Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Q196SE (page 285)

Testing for spoiled wine. Suppose that you are purchasing cases of wine (12 bottles per case) and that, periodically, you select a test case to determine the adequacy of the bottles’ seals. To do this, you randomly select and test 3 bottles in the case. If a case contains 1 spoiled bottle of wine, what is the probability that this bottle will turn up in your sample?

Short Answer

Expert verified

The probability that the spoiled bottle will turn up in the sample is 0.25.

Step by step solution

01

Given information

In a purchase case containing 12 bottles, randomly 3 bottles are selected.

02

Calculating the Probability 

Let X be the spoiled bottles in the sample.

Here, a case containing 1 spoiled bottle is taken.

By definitions, the probability that the spoiled bottle will turn up in the sample is given as

p=31121=35=0.25

Hence, the probability that the spoiled bottle will turn up in the sample is 0.25.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Waiting for a car wash. An automatic car wash takes exactly 5 minutes to wash a car. On average, 10 cars per hour arrive at the car wash. Suppose that 30 minutes before closing time, 5 cars are in line. If the car wash is in continuous use until closing time, will anyone likely be in line at closing time?

The random variable x has a normal distribution with μ=40and σ2=36. Find a value of x, call itx0, such that

a.P(x≥x0)=0.10

b.P(μ≤x≤x0)=0.40

c.P(x≤x0)=0.05

d.P(x≥x0)=0.40

e.P(x0≤x<μ)=0.45

Lead in metal shredder residue. On the basis of data collectedfrom metal shredders across the nation, the amount xof extractable lead in metal shredder residue has an approximateexponential distribution with meanθ= 2.5 milligramsper liter (Florida Shredder’s Association).

a. Find the probability that xis greater than 2 milligramsper liter.

b. Find the probability that xis less than 5 milligrams perliter.

Industrial filling process. The characteristics of an industrialfilling process in which an expensive liquid is injectedinto a container were investigated in the Journal of QualityTechnology(July 1999). The quantity injected per containeris approximately normally distributed with mean 10

units and standard deviation .2 units. Each unit of fill costs\(20 per unit. If a container contains less than 10 units (i.e.,is underfilled), it must be reprocessed at a cost of \)10. A properly filled container sells for $230.

a. Find the probability that a container is underfilled. Notunderfilled.

b. A container is initially underfilled and must be reprocessed.Upon refilling, it contains 10.60 units. Howmuch profit will the company make on thiscontainer?

c. The operations manager adjusts the mean of the fillingprocess upward to 10.60 units in order to makethe probability of underfilling approximately zero.

Under these conditions, what is the expected profit percontainer?

Find the following probabilities for the standard normal

random variable z:

a.P(z≤2.1)

b.P(z≥2.1)

c.P(z≥-1.65)

d.P(-2.13≤z≤-.41)

e.P(-1.45≤z≤2.15)

f.P(z≤-1.43)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation
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.

Sign-up for free