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Chapter 2: Problem 84

A small manufacturing company will start operating a night shift. There are 20 machinists employed by the company. a. If a night crew consists of 3 machinists, how many different crews are possible? b. If the machinists are ranked \(1,2, \ldots, 20\) in order of competence, how many of these crews would not have the best machinist? c. How many of the crews would have at least 1 of the 10 best machinists? d. If one of these crews is selected at random to work on a particular night, what is the probability that the best machinist will not work that night?

Short Answer

Expert verified
a. 1140 crews b. 969 crews c. 1020 crews d. 0.85

Step by step solution

01

Calculate Total Possible Crews

First, we need to find how many different ways we can select 3 machinists out of 20. This is a combination problem, and the formula for combinations is given by \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \(n\) is the total number of items to choose from, and \(k\) is the number of items to choose. Here, \(n = 20\) and \(k = 3\). Thus, the total number of crews is \( \binom{20}{3} = \frac{20!}{3!(20-3)!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \).
02

Calculate Crews Without the Best Machinist

We now need to determine how many crews would not include the best machinist (machinist 1). This means choosing 3 machinists from the remaining 19 (since we exclude machinist 1). Thus, using combinations again, we calculate \( \binom{19}{3} = \frac{19 \times 18 \times 17}{3 \times 2 \times 1} = 969 \).
03

Calculate Crews with At Least 1 of the Top 10 Machinists

To find the number of crews with at least 1 of the top 10 machinists, we first calculate the total number of crews with no top 10 machinists. This means selecting all 3 machinists from the bottom 10. Thus, \( \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \). Consequently, the number of crews with at least 1 from the top 10 is given by the total crews minus crews from the bottom 10, which is \(1140 - 120 = 1020 \).
04

Calculate Probability the Best Machinist is Not Working

Lastly, we want to find the probability that the best machinist is not working on a randomly selected crew. We've already found that there are 969 crews without the best machinist out of a total of 1140. Thus, the probability is \( \frac{969}{1140} \approx 0.85 \).

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Key Concepts

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

Combinations
Combinations are a fundamental concept in combinatorics, used to determine how many ways you can select items from a larger set without regard to order. The combination formula is given by \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items and \( k \) is the number of items to choose.

For example, in the context of a manufacturing company with 20 machinists, choosing a 3-person night shift team involves calculating combinations. Here, \( n = 20 \) and \( k = 3 \).

Thus, we calculate \( \binom{20}{3} = \frac{20!}{3!(20-3)!} = 1140 \).

This means there are 1,140 different possible combinations of machinists to form a night crew of 3. This concept helps solve problems where the arrangement does not matter, emphasizing selection over order.
Probability
Probability measures the likelihood of an event occurring and is calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this context, we examine the probability of forming a specific crew from machinists or certain conditions being met, such as not including the best machinist.

When calculating the likelihood that the best machinist is not included in a chosen crew, we leverage the idea that 969 crews are without this individual out of 1140 possible crews. The probability, therefore, is \( \frac{969}{1140} \approx 0.85 \).

This suggests there's an 85% chance a randomly selected crew at night won't include the best machinist, illustrating how probability provides insight into scenario expectations in varied contexts.
Permutations
Permutations are also part of combinatorics, but unlike combinations, they consider the arrangement order of items. When the sequence in which items are arranged is important, permutations come into play. The formula for calculating permutations is \( P(n, k) = \frac{n!}{(n-k)!} \).

In contrast to combinations, permutations ensure every different order counts as unique. When considering machinists, if their order within a crew mattered, you鈥檇 use permutations rather than combinations.

This distinction is crucial when solving problems where sequence is critical, emphasizing why understanding the right application between these two is necessary when dealing with different scenarios in mathematical problems.
Mathematical Statistics
Mathematical statistics plays a role in interpreting data through analysis and inference. While combinatorics focuses on counting outcomes and probability evaluates expectations, mathematical statistics takes it further by analyzing patterns and trends within data.

When examining crews of machinists through the lens of statistics, one could draw inferences about the reliability and effectiveness of chosen strategies or forecasts future patterns.

By understanding how often certain compositions of machinists lead to successful shifts based on gathered data, managers can make informed decisions. Mathematical statistics helps move beyond raw probability and combinations' counts towards a deeper understanding of data-driven strategies and operations in various fields.

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Most popular questions from this chapter

Ann and Bev have each applied for several jobs at a local university. Let \(A\) be the event that Ann is hired and let \(B\) be the event that Bev is hired. Express in terms of \(A\) and \(B\) the events a. Ann is hired but not Bev. b. At least one of them is hired. c. Exactly one of them is hired.

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?

If \(A\) and \(B\) are independent events, show that \(A^{\prime}\) and \(B\) are also independent. [Hint: First establish a relationship among \(P\left(A^{\prime} \cap B\right), P(B)\), and \(P(A \cap B)\).]

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for indepth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). a. How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift? b. What is the probability that all 6 selected workers will be from the same shift? c. What is the probability that at least two different shifts will be represented among the selected workers? d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

A particular airline has 10 a.m. flights from Chicago to New York, Atlanta, and Los Angeles. Let A denote the event that the New York flight is full and define events \(B\) and \(C\) analogously for the other two flights. Suppose \(P(A)=.6, P(B)=.5\), \(P(C)=.4\) and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?
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