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Chapter 4: Problem 29

A professional proofreader has a \(98 \%\) chance of detecting an error in a piece of written work (other than misspellings, double words, and similar errors that are machine detected). A work contains four errors. a. Find the probability that the proofreader will miss at least one of them. b. Show that two such proofreaders working independently have a \(99.96 \%\) chance of detecting an error in a piece of written work. c. Find the probability that two such proofreaders working independently will miss at least one error in a work that contains four errors.

Short Answer

Expert verified
a: 0.0794, b: 99.96%, c: 2.56e-13

Step by step solution

01

Calculate the Probability of Missing One Error

A proofreader has a 98% chance of detecting an error, so there is a 2% probability (which can be expressed as 0.02) that they will miss an error.
02

Calculate the Probability of Missing All Errors by One Proofreader

Since there are four errors and the probability of missing each error is 0.02, the probability of missing all four errors is \( 0.02^4 \). Calculate this as:\[0.02^4 = 0.02 \times 0.02 \times 0.02 \times 0.02 = 1.6 \times 10^{-7}\]
03

Calculate the Probability of Missing At Least One Error by One Proofreader

The probability of not missing all the errors is the complement of the probability calculated in Step 2. So, calculate:\[1 - 1.6 \times 10^{-7} \approx 0.99999984\]Then, the probability of missing at least one error is:\[1 - 0.99999984 = 1.6 \times 10^{-7}\]
04

Calculate the Probability of Detecting an Error by Two Proofreaders

Each proofreader independently has a 98% chance of detecting an error. Therefore, the probability both detect it is:\[1 - (0.02 \times 0.02) = 1 - 0.0004 = 0.9996\]
05

Calculate the Probability of Missing At Least One Error by Two Proofreaders

With a 0.9996 probability of detecting one error, for four errors, calculate:\[(1 - 0.9996)^4 = 0.0004^4 = 2.56 \times 10^{-13}\]Thus, the probability of at least one error being missed is:\[1 - (1 - 2.56 \times 10^{-13}) \approx 2.56 \times 10^{-13}\]

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Key Concepts

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

Error Detection
In probability theory, error detection refers to the process of identifying inaccuracies or mistakes in a task or a dataset. In this context, the professional proofreader has an impressive 98% accuracy rate. However, it's crucial to understand that even with this high rate, there is still a chance some errors can slip through.When calculating the likelihood of missing an error, we must consider the complement of the detection probability, which is 2% in this case (100% - 98%). This means a proofreader has a 2% chance of missing any individual error in the document.Consider a document with multiple (in this case, four) errors. To find out the probability that a proofreader will ultimately miss at least one of these errors, we initially calculate the probability that they miss all errors. Use the formula for independent events, multiplying the probability of missing each single error:
  • Missing all errors: \( 0.02^4 \)
  • This yields: \( 1.6 \times 10^{-7} \), which is quite small, indicating a high likelihood that at least one error will be caught.
Proofreading Statistics
Proofreading statistics explore the reliability and effectiveness of proofreading processes, often using probability theory to provide insight. In this exercise, we are asked to see what happens when two proofreaders review a piece of work independently.The probability of both proofreaders missing the same error simultaneously is derived from their independent probabilities. This can be seen as the product of the probabilities of each proofreader missing an error, since they do not influence each other. Therefore, 0.02 * 0.02 gives 0.0004, or 0.04% chance both will simultaneously miss an error. Subtracting this from one gives us the probability that at least one will catch the error, which is useful:
  • Chance of detection by at least one: \( 0.9996 \)
To visualize, consider how this relates to error detection across more errors. When documents have four errors, calculating based on one error detection probability shows us how many errors might slip through with two independent proofreaders working together.
Independent Events
In probability theory, independent events are those whose outcomes do not affect each other. This is a fundamental concept for this exercise as each proofreader operates independently from the other.In practice, this independence implies the choices or actions of one proofreader do not influence the other's decisions. Each proofreader independently has a 98% chance of detecting an error and a 2% chance of missing it, meaning their individual performances do not correlate, and one proofreader's success or failure does not impact the other.This is particularly important when considering the collective probability of missing at least one error by both proofreaders. The independence allows us to multiply their chances of missing errors to find a cumulative probability. By leveraging their independent probabilities, we can compute the likelihood of both proofreaders either missing an error or catching it:

  • The probability that two proofreaders miss at least one error in a work filled with four potential mistakes is calculated by multiplying the probability of missing all errors together.
  • As calculated: \( 0.02^8 \) results in an extremely small number, ensuring a high probability of recognizing most errors when two qualified professionals check independently.

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Most popular questions from this chapter

In spite of the requirement that all dogs boarded in a kennel be inoculated, the chance that a healthy dog boarded in a clean, well-ventilated kennel will develop kennel cough from a carrier is 0.008 . a. If a carrier (not known to be such, of course) is boarded with three other dogs, what is the probability that at least one of the three healthy dogs will develop kennel cough? b. If a carrier is boarded with four other dogs, what is the probability that at least one of the four healthy dogs will develop kennel cough? c. The pattern evident from parts (a) and (b) is that if \(K+1\) dogs are boarded together, one a carrier and \(K\) healthy dogs, then the probability that at least one of the healthy dogs will develop kennel cough is \(P(X \geq 1)=1-(0.992)^{x},\) where \(X\) is the binomial random variable that counts the number of healthy dogs that develop the condition. Experiment with different values of \(K\) in this formula to find the maximum number \(K+1\) of dogs that a kennel owner can board together so that if one of the dogs has the condition, the chance that another dog will be infected is less than \(0.05 .\)

A fair coin is tossed repeatedly until either it lands heads or a total of five tosses have been made, whichever comes first. Let \(X\) denote the number of tosses made. a. Construct the probability distribution for \(X\). b. Compute the mean \(\mu\) of \(X\). c. Compute the standard deviation \(\sigma\) of \(X\).

A roulette wheel has 38 slots. Thirty-six slots are numbered from 1 to \(36 ;\) half of them are red and half are black. The remaining two slots are numbered 0 and 00 and are green. In a \(\$ 1\) bet on red, the bettor pays \(\$ 1\) to play. If the ball lands in a red slot, he receives back the dollar he bet plus an additional dollar. If the ball does not land on red he loses his dollar. Let \(X\) denote the net gain to the bettor on one play of the game. a. Construct the probability distribution of \(X\). b. Compute the expected value \(E(X)\) of \(X\), and interpret its meaning in the context of the problem. c. Compute the standard deviation of \(X\).

An appliance store sells 20 refrigerators each week. Ten percent of all purchasers of a refrigerator buy an extended warranty. Let \(X\) denote the number of the next 20 purchasers who do so. a. Verify that \(X\) satisfies the conditions for a binomial random variable, and find \(n\) and \(p\). b. Find the probability that \(X\) is zero. c. Find the probability that \(X\) is two, three, or four. d. Find the probability that \(X\) is at least five.

Let \(X\) denote the number of boys in a randomly selected three-child family. Assuming that boys and girls are equally likely, construct the probability distribution of \(X\).
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