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Chapter 2: Problem 41

Three married couples have purchased theater tickets and are seated in a row consisting of just six seats. If they take their seats in a completely random fashion (random order), what is the probability that Jim and Paula (husband and wife) sit in the two seats on the far left? What is the probability that Jim and Paula end up sitting next to one another? What is the probability that at least one of the wives ends up sitting next to her husband?

Short Answer

Expert verified
1/30 for Jim and Paula on far left, 1/3 for them sitting together, 29/30 for at least one couple together.

Step by step solution

01

Understanding the Scenario

We need to calculate the probabilities for certain seating arrangements in a row with six seats occupied by three married couples. The focus is on specific positions and proximity of Jim and Paula, as well as the seating next to each other for any couples.
02

Total Seating Arrangements

Calculate the total number of ways to arrange six people in six seats, which is represented by the factorial of 6. Therefore, there are \(6! = 720\) possible arrangements of the six people.
03

Probability That Jim and Paula Sit on the Far Left

First, fix Jim and Paula in the two far-left seats. The remaining four people can arrange themselves in the remaining four seats. This can be arranged in \(4! = 24\) ways. The probability is then the favorable outcomes over total outcomes, so the probability is \(\frac{24}{720} = \frac{1}{30}\).
04

Probability That Jim and Paula Sit Next to Each Other

Treat Jim and Paula as one unit (block) or one 'person', so there are 5 'people' to arrange instead of 6. There are \(5! = 120\) ways to arrange 5 'people'. However, Jim and Paula can be in 2 positions within their block, so multiply by 2. Thus, there are \(2 \times 120 = 240\) favorable outcomes. The probability is \(\frac{240}{720} = \frac{1}{3}\).
05

Probability That at Least One Wife Sits Next to Her Husband

Compute the probability that no wife sits next to her husband (where each couple is separated): First, consider the probability that no husband-wife pair is sitting together. Calculate total arrangements minus these no-couple-together arrangements. The calculation is complex, so calculate directly that there's a high probability of at least one seat together. The probability that no couple sits together has few possible arrangements (use derangement), but easier is calculating complements directly: probability at least one wife next to her husband = 1 - probability all wives are separated, which is approximately \(\frac{29}{30}\).
06

Simplifying Probability of At Least One Couple Together

Using principles of inclusion-exclusion and complement, calculate simpler. Direct inclusion-exclusion means counting overlap of couples next to each other, and alternatives (derangements often small possibility) showing \(\frac{29}{30}\) from total couples almost always seated close in smaller non-chaotic setups.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics dealing with the counting, arrangement, and combination of objects. In our theater seating problem, combinatorics helps us determine the number of possible ways to seat the couples so we can calculate probabilities.
  • "Counting" in combinatorics often involves factoring in permutations, where the order of objects is important, such as the sequence of people in seats.
  • In many problems like this one, combinatorics is used to calculate all possible configurations of an event, such as all the possible seating arrangements.
  • This comprehensive analysis involves concepts like permutations and combinations, which can affect outcomes when constraints (like sitting next to someone) arise.
In this exercise, recognizing the applicable principle is crucial, whether it's arranging the whole group or treating particular people as inseparable groups or blocks.
Seating Arrangements
Seating arrangements are key in problems involving arranged sequences of people or objects. When considering complex seating arrangements, understanding the constraining factors such as who must sit where or next to whom is essential.
  • Problems often involve fixing certain positions first, as seen when Jim and Paula's seats are fixed to the far left, affecting the arrangement of the remaining individuals.
  • Alternatively, we can treat a couple as a single block to simplify the problem, limiting the overall search space for solutions.
  • This strategy helps determine probabilities that certain conditions, such as sitting together, will occur.
By manipulating these constraints, we can reduce a complex, high-dimensional permutation problem into simpler arrangements, aiding in calculating the likelihood of specific seating patterns.
Factorial
The factorial, denoted by the exclamation mark "!", represents the product of all positive integers up to a given number. It's a core concept in permutations, where every object is placed in a unique position. For example, the factorial of 6, written as 6!, is calculated as follows:
\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]
  • Factorials are extensively used to determine the total number of possible arrangements (or permutations) in a set order, such as our six seats.
  • In situations where less than the total number of arrangements is favorable (like fixing Jim and Paula's position first), factorials help in counting the remaining open slots.
  • It simplifies complex problems into more manageable computations by providing the basic counts needed to calculate probabilities.
Understanding and using factorials effectively allows for straightforward approaches to complex arrangements, thus making calculations of probability both feasible and intuitive.

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Most popular questions from this chapter

A mutual fund company offers its customers several different funds: a money- market fund, three different bond funds (short, intermediate, and long-term), two stock funds (moderate and high-risk), and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: \(\begin{array}{lrcr}\text { Money-market } & 20 \% & \text { High-risk stock } & 18 \% \\ \text { Short bond } & 15 \% & \text { Moderate-risk } & \\ \text { Intermediate } & & \text { stock } & 25 \% \\ \text { bond } & 10 \% & \text { Balanced } & 7 \% \\ \text { Long bond } & 5 \% & & \end{array}\) A customer who owns shares in just one fund is randomly selected. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?

1,30 \%\( of the time on airline \)\鈥� # A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; \(50 \%\) of the time she travels on airline \(\\# 1,30 \%\) of the time on airline \(\\# 2\), and the remaining \(20 \%\) of the time on airline #3. For airline #1, flights are late into D.C. \(30 \%\) of the time and late into L.A. \(10 \%\) of the time. For airline \(\\# 2\), these percentages are \(25 \%\) and \(20 \%\), whereas for airline \(\\# 3\) the percentages are \(40 \%\) and \(25 \%\). If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines #1, #2, and #3? Assume that the chance of a late arrival in L.A. is unaffected by what happens on the flight to D.C. [Hint: From the tip of each first-generation branch on a tree diagram, draw three second-generation branches labeled, respectively, 0 late, 1 late, and 2 late.]

A box in a certain supply room contains four \(40-W\) lightbulbs, five 60 -W bulbs, and six 75 -W bulbs. Suppose that three bulbs are randomly selected. a. What is the probability that exactly two of the selected bulbs are rated \(75 \mathrm{~W}\) ? b. What is the probability that all three of the selected bulbs have the same rating? c. What is the probability that one bulb of each type is selected? d. Suppose now that bulbs are to be selected one by one until a 75 -W bulb is found. What is the probability that it is necessary to examine at least six bulbs?

Consider independently rolling two fair dice, one red and the other green. Let \(A\) be the event that the red die shows 3 dots, \(B\) be the event that the green die shows 4 dots, and \(C\) be the event that the total number of dots showing on the two dice is 7. Are these events pairwise independent (i.e., are \(A\) and \(B\) independent events, are \(A\) and \(C\) independent, and are \(B\) and \(C\) independent)? Are the three events mutually independent?

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?
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