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Four engineers, \(A, B, C\), and \(D\), have been scheduled for job interviews at \(10 \mathrm{~A} . \mathrm{M}\). on Friday, January 13 , at Random Sampling, Inc. The personnel manager has scheduled the four for interview rooms \(1,2,3\), and 4 , respectively. However, the manager's secretary does not know this, so assigns them to the four rooms in a completely random fashion (what else!). What is the probability that a. All four end up in the correct rooms? b. None of the four ends up in the correct room?

Short Answer

Expert verified
a. \(\frac{1}{24}\); b. \(\frac{3}{8}\).

Step by step solution

01

Calculate Total Permutations

To find the total number of ways to arrange four engineers in four rooms, we calculate the permutations of 4 items: \(4!\). \[4! = 4 \times 3 \times 2 \times 1 = 24\]So, there are 24 possible ways to assign the engineers to the rooms.
02

Calculate Probability of All Correct

For all engineers to end up in the correct rooms (rooms 1, 2, 3, and 4 for A, B, C, and D respectively), there is exactly 1 correct arrangement. Therefore, the probability that all four engineers end up in the correct rooms is:\[P(\text{All correct}) = \frac{1}{24}\]
03

Calculate Derangements for None Correct

A derangement is a permutation where no element appears in its original position. For four rooms, we use the formula for the number of derangements \(!n\):\[!4 = 4! \sum_{k=0}^{4} \frac{(-1)^k}{k!}\]Calculating each term gives:- For \(k=0\): \(\frac{(-1)^0}{0!} = 1\)- For \(k=1\): \(\frac{(-1)^1}{1!} = -1\)- For \(k=2\): \(\frac{(-1)^2}{2!} = \frac{1}{2}\)- For \(k=3\): \(\frac{(-1)^3}{3!} = -\frac{1}{6}\)- For \(k=4\): \(\frac{(-1)^4}{4!} = \frac{1}{24}\)Adding these gives:\[1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} = \frac{9}{24}\]Thus, there are \(!4 = 9\) derangements.
04

Calculate Probability of None Correct

Using the number of derangements calculated in the previous step, the probability that none of the engineers ends up in their correct rooms is:\[P(\text{None correct}) = \frac{!4}{4!} = \frac{9}{24}\]Simplified:\[P(\text{None correct}) = \frac{3}{8}\]

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Key Concepts

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

Permutations
Permutations are an essential concept in probability theory and combinatorics. They refer to the different ways in which a set of items can be organized or arranged. When we speak of permutations, we are usually concerned with the order of arrangement of items. For instance, when you have four engineers and four rooms, there are multiple ways to assign these engineers to the rooms.

To calculate the number of permutations, we use the factorial notation, denoted by an exclamation mark. The formula for finding the number of permutations of a set of "n" items is given by:
  • \[ n! = n \times (n-1) \times (n-2) \times \, ... \, \times 1 \]
For our specific problem, where four engineers need to be assigned to four rooms, the total permutations (\(4!\)) equals 24. This means there are 24 possible ways of arranging the engineers in the rooms, considering every permutation puts a unique order to the engineers and the rooms they are assigned.
Derangements
Derangements are a special kind of permutation where none of the items appear in their initial positions. This concept becomes particularly interesting when items or people need to be mismatched entirely, such as in the case of arranging engineers where none should be in their initially designated room.

The formula for finding the number of derangements of "n" items is:
  • \[ !n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} \]
Let's break this down for four engineers. We calculate each component of the series to find the total number of derangements, \(!4\), which amounts to 9. Here:
  • For \(k=0\) yields 1,
  • \(k=1\) contributes a \(-1\),
  • \(k=2\) gives us \(\frac{1}{2}\),
  • \(k=3\) is \(-\frac{1}{6}\),
  • and \(k=4\) gives \(\frac{1}{24}\).
Adding them yields \(!4 = 9\), demonstrating there are 9 ways the engineers can be completely mismatched to their original rooms, making none end up in their initially assigned rooms.
Combinatorics
Combinatorics is the study of counting, arrangement, and combination of elements within a set. It encompasses concepts like permutations and derangements as part of its wider approach to analyzing various ways to count and organize items.

In the given exercise, combinatorics helps determine how four engineers can be assigned to four different rooms. Through this field, one learns about permutations, which provides the baseline number of possible arrangements (24 here), and about derangements, which further delves into specific conditions that disrupt typical arrangements by ensuring none of the engineers go to their pre-designated place.

Methods like factorial calculations and inclusion-exclusion principles, which are central in this discipline, allow us to solve larger, more complex problems involving restricted arrangements and combinations. By understanding and using combinatorial principles, one can easily calculate and reason about probability in structured and systematic ways.

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Most popular questions from this chapter

For customers purchasing a refrigerator at a certain appliance store, let \(A\) be the event that the refrigerator was manufactured in the U.S., \(B\) be the event that the refrigerator had an icemaker, and \(C\) be the event that the customer purchased an extended warranty. Relevant probabilities are $$ \begin{aligned} &P(A)=.75 \quad P(B \mid A)=.9 \quad P\left(B \mid A^{\prime}\right)=.8 \\ &P(C \mid A \cap B)=.8 \quad P\left(C \mid A \cap B^{\prime}\right)=.6 \\ &P\left(C \mid A^{\prime} \cap B\right)=.7 \quad P\left(C \mid A^{\prime} \cap B^{\prime}\right)=.3 \end{aligned} $$ a. Construct a tree diagram consisting of first-, second-, and third- generation branches, and place an event label and appropriate probability next to each branch. b. Compute \(P(A \cap B \cap C)\). c. Compute \(P(B \cap C)\). d. Compute \(P(C)\). e. Compute \(P(A \mid B \cap C)\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased.

A box contains six \(40-\mathrm{W}\) bulbs, five \(60-\mathrm{W}\) bulbs, and four \(75-\mathrm{W}\) bulbs. If bulbs are selected one by one in random order, what is the probability that at least two bulbs must be selected to obtain one that is rated \(75 \mathrm{~W}\) ?

Each contestant on a quiz show is asked to specify one of six possible categories from which questions will be asked. Suppose \(P(\) contestant requests category \(i)=\frac{1}{6}\) and successive contestants choose their categories independently of one another. If there are three contestants on each show and all three contestants on a particular show select different categories, what is the probability that exactly one has selected category 1 ?

Show that \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\left(\begin{array}{c}n \\ n-k\end{array}\right)\). Give an interpretation involving subsets.

A family consisting of three persons- \(A, B\), and \(C-\) goes to a medical clinic that always has a doctor at each of stations 1 , 2 , and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. One outcome is \((1,2,1)\) for \(A\) to station \(1, B\) to station 2 , and \(C\) to station 1 . a. List the 27 outcomes in the sample space. b. List all outcomes in the event that all three members go to the same station. c. List all outcomes in the event that all members go to different stations. d. List all outcomes in the event that no one goes to station 2 .

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