/*! 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 25 Reese Prosser never puts money i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Reese Prosser never puts money in a 10 -cent parking meter in Hanover. He assumes that there is a probability of .05 that he will be caught. The first offense costs nothing, the second costs 2 dollars, and subsequent offenses cost 5 dollars each. Under his assumptions, how does the expected cost of parking 100 times without paying the meter compare with the cost of paying the meter each time?

Short Answer

Expert verified
Paying the meter costs $10, while the expected fine costs $17 for parking 100 times.

Step by step solution

01

Determine Meter Costs

If Reese pays the 10-cent meter every time he parks, the cost per parking is $0.10. For 100 times parking, the cost is \( 100 \times 0.10 = 10 \) dollars.
02

Understand Fine Mechanism

The first parking without paying is free if caught; the second incurs a $2 penalty, and subsequent offenses incur $5 each. Reese is caught with a probability of 0.05 each time he parks without paying.
03

Estimate Catch Frequency

The expected number of times Reese is caught over 100 parkings is \( 100 \times 0.05 = 5 \). This means on average, he will have 1 free offense, 1 offense with a \(2 fine, and 3 offenses with a \)5 fine each.
04

Calculate Expected Fine Costs

For the first free offense, there's no charge. For the second offense, the fine is \(2. For the next three offenses, the fine is \)5 each. Total expected fines are calculated as \(0 \times 1 + 2 \times 1 + 5 \times 3 = 0 + 2 + 15 = 17 \) dollars.
05

Compare Expected Costs

Comparing the cost of paying metered parking ($10) with the expected fine cost $17, parking without paying the meter is on average more expensive.

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.

Probability
Probability is a mathematical concept used to measure the likelihood of an event occurring. In Reese's situation, he relies on the probability of 0.05 when considering the chances of being caught without paying at the parking meter.

This probability can be expressed as a fraction, 5 out of every 100 instances of parking.

Understanding probability helps us to estimate possible outcomes in uncertain situations, just like Reese's decision to skip paying the meter.

By calculating the expected number of times he might get caught, Reese utilizes probability to make a cost-effective decision on whether or not to pay for parking.
Parking Fines
Parking fines are penalties enforced when a person parks a vehicle in a restricted area or violates other parking rules. In Reese's case, parking fines are implemented based on repeated offenses:
  • The first offense is waived with no charge.
  • The second offense results in a $2 fine.
  • Any offense thereafter leads to a $5 fine per occurrence.

Reese's approach to parking without paying the meter means each instance of getting caught is progressively costlier.

Understanding how fines accumulate assists in assessing risks and eventual financial consequences tied to actions like repetitive, unpaid meter parking.
Cost Comparison
Cost comparison involves analyzing the financial implications of different options to determine the most economical choice. For Reese, there are two potential costs:
  • Paying the meter fees consistently: $0.10 per parking instance.
  • Potential fines for avoiding meter fees.

Over 100 parking instances, paying the meter would sum to $10.

In comparison, not paying and facing fines has an expected cost of $17.

This comparison shows it is less expensive for Reese to consistently pay for parking, avoiding the higher potential costs associated with fines due to being caught.
Expected Number of Offenses
The expected number of offenses over a series is a key concept in evaluating potential outcomes. Here, it refers to how many times Reese anticipates being caught if he parks without paying across 100 instances.

Calculated by multiplying the total number of parking events by the probability of getting caught (0.05), it results in an expectation of being caught 5 times.

From these expected 5 catches, Reese would experience different fines:
  • 1 free-of-charge catch
  • 1 occurrence of a $2 fine
  • remaining 3 with $5 fines each

Calculating expected numbers is invaluable in making informed decisions, embracing potential risks, and understanding their eventual costs.

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

Assume that, during each second, a Dartmouth switchboard receives one call with probability .01 and no calls with probability .99. Use the Poisson approximation to estimate the probability that the operator will miss at most one call if she takes a 5 -minute coffee break.

You are presented with four different dice. The first one has two sides marked 0 and four sides marked \(4 .\) The second one has a 3 on every side. The third one has a 2 on four sides and a 6 on two sides, and the fourth one has a 1 on three sides and a 5 on three sides. You allow your friend to pick any of the four dice he wishes. Then you pick one of the remaining three and you each roll your die. The person with the largest number showing wins a dollar. Show that you can choose your die so that you have probability \(2 / 3\) of winning no matter which die your friend picks. (See Tenney and Foster. \(\left.^{8}\right)\)

An airline finds that 4 percent of the passengers that make reservations on a particular flight will not show up. Consequently, their policy is to sell 100 reserved seats on a plane that has only 98 seats. Find the probability that every person who shows up for the flight will find a seat available.

Feller \(^{5}\) discusses the statistics of flying bomb hits in an area in the south of London during the Second World War. The area in question was divided into \(24 \times 24=576\) small areas. The total number of hits was 537 . There were 229 squares with 0 hits, 211 with 1 hit, 93 with 2 hits, 35 with 3 hits, 7 with 4 hits, and 1 with 5 or more. Assuming the hits were purely random, use the Poisson approximation to find the probability that a particular square would have exactly \(k\) hits. Compute the expected number of squares that would have \(0,1,2,3,4,\) and 5 or more hits and compare this with the observed results.

Choose a number \(U\) from the interval [0,1] with uniform distribution. Find the cumulative distribution and density for the random variables (a) \(Y=1 /(U+1)\). (b) \(Y=\log (U+1)\).

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.