/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 A corporate Web site contains er... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A corporate Web site contains errors on 50 of 1000 pages. If 100 pages are sampled randomly, without replacement, approximate the probability that at least 1 of the pages in error is in the sample.

Short Answer

Expert verified
The probability is approximately 99.4%.

Step by step solution

01

Understanding the Problem

We have a corporate website with a total of 1000 pages and 50 of these pages contain errors. We need to find the probability that, when sampling 100 pages randomly without replacement, at least one of the sampled pages contains an error.
02

Complementary Probability Approach

To find the probability that at least one of the sampled pages has an error, it's often easier to first find the probability that none of the pages has an error, and then subtract that from 1. This complementary approach can be more straightforward.
03

Calculating Probability of No Errors

The probability that a single sampled page is not in error is \(\frac{950}{1000}\), since 950 out of 1000 pages are error-free. For two pages, it would be \(\frac{949}{999}\) for the second page, and generally \(\frac{950 - x}{1000 - x}\) for the x-th page. Thus, the probability of sampling 100 pages with no errors is given by the product \[ \frac{950}{1000} \times \frac{949}{999} \times \dots \times \frac{851}{901} \].
04

Approximating with Binomial Probability

Since exact calculation is cumbersome, we use the binomial approximation. The probability of a page being error-free is \(p = \frac{950}{1000} = 0.95\) and the number of trials \(n = 100\). Then, the approximate probability that all 100 pages are error-free (none are errors) is given by \[(0.95)^{100}\].
05

Calculating Complementary Probability using Binomial

Calculate \((0.95)^{100}\) to estimate the probability of no errors in the sample. Subtract this result from 1 to find the probability that at least one page in the sample has an error: \[ P(\text{at least one error}) = 1 - (0.95)^{100} \].
06

Final Calculation

Using a calculator, evaluate \((0.95)^{100} \approx 0.0060\). Thus, \[ P(\text{at least one error}) = 1 - 0.0060 = 0.994\]. This means there is approximately a 99.4% chance of finding at least one error in the sample.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Complementary Probability
Complementary probability is a useful method in probability theory that helps simplify problems by focusing on the opposite event. Instead of calculating the probability of an event directly, we often find it easier to calculate the probability that the event does not happen, and then subtract that from one.
For instance, in the context of the given exercise, it's easier to find the probability that none of the sampled pages contain errors first. By calculating the likelihood of this opposite scenario, we leverage complementary probability.
Here's why it's helpful:
  • Complex calculations can be simplified.
  • Focuses on a single, often more manageable, scenario.
  • Provides a clearer path to the final probability calculation.
In our specific case, finding the chance of zero errors among the pages involves consecutive multiplication of probabilities (as shown in Step 3). This is often much more straightforward than trying to directly compute the probability of at least one error occurring among 100 pages.
Binomial Approximation
Binomial approximation is a technique that simplifies probability calculations, especially when dealing with large numbers. This approach is invaluable when the classical probability calculations become too tedious.
In our exercise, after calculating complementary probability, we apply the binomial approximation to synthesize results efficiently. The binomial approximation assumes a binary outcome (like error or no error) in a series of trials, making it ideal here.
The important factors include:
  • Identifying the number of trials, which here is 100 sampled pages.
  • Estimating the probability of success (error-free page), calculated as 0.95.
  • Applying the approximation formula, shown as \((0.95)^{100}\).
This approximation provides an estimate of the probability that all pages are error-free. It is substantially less computationally intense than calculating each probability turn by turn, which can be crucial for handling computational complexity efficiently.
Sampling Without Replacement
Sampling without replacement is a common situation in statistical exercises, where the sample is drawn and not returned to the pool before the next draw. This approach means the total number of items decreases with each sample taken.
In the context of the corporate website example, pages are removed from the pool once they are sampled. This impacts the probabilities being calculated, as each draw affects the subsequent probability calculations.
Key points to consider:
  • Each selected page influences the pool for upcoming selections.
  • The probability of drawing error-free pages changes with every draw.
  • It requires recalculation of probabilities, making the process more dynamic compared to sampling with replacement.
This "without replacement" nature crucially impacts the setup of the complementary probability process and the need for binomial approximation, as exact calculations would become progressively intricate.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A bearing assembly contains 10 bearings. The bearing diameters are assumed to be independent and normally distributed with a mean of 1.5 millimeters and a standard deviation of 0.025 millimeter. What is the probability that the maximum diameter bearing in the assembly exceeds 1.6 millimeters?

The volume of a shampoo filled into a container is uniformly distributed between 374 and 380 milliliters. (a) What are the mean and standard deviation of the volume of shampoo? (b) What is the probability that the container is filled with less than the advertised target of 375 milliliters? (c) What is the volume of shampoo that is cxceeded by \(95 \%\) of the containers? (d) Iivery milliliter of shampoo costs the producer \(\$ 0.002\). Any more shampoo in the container than 375 milliliters is an cxtra cost to the produecr. What is the mean extra cost?

A show is scheduled to start at 9: 00 A.M., 9: 30 A.M. and 10: 00 A.M. Once the show starts, the gate will be closed. A visitor will arrive at the gate at a time uniformly distributed between 8: 30 A.M. and 10: 00 A.M. Determine (a) The cumulative distribution function of the time (in minutes) between arrival and 8: 30 A.M. (b) The mean and variance of the distribution in the previous part. (c) The probability that a visitor waits less than 10 minutes for a show (d) The probability that a visitor waits more than 20 minutes

Suppose the length of stay (in hours) at an emergency department is modeled with a lognormal random variable \(X\) with \(\theta=1.5\) and \(\omega=0.4\). Determine the following: (a) mean and variance (b) \(P(X<8)\) (c) Comment on the difference between the probability \(P(X<0)\) calculated from this lognormal distribution and a normal distribution with the same mean and variance.

The maximum time to complete a task in a project is 2.5 days. Suppose that the completion time as a proportion of this maximum is a beta random variable with \(\alpha=2\) and \(\beta=\) 3\. What is the probability that the task requires more than two days to complete?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.