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Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to order)? b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect? c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect?

Short Answer

Expert verified
a. 53,130 ways b. 14,535 ways c. Probability ≈ 0.6566

Step by step solution

01

Calculate Total Ways to Select 5 Keyboards

We need to find the total number of ways to select 5 keyboards from the 25 available keyboards. This is a combination problem because the order does not matter. The formula to calculate combinations is given by:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \(n\) is the total number of keyboards, and \(k\) is the number of keyboards to select. Substituting the values:\[ \binom{25}{5} = \frac{25!}{5!(25-5)!} = \frac{25!}{5! \times 20!} = 53130 \]
02

Calculate Ways to Select 2 Electrical and 3 Mechanical Keyboards

To find how many ways to select 5 keyboards with exactly 2 electrical defects, we find combinations for selecting 2 electrical keyboards from 6 and 3 mechanical keyboards from 19. Using the combination formula:\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = 15 \]\[ \binom{19}{3} = \frac{19!}{3!(19-3)!} = 969 \]Now, multiply these two results to get the total ways:\[ 15 \times 969 = 14535 \]
03

Calculate Probability of At Least 4 Mechanical Defects

We need to find the probability that at least 4 of the 5 selected keyboards have a mechanical defect. This means calculating the cases where there are exactly 4 or all 5 mechanical defects.**Case 1: Exactly 4 Mechanical Defects:**Select 4 mechanical from 19:\[ \binom{19}{4} = 3876 \]Select 1 electrical from 6:\[ \binom{6}{1} = 6 \]Total ways for 4 mechanicals:\[ 3876 \times 6 = 23256 \]**Case 2: Exactly 5 Mechanical Defects:**Select 5 mechanical from 19:\[ \binom{19}{5} = 11628 \]**Add Both Cases:**Total favorable ways:\[ 23256 + 11628 = 34884 \]**Compute Probability:**\[ \text{Probability} = \frac{34884}{53130} \approx 0.6566 \]
04

Final Answer Compilation

a. The number of ways to select 5 sample keyboards is 53,130. b. The number of ways to select a sample with exactly 2 keyboards having electrical defects is 14,535. c. The probability that at least 4 keyboards will have a mechanical defect is approximately 0.6566.

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Key Concepts

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

Probability
Probability is the branch of mathematics that deals with the likelihood of an event occurring. In simpler terms, it is the measure of how likely something is to happen.
The probability of an event is calculated by taking the number of favorable outcomes and dividing it by the total number of possible outcomes. An important fact to remember is that the probability value will always fall between 0 and 1. A probability of 0 means the event will not happen, while a probability of 1 means the event is certain to happen.
In the keyboard selection problem, part of the question asks for the probability that at least 4 out of 5 randomly selected keyboards have mechanical defects. To solve this, we calculated the total number of possible selections and then counted the selections where at least 4 keyboards have a mechanical defect. By dividing the favorable outcomes by the total outcomes, we find the probability is approximately 0.6566.
Combinations
Combinations refer to a selection of items from a larger set where the order does not matter. This is different from permutations, where order does matter.
The basic formula used to find the number of combinations is: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Here, \( n \) represents the total number of items, \( k \) represents the number of items to select, and \( ! \) is the factorial function, meaning the product of all positive integers up to that number.
In our context, we used combinations to determine how many ways we can select 5 keyboards from a total of 25. Since the order in which we select the keyboards isn't important, this problem is solved using combinations rather than permutations. Further, we applied it to calculate different scenarios, like selecting keyboards with certain defects. Understanding combinations can be very helpful in situations where the arrangement of selected items doesn't matter.
Defects Analysis
Defects analysis involves the examination of faulty items to determine the nature and frequency of defects. In the context of the keyboard problem, we classify defects into electrical and mechanical.
The exercise specifically looks at how to analyze combinations of defects within a randomly selected sample of keyboards. For example, we calculated the number of ways to select a sample of keyboards with strictly 2 electrical defects out of a total of 6 and 3 mechanical defects out of a total of 19.
These analyses are vital in understanding product reliability and helping manufacturers address and improve production processes. It’s an essential aspect for quality assurance as it provides statistics that can be used to prioritize issues based on how frequently they occur and how severe they are.
Random Selection
Random selection is used when you need to choose a sample from a larger population where each item has an equal probability of being selected. This concept is foundational in statistical experiments to ensure unbiased results.
For example, when selecting 5 keyboards out of 25, using random selection ensures that each keyboard has an equal chance of being selected, minimizing the potential for bias.
This concept is crucial in probability and statistics as it allows for the generalization of outcomes over larger sets. In the problem, by randomly selecting keyboards, we can use the calculated probabilities and combinations to predict defect occurrences on average, giving insight into potential manufacturing problems without examining the entire population.
  • Ensures fairness in sample selection
  • Enables representative sampling from a larger group
  • Helps gather data for statistical analysis

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Most popular questions from this chapter

A wallet contains five \(\$ 10\) bills, four \(\$ 5\) bills, and six \(\$ 1\) bills (nothing larger). If the bills are selected one by one in random order, what is the probability that at least two bills must be selected to obtain a first \(\$ 10\) bill?

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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 (e.g., this event occurs if the selected ordering is \(A A B A B\), but not for \(A B B A A)\) ?

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