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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
\(\frac{1}{30}\) for Jim and Paula on far left, \(\frac{1}{3}\) for Jim and Paula together, \(\frac{5}{6}\) for at least one wife next to her husband.

Step by step solution

01

Total Arrangements of Couples

First, calculate the total number of ways to arrange the six individuals (three married couples or six people) in six seats. Since there are 6 people and each can occupy one seat, there are 6 factorial (\(6!\)) possible arrangements. This is equal to \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\).
02

Probability for Jim and Paula on Far Left

To find the probability that Jim and Paula sit in the two far-left seats, consider fixing Jim in the first seat and Paula in the second. Then, the remaining four people can occupy the remaining four seats in \(4!\) ways. This is \(4 \times 3 \times 2 \times 1 = 24\). Therefore, the probability is the number of favorable outcomes over the total outcomes, or \(\frac{24}{720} = \frac{1}{30}\).
03

Probability for Jim and Paula Sitting Together

For Jim and Paula to sit next to each other in any of the six seats, treat them as a single unit or 'block'. This 'block' can occupy 5 positions along the row (positions 1&2, 2&3, 3&4, 4&5, or 5&6). Within the block, they can switch positions with one another, giving 2 arrangements for each position of the block. The other four people can be arranged in the remaining 4 positions. Thus, the total arrangements are \(5 \times 2 \times 4! = 240\). The probability is \(\frac{240}{720} = \frac{1}{3}\).
04

Probability for At Least One Wife Sitting Next to Her Husband

Consider the complementary event: no wife sits next to her husband. Calculate the ways for three couples to sit without a wife next to her respective husband. The formula derived from inclusion-exclusion principle results in 120 such arrangements (this can be calculated or verified through perspective of considering restrictions). So, the probability of no wife next to her husband is \(\frac{120}{720} = \frac{1}{6}\). Therefore, the probability of at least one wife sitting next to her husband is the complement, which is \(1 - \frac{1}{6} = \frac{5}{6}\).

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Key Concepts

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

Combinatorial Probability
Combinatorial probability involves finding the likelihood of various outcomes based on the number of ways events can occur. In problems where different arrangements of items or people are possible, combinatorial methods are used to determine probabilities.
In our exercise involving theatre seating, we're dealing with permutations because the order in which people sit matters. Six theater-goers can be arranged in the six seats in a certain number of ways, which is calculated by finding the factorial of six (\(6!\)). This gives us 720 possible arrangements.
  • Permutations account for the sequence of elements where each order is unique.
  • A fundamental aspect is understanding factorial notation, where \( n! \) represents the product of all positive integers up to \( n \).
By counting these permutations and limiting them with conditions, like Jim and Paula sitting in specific seats, we derive combinatorial probabilities. Understanding these concept-based calculations helps solve arrangement problems easily.
Probability Calculations
Probability calculations help measure the chance of certain events occurring. We calculate probabilities by dividing the number of favorable outcomes by the total number of possible outcomes.
In the problem example, to find the probability of Jim and Paula sitting in the two far-left seats, we calculated how fixing their positions led to a certain number of arrangements for the remaining four people, which was \(4!\). Thus, only 24 favorable outcomes existed, so the probability was calculated as \(\frac{24}{720}\) or \(\frac{1}{30}\).
  • Each probability calculation begins by identifying all possible outcomes.
  • Understanding 'favorable outcomes' involves focusing only on situations that satisfy the problem's conditions.
The power of probability calculations lies in their ability to simplify complex events into manageable, numerical expressions.
Arrangement Problems in Probability
Arrangement problems in probability focus on the sequences in which objects or individuals can be organized to satisfy specific conditions. These problems often involve permutations where order matters, just as in our seating exercise.
One common scenario is to find the probability that individuals are seated in a way that adheres to given constraints, like Jim and Paula sitting together, which was solved by treating them as 'blocks'. Such 'block' formation adapts the problem into simpler components, allowing us to calculate using modified permutations of the remaining elements. The total arrangement here came to \(5 \times 2 \times 4!\).
  • Understanding this involves recognizing patterns and breaking larger problems into smaller, manageable tasks.
  • These arrangements often utilize the principle of treating a set of items as a single unit where advantageous.
Solving arrangement problems helps in visualizing the structure of possible outcomes quickly and efficiently, further easing complex probability questions.

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Most popular questions from this chapter

At a large university, in the never-ending quest for a satisfactory textbook, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used the text by Professor Mean; during the winter quarter, 300 students used the text by Professor Median; and during the spring quarter, 200 students used the text by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with Mean's book, 150 were satisfied with Median's book, and 160 were satisfied with Mode's book. If a student who took statistics during one of these quarters is selected at random and admits to having been satisfied with the text, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? [Hint: Draw a tree- diagram or use Bayes' theorem.]

An academic department with five faculty members narrowed its choice for department head to either candidate \(A\) or candidate \(B\). Each member then voted on a slip of paper for one of the candidates. Suppose there are actually three votes for \(A\) and two for \(B\). If the slips are selected for tallying in random order, what is the probability that \(A\) remains ahead of \(B\) throughout the vote count (for example, this event occurs if the selected ordering is \(A A B A B\), but not for \(A B B A A)\) ?

A particular iPod playlist contains 100 songs, of which 10 are by the Beatles. Suppose the shuffle feature is used to play the songs in random order (the randomness of the shuffling process is investigated in "Does Your iPod Really Play Favorites?" (The Amer. Statistician, 2009: 263 - 268)). What is the probability that the first Beatles song heard is the fifth song played?

Allan and Beth currently have \(\$ 2\) and \(\$ 3\), respectively. A fair coin is tossed. If the result of the toss is \(\mathrm{H}\), Allan wins \(\$ 1\) from Beth, whereas if the coin toss results in \(\mathrm{T}\), then Beth wins \(\$ 1\) from Allan. This process is then repeated, with a coin toss followed by the exchange of \(\$ 1\), until one of the two players goes broke (one of the two gamblers is ruined). We wish to determine \(a_{2}=P\) (Allan is the winner \(\mid\) he starts with \(\$ 2\) ) To do so, let's also consider \(a_{i}=P\) (Allan wins | he starts with \(\$ i\) ) for \(i=0,1,3,4\), and 5 . a. What are the values of \(a_{0}\) and \(a_{5}\) ? b. Use the law of total probability to obtain an equation relating \(a_{2}\) to \(a_{1}\) and \(a_{3}\). [Hint: Condition on the result of the first coin toss, realizing that if it is a \(\mathrm{H}\), then from that point Allan starts with \$3.] c. Using the logic described in (b), develop a system of equations relating \(a_{i}(i=1,2,3,4)\) to \(a_{i-1}\) and \(a_{i+1}\). Then solve these equations. [Hint: Write each equation so that \(a_{i}-a_{i-1}\) is on the left hand side. Then use the result of the first equation to express each other \(a_{i}-a_{i-1}\) as a function of \(a_{1}\), and add together all four of these expressions \((i=2,3,4,5)\).] d. Generalize the result to the situation in which Allan's initial fortune is \(\$ a\) and Beth's is \(\$ b\). Note: The solution is a bit more complicated if \(p=P(\) Allan wins \(\$ 1) \neq .5 .\)

A boiler has five identical relief valves. The probability that any particular valve will open on demand is 95 . Assuming independent operation of the valves, calculate \(P\) (at least one valve opens) and \(P\) (at least one valve fails to open).

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