/*! 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 41 Many fire stations handle emerge... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is \(.85 .\) This can be expressed as \(P(\) call is for medical assistance \()=.85\). a. Give a relative frequency interpretation of the given probability. b. What is the probability that a call is not for medical assistance? c. Assuming that successive calls are independent of one another, calculate the probability that two successive calls will both be for medical assistance. d. Still assuming independence, calculate the prohahility that for two successive calls, the first is for medical assistance and the second is not for medical assistance. e. Still assuming independence, calculate the probability that exactly one of the next two calls will be for medical assistance. (Hint: There are two different possibilities. The one call for medical assistance might be the first call, or it might be the second call.) f. Do you think that it is reasonable to assume that the requests made in successive calls are independent? Explain.

Short Answer

Expert verified
a. Out of 100 calls, 85 are expected to be for medical assistance. b. The probability that a call is not for medical assistance is 0.15. c. The probability that two successive calls are both for medical assistance is 0.7225. d. For two successive calls, the probability the first is for medical assistance and the second is not is 0.1275. e. The probability that exactly one of the next two calls will be for medical assistance is 0.255. f. The independence of successive calls depends on more factors, but for this calculation, they have been considered independent.

Step by step solution

01

Interpreting the given probability

According to the problem, the probability that an incoming call is for medical assistance, denoted as \(P(M)\), is 0.85. This means that, out of 100 calls, we would expect approximately 85 calls to be for medical assistance.
02

Calculating the probability that a call is not for medical assistance

The probability that a call is not for medical assistance, denoted as \(P(NM)\), is the complement of the probability that a call is for medical assistance. We can use the Complement Rule here, which states that \(P(A^C) = 1- P(A)\), where \(A^C\) denotes the complement of event A. Therefore, \(P(NM) = 1 - P(M) = 1 - 0.85 = 0.15.\
03

Calculating the probability of two successive calls for medical assistance

If we're to assume statistical independence of successive calls, the probability of getting two calls for medical assistance, denoted as \(P(M_1 ∩ M_2)\), is the product of their individual probabilities. The Multiplication Rule for independent events states that \(P(A ∩ B) = P(A)P(B)\). Therefore, \(P(M_1 ∩ M_2) = P(M_1)P(M_2) = 0.85 * 0.85 = 0.7225.\
04

Calculating the probability that the first is for medical assistance and the second is not for medical assistance

The probability the first call is for medical assistance and the second is not, denoted as \(P(M ∩ NM)\), can again be calculated using the Multiplication Rule. Therefore, \(P(M ∩ NM) = P(M)P(NM) = 0.85 * 0.15 = 0.1275.\
05

Calculating the probability that exactly one of the next two calls will be for medical assistance

There are two different possibilities: the first call is for medical assistance and the second is not, or the first call is not for medical assistance and the second is. We sum these probabilities, each calculated using the Multiplication Rule. Therefore, \(P(One M) = P(M ∩ NM) + P(NM ∩ M) = 0.1275 + 0.1275 = 0.255.\
06

Analyzing the independence assumption

Whether or not it's reasonable to assume that the requests made in successive calls are independent would depend on the nature of emergencies and their root causes. However, for this calculation, we have assumed they are independent. A more in-depth study may be required to validate this assumption.

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.

Relative Frequency Interpretation
When we talk about the relative frequency interpretation of probability, we're focusing on how likely an event is to occur based on the long-term behavior observed through experimentation or historical data. This interpretation connects the theoretical aspect of probability with real-world frequencies. Let's take the example from the exercise: the probability that an incoming call is for medical assistance is given as 0.85, or 85%. This percentage is derived from observing a large number of calls and noting that approximately 85 out of every 100 calls are for medical assistance. This interpretation is practical because it offers a concrete way to understand probabilities by relating them to observable frequencies.
Complement Rule
Understanding the complement rule is essential in probability. It simply states that the probability of an event not occurring is 1 minus the probability of the event occurring. This is represented mathematically as \(P(A^C) = 1 - P(A)\). In our scenario, if the probability of a call being for medical assistance (M) is 0.85, then the probability of a call not being for medical assistance (NM) is \(1 - 0.85 = 0.15\), or 15%. Remember, the probabilities of an event and its complement always add up to 1, which makes sense since an event either happens or does not happen.
Multiplication Rule for Independent Events
Another crucial concept is the multiplication rule for independent events. This rule states that if you have two independent events, the probability of both occurring is the product of their individual probabilities. For independent events A and B, this is given by \(P(A \cap B) = P(A) \times P(B)\). As applied to our exercise, the chance that two successive calls are for medical assistance is the probability of the first call being for medical assistance multiplied by the probability of the second call being for medical assistance (since we're assuming they are independent). This calculation gives us 0.85 times 0.85, which equals 0.7225, or 72.25%.
Statistical Independence
Finally, let's unpack the idea of statistical independence. Two events are considered to be statistically independent if the occurrence of one event does not affect the probability of the other event occurring. This assumption is critical when applying the multiplication rule. In the context of our exercise, we're assuming that one call requesting medical assistance is independent of the subsequent call requesting medical assistance. However, this assumption may not always hold in real situations, as the nature of the calls may be influenced by external factors like time of day, public health issues, etc. Questioning and validating the independence of events is an important part of statistical analysis and can influence the interpretation and accuracy of probability calculations.

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

The Los Angeles Times (June 14,1995 ) reported that the U.S. Postal Service is getting speedier, with higher overnight on-time delivery rates than in the past. The Price Waterhouse accounting firm conducted an independent audit by seeding the mail with letters and recording ontime delivery rates for these letters. Suppose that the results were as follows (these numhers are fictitions hut are compatible with summary values given in the article): $$ \begin{array}{lcc} & \begin{array}{l} \text { Number } \\ \text { of Letters } \\ \text { Mailed } \end{array} & \begin{array}{l} \text { Number of } \\ \text { Lefters Arriving } \\ \text { on Time } \end{array} \\ \hline \text { Los Angeles } & 500 & 425 \\ \text { New York } & 500 & 415 \\ \text { Washington, D.C. } & 500 & 405 \\ \text { Nationwide } & 6000 & 5220 \\ & & \\ \hline \end{array} $$ Use the given information to estimate the following probabilities: a. The probability of an on-time delivery in Los Angeles b. The probability of late delivery in Washington, D.C. c. The probability that two letters mailed in New York are both delivered on time d. The probability of on-time delivery nationwide

Approximately \(30 \%\) of the calls to an airline reservation phone line result in a reservation being made. a. Suppose that an operator handles 10 calls. What is the probability that none of the 10 calls result in a reservation? b. What assumption did you make to calculate the probability in Part (a)? c. What is the probability that at least one call results in a reservation being made?

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?

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

Refer to the following information on births in the United States over a given period of time: $$ \begin{array}{lr} \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 41,500,000 \\ \text { Twins } & 500,000 \\ \text { Triplets } & 5000 \\ \text { Quadruplets } & 100 \\ & \\ \hline \end{array} $$ Use this information to approximate the probability that a randomly selected pregnant woman who reaches full term a. Delivers twins b. Delivers quadruplets c. Gives birth to more than a single child

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.