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Chapter 4: Problem 112

Four union men, two from a minority group, are assigned to four distinctly different one-man jobs, which can be ranked in order of desirability. a. Define the experiment. b. List the simple events in \(S\). c. If the assignment to the jobs is unbiased-that is, if any one ordering of assignments is as probable as any other-what is the probability that the two men from the minority group are assigned to the least desirable jobs?

Short Answer

Expert verified
Answer: The probability is \(\frac{1}{4}\) or 25%.

Step by step solution

01

Define the experiment

The experiment consists of assigning four union men, among which two are from a minority group, to four distinctly different one-man jobs that can be ranked in order of desirability.
02

List the simple events in \(S\)

Let's represent the union men as A, B (minority group), C, and D, and the jobs as J1 (most desirable), J2, J3, and J4 (least desirable). The simple events in the sample space \(S\) can be listed as the possible ways the four men can be assigned to four jobs: 1. ABCD 2. ABDC 3. ACBD 4. ACDB 5. ADBC 6. ADCB 7. BACD 8. BADC 9. BCAD 10. BCDA 11. BDAC 12. BDCA 13. CABD 14. CADB 15. CBAD 16. CBDA 17. CDAB 18. CDBA 19. DABC 20. DACB 21. DBAC 22. DBCA 23. DCAB 24. DCBA In total, there are 4! (24) simple events in the sample space \(S\).
03

Determine the probability of assigning the two men from the minority group to the least desirable jobs

In the unbiased assignment scenario, the probability of assigning the two men from the minority group to the least desirable jobs (J3 and J4) can be calculated by counting the number of times this happens and dividing it by the total number of possible assignments in the sample space. The favorable cases where the two men from the minority group are assigned to J3 and J4 are as follows: 1. ACBD 2. ACDB 3. CADB 4. CBDA 5. CDAB 6. CDBA There are 6 favorable events, so the probability of the two men from the minority group getting the least desirable jobs is given by: $$P(\text{two men from minority group assigned to J3 and J4}) = \frac{\text{Number of favorable events}}{\text{Total number of events}} = \frac{6}{24} = \frac{1}{4}$$ Hence, the probability that the two men from the minority group are assigned to the least desirable jobs is \(\frac{1}{4}\) or 25%.

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Key Concepts

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

Sample Space
In probability theory, the sample space of an experiment refers to the set of all possible outcomes of that experiment. It is important to identify the elements in the sample space correctly, as they form the foundation of probability calculations.
For our exercise, we are dealing with the assignment of four union men to four different jobs. Let's name the men A, B, C, and D, where A and B are from a minority group, and the jobs are J1 (most desirable) to J4 (least desirable).
  • The total number of possible assignments of these men to jobs is represented by the permutations of the four individuals, which is 4! (factorial), equating to 24 different outcomes.
  • Each different ordering of men to jobs, such as ABCD or BACD, is a simple event in the sample space.
Thus, the sample space consists of all these unique permutations, making it essential to list and understand to solve any probability problem regarding this scenario.
Probability Calculation
Probability is all about determining how likely an event is to occur. In our exercise, it's about finding the likelihood that the two minority union men, A and B, end up in the least desirable jobs, J3 and J4.
Once the sample space is identified, calculating probability often means counting the favorable outcomes for the event and dividing by the total number of possible outcomes.
  • There are 6 permutations out of the total 24 where A and B are assigned to J3 and J4. These are ACBD, ACDB, CADB, CBDA, CDAB, and CDBA.
  • The probability is calculated as the ratio of favorable outcomes (6) to total outcomes (24): \[ P = \frac{\text{Number of favorable events}}{\text{Total number of events}} = \frac{6}{24} = \frac{1}{4} \]
This means there is a 25% chance that A and B will find themselves in the least desirable positions. Probability calculations like this are foundational in evaluating how likely certain outcomes will occur in defined scenarios.
Combinatorics
Combinatorics is a branch of mathematics that studies how to count, arrange, and structure items. In the context of our problem, it helps us understand the different ways individuals can be assigned to positions.Permutations and combinations are the two primary operations in combinatorics used to solve such problems.
  • Permutations focus on the arrangement of items, where order matters, which was applied here to calculate the total sample space by arranging four men for four positions.
  • The factorial notation, such as \(4!\), means multiplying a stream of descending natural numbers to find out how many different sequences can be achieved with all four positions filled.
Using combinatorial logic allows us to break down scenarios into easily predictable sets, identifying all possible outcomes. This systematic approach ensures that all options are considered, which is crucial for accurate probability determination.

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Most popular questions from this chapter

An experiment can result in one of five equally likely simple events, \(E_{1}, E_{2}, \ldots, E_{5} .\) Events \(A, B,\) and \(C\) are defined as follows: $$A: E_{1}, E_{3} \quad P(A)=.4$$ $$B: E_{1}, E_{2}, E_{4}, E_{5} \quad P(B)=.8$$ $$C: E_{3}, E_{4} \quad P(C)=.4$$. Find the probabilities associated with these compound events by listing the simple events in each. a. \(A^{c}\) b. \(A \cap B\) c. \(B \cap C\) d. \(A \cup B\) e. \(B \mid C\) f. \(A \mid B\) g. \(A \cup B \cup C\) h. \((A \cap B)^{c}\)

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