/*! 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} Q47E Cell phone handoff behaviour. A ... [FREE SOLUTION] | 91影视

91影视

Cell phone handoff behaviour. A 鈥渉andoff鈥 is a term used in wireless communications to describe the process of a cell phone moving from the coverage area of one base station to that of another. Each base station has multiple channels (called color codes) that allow it to communicate with the cell phone. The Journal of Engineering, Computing and Architecture (Vol. 3., 2009) published a cell phone handoff behavior study. During a sample driving trip that involved crossing from one base station to another, the different color codes accessed by the cell phone were monitored and recorded. The table below shows the number of times each color code was accessed for two identical driving trips, each using a different cell phone model. (Note: The table is similar to the one published in the article.) Suppose you randomly select one point during the combined driving trips.

Color code

0

5

b

c

Total

Model 1

20

35

40

0

85

Model 2

15

50

6

4

75

Total

35

85

46

4

160

a. What is the probability that the cell phone was using color code 5?

b. What is the probability that the cell phone was using color code 5 or color code 0?

c. What is the probability that the cell phone used was Model 2 and the color code was 0?

Short Answer

Expert verified
  1. 0.53
  2. 0.75
  3. 0.94

Step by step solution

01

Introduction

The probability of an occurrence refers to the possibility that the event will occur. The formula represents as:

P(E)=FavourableoutcomeTotaloutcome

02

Find the probability of color code 5

P(Colorcode5)=FavorableoutcomeTotaloutcome=85160=0.53

Hence, the probability of color code 5 is 0.53.

03

Find the probability of color code 5 or 0

P(Colorcode5or0)=P(colorcode5)+P(colorcode0)-P(colorcode5and0)

P(colorcode0)=FavorableoutcomeTotaloutcome=35160=0.22

P(colorcode5and0)=0

P(Colorcode5or0)=0.53+0.220

Hence, the probability of color code 5 or 0 is 0.75.

04

Find the probability of color code Model 2 and color code 0

P(Model2andcolorcode0)=15160=0.94

Hence, the probability of Model 2 and color code 0 is 0.94.

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

Two fair dice are tossed, and the following events are defined:

A: {Sum of the numbers showing is odd.}

B: {Sum of the numbers showing is 9, 11, or 12.}

Are events A and B independent? Why?

Most likely coin-tossing sequence. In Parade Magazine鈥檚 (November 26, 2000) column 鈥淎sk Marilyn,鈥 the following question was posed: 鈥淚 have just tossed a [balanced] coin 10 times, and I ask you to guess which of the following three sequences was the result. One (and only one) of the sequences is genuine.鈥

(1) H HHHHHHHHH

(2) H H T T H T T H HH

(3) T TTTTTTTTT

  1. Demonstrate that prior to actually tossing the coins, thethree sequences are equally likely to occur.
  2. Find the probability that the 10 coin tosses result in all heads or all tails.
  3. Find the probability that the 10 coin tosses result in a mix of heads and tails.
  4. Marilyn鈥檚 answer to the question posed was 鈥淭hough the chances of the three specific sequences occurring randomly are equal . . . it鈥檚 reasonable for us to choose sequence (2) as the most likely genuine result.鈥 If you know that only one of the three sequences actually occurred, explain why Marilyn鈥檚 answer is correct. [Hint: Compare the probabilities in parts b and c.]

A number between 1 and 10, inclusive, is randomly chosen, and the events A and B are defined as follows:

A: [The number is even.]

B: [The number is less than 7.]

a. Identify the sample points in the event AB.

b. Identify the sample points in the event AB.

c. Which expression represents the event that the number is even or less than 7 or both?

d. Which expression represents the event that the number is both even and less than 7?

Advertising proposals. The manager of an advertising department has asked her creative team to propose six new ideas for an advertising campaign for a major client. She will choose three of the six proposals to present to the client. The proposals were named A, B, C, D, E, and F, respectively.

a. In how many ways can the manager select the three proposals? List the possibilities.

b. It is unlikely that the manager will randomly select three of the six proposals, but if she does, what is the probability that she will select proposals A, D, and E?

Risk of a natural gas pipeline accident. Process Safety Progress (December 2004) published a risk analysis for a natural gas pipeline between Bolivia and Brazil. The most likely scenario for an accident would be natural gas leakage from a hole in the pipeline. The probability that the leak ignites immediately (causing a jet fire) is .01. If the leak does not immediately ignite, it may result in a delayed ignition of a gas cloud. Given no immediate ignition, the probability of delayed ignition (causing a flash fire) is .01. If there is no ignition, the gas cloud will harmlessly disperse. Suppose a leak occurs in the natural gas pipeline. Find the probability that either a jet or flash fire will occur. Illustrate with a tree diagram.

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.