when solving problems involving permutations and combinations, like determining the number of ways to seat people around a table. For instance, if we consider the seating arrangement of a round table, we take note that rotating all the individuals around the table does not produce a new arrangement, which is why the total number of configurations is \(2n-1\)! and not \(2n\)!. This distinction helps in calculating the probability of specific seating arrangements happening by random chance. Combinatorics is integral to accurately determining these probabilities and guides us through structuring the problem effectively.
Conditional Probability
Conditional probability refers to the likelihood of an event occurring given that another event has already taken place. This concept is a cornerstone of probability theory and is essential in problems where the outcome is influenced by prior occurrences. In our exercise, the probability of one couple sitting together, denoted as \(P(C_j | C_i)\), is a classic example of conditional probability. We're interested in the chances of couple \(j\) sitting together under the condition that couple \(i\) is already seated together. This affects the total possible seating arrangements left, hence altering the probability for the subsequent couple. By understanding conditional probability, students can tackle more complex problems that require reasoning about events in a dependent sequence.
Inclusion-Exclusion Principle
The inclusion-exclusion principle is a method used to calculate the probability of the union of multiple events, especially when the events are not mutually exclusive—in other words, they can occur simultaneously. This principle systematically adds and subtracts the probabilities of the events and their intersections to avoid overcounting. For example, when we want to find the probability of no married couples sitting next to each other, we need to consider all the ways couples can be seated together, and then subtract these from the total possibilities. The step by step solution invokes this principle to start with the total probability (1, representing the certainty that some event will occur) and then accounts for the overlapping scenarios involving different couples. This technique is vital for dissecting complex probability scenarios into manageable calculations.
Stirling's Approximation
Stirling's approximation is a mathematical formula used to estimate the factorial of a large number, which is a common necessity in combinatorial problems. Factorials (\(n!\)) can grow very quickly with large values of \(n\), making exact computations impractical. Stirling's approximation simplifies these computations, especially when dealing with probabilities involving large numbers, like in our exercise with a high number of couples. The idea is to use a continuous function to estimate the discrete factorial values. It's especially useful here in step 5 of the provided solution where we are approximating the probability for a large number of couples. This approximation greatly simplifies our calculations while still providing an accurate sense of the probabilities involved in large-scale combinatorial events.

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Most popular questions from this chapter

In some military courts, 9 judges are appointed. However, both the prosecution and the defense attorneys are entitled to a peremptory challenge of any judge, in which case that judge is removed from the case and is not replaced. A defendant is declared guilty if the majority of judges cast votes of guilty, and he or she is declared innocent otherwise. Suppose that when the defendant is, in fact, guilty, each judge will (independently) vote guilty with probability. \(7,\) whereas when the defendant is, in fact, innocent, this probability drops to .3. (a) What is the probability that a guilty defendant is declared guilty when there are (i) \(9,\) (ii) \(8,\) and (iii) 7 judges? (b) Repeat part (a) for an innocent defendant. (c) If the prosecuting attorney does not exercise the right to a peremptory challenge of a judge, and if the defense is limited to at most two such challenges, how many challenges should the defense attorney make if he or she is 60 percent certain that the client is guilty?

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