/*! 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 85 A certain company sends \(40 \%\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 85

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?

Short Answer

Expert verified
The probability that the parcel went by means of \(A_{1}\) and was late is \(0.008\) or \(0.8\%\).

Step by step solution

01

Identify the Probabilities of Individual Events

From the problem, it's known that the probability of a parcel being sent by means of service A1, denoted as \(P(A1)\), is \(40\%\), or \(0.4\) in decimal terms. The probability of a parcel arriving late given that it was sent by A1, denoted as \(P(L | A1)\), is \(2\%\), or \(0.02\) in decimal terms.
02

Compute the Intersection of the Two Events

The probability of intersection of two events (i.e., the parcel being sent by A1 and arriving late) can be calculated as: \(P(A1 \cap L) = P(A1) \times P(L | A1)\).
03

Calculate the Result

Substituting the values into the formula, we get: \(P(A1 \cap L) = 0.4 \times 0.02 = 0.008\).

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 91影视!

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 deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are \(2,598,960\) possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented? c. Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits?

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let not \(E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(\left.E_{1}\right)\) ? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: a. The probability that grip size is \(4 \frac{1}{2}\) in. (event \(\mathrm{A}\) ) is $$ P(A)=P\left(O_{2} \text { or } O_{5}\right)=.20+.15=.35 $$ How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) in. c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) in.?

According to a study released by the federal Substance Abuse and Mental Health Services Administration (Knight Ridder Tribune, September 9,1999 ), approximately \(8 \%\) of all adult full-time workers are drug users and approximately \(70 \%\) of adult drug users are employed full-time. a. Is it possible for both of the reported percentages to be correct? Explain. b. Define the events \(D\) and \(E\) as \(D=\) event that a randomly selected adult is a drug user and \(E=\) event that a randomly selected adult is employed full- time. What are the estimated values of \(P(D \mid E)\) and \(P(E \mid D) ?\) c. Is it possible to determine \(P(D)\), the probability that a randomly selected adult is a drug user, from the information given? If not, what additional information would be needed?

A Gallup survey of 2002 adults found that \(46 \%\) of women and \(37 \%\) of men experience pain daily (San Luis Obispo Tribune, April 6, 2000). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the events selected adult is male and selected adult experiences pain daily independent or dependent? Explain.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

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.