/*! 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 22 The Denver Post reported that a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Denver Post reported that a recent audit of Los Angeles 911 calls showed that \(85 \%\) were not emergencies. Suppose the 911 operators in Los Angeles have just received four calls. (a) What is the probability that all four calls are, in fact, emergencies? (b) What is the probability that three or more calls are not emergencies? (c) How many calls \(n\) would the 911 operators need to answer to be \(96 \%\) (or more) sure that at least one call is, in fact, an emergency?

Short Answer

Expert verified
(a) 0.0005, (b) 0.8906, (c) 20 calls.

Step by step solution

01

Define Non-Emergency and Emergency Probabilities

The probability of a call being a non-emergency is 85%, or 0.85. Therefore, the probability of a call being an emergency is \[ P(E) = 1 - 0.85 = 0.15. \]
02

Calculate Probability for Part (a)

We need the probability that all four calls are emergencies. Each call is an independent event, so we use \[ P( ext{4 emergencies}) = (0.15)^4. \]Compute this value to find the probability.
03

Compute for Part (a)

Calculating the probability, we have \[ (0.15)^4 = 0.00050625. \] Hence, the probability that all four calls are emergencies is 0.00050625.
04

Calculate Probability for Part (b)

We need the probability that three or more calls are not emergencies. First, calculate the probabilities of exactly 3 and 4 calls being non-emergencies: - Exactly 3 calls: \[ \binom{4}{3} (0.85)^3 (0.15)^1 = 4 \times 0.614125 \times 0.15. \]- Exactly 4 calls: \[ (0.85)^4. \]Add these probabilities together.
05

Compute for Part (b)

- For exactly 3 non-emergencies: - Compute \( 4 \times 0.614125 \times 0.15 = 0.36855 \).- For exactly 4 non-emergencies: - Compute \( (0.85)^4 = 0.52200625 \).- Add them together: \[ 0.36855 + 0.52200625 = 0.89055625. \]Thus, the probability that 3 or more calls are not emergencies is 0.8906 (rounded to four decimal places).
06

Solving for Part (c) using the Complement Rule

To be 96% sure that at least one call is an emergency, the probability that all calls are non-emergencies should be less than 4%.Write the inequality: \[ (0.85)^n < 0.04. \]
07

Solve the Inequality for Part (c)

By taking the natural logarithm on both sides: \[ \ln((0.85)^n) < \ln(0.04). \]This simplifies using logarithmic properties to:\[ n \cdot \ln(0.85) < \ln(0.04). \]Solve for \(n\):\[ n > \frac{\ln(0.04)}{\ln(0.85)}. \]
08

Calculate Minimum n for Part (c)

Calculate this using a calculator:\[ n > \frac{-3.21888}{-0.1625} \approx 19.8. \]Since \(n\) must be an integer, round up to 20. Therefore, operators need to answer at least 20 calls to be 96% sure one is an emergency.

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.

Binomial Distribution
In probability theory, a binomial distribution is a useful tool for modeling scenarios with two possible outcomes: success or failure. In our 911 call scenario, an 'emergency' is considered a success, while a 'non-emergency' represents a failure. Each call is addressed as an independent trial.
The binomial distribution is characterized by two parameters: the number of trials (denoted by 'n') and the probability of success in each trial (denoted by 'p'). The formula for finding the probability of exactly 'k' successes in 'n' trials is given by:
  • \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Here, \( \binom{n}{k} \) represents the binomial coefficient, which counts the number of ways to choose 'k' successes out of 'n' trials.
This concept is crucial when you need to determine the probability of certain numbers of 'successes' or 'failures' in a fixed number of trials, as seen when calculating the probability of exactly 3 or more non-emergency calls.
Complement Rule
The complement rule is a key technique in probability that provides an alternative way to find the probability of an event. Instead of calculating the probability of the event directly, you can calculate the probability of the event not happening and then subtract from one.
In probability terms, if \( P(A) \) is the probability of an event, the complement rule states that:
  • \( P(A') = 1 - P(A) \)
Here, \( P(A') \) is the probability that event 'A' does not occur.
This rule is especially helpful in our problem when determining how many calls the operators need to receive to be 96% certain that at least one is an emergency. Instead of directly finding the number of required emergency calls, we calculate the number of non-emergency calls that would lead to the opposite situation being less likely.Using the inequality \( (0.85)^n < 0.04 \), the complement rule allows us to effectively determine the minimum number of trials needed to achieve a desired probability threshold.
Independent Events
In probability, two events are independent if the occurrence of one does not affect the occurrence of another. This concept is pivotal when you are dealing with repeated trials like those in the 911 call scenario. Each call being an emergency or not is considered an independent event, meaning the result of one call does not inform or alter the result of another.
Understanding independent events is crucial when applying the Binomial Distribution. When each trial is independent, the probability of getting a certain result on one trial is consistent, unaffected by what happened in previous trials.
In mathematical terms, if two events A and B are independent, then the probability of both events occurring, denoted \( P(A \cap B) \), is:
  • \( P(A \cap B) = P(A) \times P(B) \)
This notion simplifies calculating the occurrences of multiple independent events, such as when employing the probability of all four 911 calls being emergencies (which would be rare, given 85% are non-emergencies). This understanding assists in accurately predicting outcomes and required conditions in statistical problems and probability exercises.

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

USA Today reported that the U.S. (annual) birthrate is about 16 per 1000 people, and the death rate is about 8 per 1000 people. (a) Explain why the Poisson probability distribution would be a good choice for the random variable \(r=\) number of births (or deaths) for a community of a given population size. (b) In a community of 1000 people, what is the (annual) probability of 10 births? What is the probability of 10 deaths? What is the probability of 16 births? 16 deaths? (c) Repeat part (b) for a community of 1500 people. You will need to use a calculator to compute \(P(10 \text { births) and } P(16\text { births). }\) (d) Repeat part (b) for a community of 750 people.

A large bank vault has several automatic burglar alarms. The probability is 0.55 that a single alarm will detect a burglar. (a) How many such alarms should be used for \(99 \%\) certainty that a burglar trying to enter will be detected by at least one alarm? (b) Suppose the bank installs nine alarms. What is the expected number of alarms that will detect a burglar?

USA Today reports that about \(25 \%\) of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability \(P(r)\) of \(r\) successes ranging from 0 to 4 (b) Make a histogram for the probability distribution of part (a). (c) What is the expected number of parolees in Alice's group who will not be repeat offenders? What is the standard deviation? (d) How large a group should Alice counsel to be about \(98 \%\) sure that three or more parolees will not become repeat offenders?

Let \(r\) be a binomial random variable representing the number of successes out of \(n\) trials. (a) Explain why the sample space for \(r\) consists of the set \(\\{0,1,2, \ldots, n\\}\) and why the sum of the probabilities of all the entries in the entire sample space must be 1. (b) Explain why \(P(r \geq 1)=1-P(0)\) (c) Explain why \(P(r \geq 2)=1-P(0)-P(1)\) (d) Explain why \(P(r \geq m)=1-P(0)-P(1)-\cdots-P(m-1)\) for \(1 \leq m \leq n\).

Approximately \(75 \%\) of all marketing personnel are extroverts, whereas about \(60 \%\) of all computer programmers are introverts (Source: \(A\) Guide to the Development and Use of the Myers-Briggs Type Indicator, by Mycrs and McCaulley). (a) At a mecting of 15 marketing personnel, what is the probability that 10 or more are extroverts? What is the probability that 5 or more are extroverts? What is the probability that all are extroverts? (b) In a group of 5 computer programmers, what is the probability that none are introverts? What is the probability that 3 or more are introverts? What is the probability that all are introverts?

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.