/*! 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 43 Many fire stations handle emerge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 43

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 probability 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 every 100 calls, about 85 are expected to be for medical assistance. b. 0.15 c. 0.7225 d. 0.1275 e. 0.255 f. Assumptions about independence of successive calls may not always be reasonable.

Step by step solution

01

Interpretation of Probability

Since the probability that an incoming call is for medical assistance is .85, it means that out of every 100 calls, about 85 would be expected to be for medical assistance.
02

Probability of a Call Not for Medical Assistance

The probability that an event does not occur is 1 minus the probability that it does occur. So, the probability that a call is not for medical assistance is \(1-0.85=0.15\)
03

Probability of Successive Calls for Medical Assistance

Assuming that successive calls are independent of one another, the probability that both will be for medical assistance is calculated by multiplying the probabilities together, i.e. \(0.85 * 0.85 = 0.7225\)
04

Probability of First Call for Medical Assistance and Second Not

The probability that the first call is for medical assistance and the second is not is calculated by multiplying the probabilities of the respective events together, i.e. \(0.85 * 0.15 = 0.1275\)
05

Probability of Exactly One Call for Medical Assistance

Since exactly one of the two calls is for medical assistance, there are two ways this can occur: 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. Each scenario has a probability of 0.1275. Adding these gives a total probability of \(0.1275 + 0.1275 = 0.255\)
06

Analysis of Independence Assumption

It may not be reasonable to assume that requests made in successive calls are independent. In a real-world scenario, there may be factors which influence the type of requests made in successive calls. For example, if there's an accident causing multiple injuries, several successive calls could be for medical assistance.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relative Frequency
Relative frequency is a way of interpreting probability based on the long-term observation of events. In our exercise with the fire station, the probability that an incoming call is for medical assistance is given as 0.85. This probability can be understood through relative frequency by saying that, if you were to observe many calls, approximately 85 out of every 100 calls will be for medical assistance.

It's important to understand that relative frequency is derived from actual data over a period of time. So, if this pattern holds consistently, we can use this information to predict future occurrences.

This interpretation is particularly useful in real-world scenarios where patterns emerge from experiences over time.
Independent Events
In probability, events are considered independent if the occurrence of one event does not affect the probability of another. In this context, when the problem assumes that successive calls to the fire station are independent, it means that the nature of one call (whether it is for medical assistance or not) does not change the likelihood of the next call鈥檚 nature.

For example, calculating the probability of both calls being for medical assistance involves multiplying the probabilities: 0.85 (first call is for medical) 脳 0.85 (second call is for medical), resulting in 0.7225. The independence assumption simplifies this calculation.
  • If successive calls influence each other, such as during a large-scale event, this assumption may not hold.
  • Understanding independence helps in precise probability calculations, especially in sequences of events.
Complement Rule
The complement rule helps us find the probability of an event not occurring. The essence of this rule is that the probability of an event not occurring is 1 minus the probability of that event occurring.

In the exercise, the probability that a call is for medical assistance is 0.85. Therefore, using the complement rule, the probability that a call is not for medical assistance is calculated as 1 - 0.85 = 0.15.

This rule is fundamental in probability as it provides an easy way to calculate probabilities of complementary events, which are often easier to determine in many practical scenarios.
Joint Probability
Joint probability refers to the probability of two or more events happening simultaneously. In our problem involving the fire station, we calculate joint probabilities under the assumption of independence.

For instance, to find the probability that the first call is for medical assistance and the second is not, we multiply their individual probabilities: 0.85 (first call) 脳 0.15 (second call not for medical) = 0.1275.

Each potential sequence of calls is considered separately, and their probabilities are added to solve for specific scenarios, such as exactly one call being for medical assistance. Hence, joint probability helps us understand and predict the combination of events.
  • Consider all possible outcomes to ensure accurate probability computation.
  • Joint probability requires careful attention to whether events are independent or not.

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 medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by 2 . For example, they might observe the results in the table for Exercise \(6.82\) given below. The chance experiment would stop after the sixth pair, because Treatment 1 has 2 more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment \(2 .\) Suppose that Treatment 1 has a success rate of \(.7\) (that is, \(P(\) success \()=.7\) for Treatment 1\()\) and that Treatment 2 has a success rate of \(.4 .\) Use simulation to estimate the probabilities in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a success and 8,9 , and 0 to indicate a failure. Let the second digit represent Treatment 2, with 1-4 representing a success. For example, if the two digits selected to represent a pair were 8 and 3 , you would record failure for Treatment 1 and success for Treatment 2\. Continue to select pairs, keeping track of the total number of successes for each treatment. Stop the trial as soon as the number of successes for one treatment exceeds that for the other by \(2 .\) This would complete one trial. Now repeat this whole process until you have results for at least 20 trials [more is better]. Finally, use the simulation results to estimate the desired probabilities.) a. Estimate the probability that more than five pairs must be treated before a conclusion can be reached. (Hint: \(P(\) more than 5\()=1-P(5\) or fewer \() .\) ) b. Estimate the probability that the researchers will incorrectly conclude that Treatment 2 is the better treatment.

Delayed diagnosis of cancer is a problem because it can delay the start of treatment. The paper "Causes of Physician Delay in the Diagnosis of Breast Cancer" (Archives of Internal Medicine \([2002]: 1343-1348)\) examined possible causes for delayed diagnosis for women with breast cancer. The accompanying table summarizes data on the initial written mammogram report (benign or suspicious) and whether or not diagnosis was delayed for 433 women with breast cancer. $$ \begin{array}{l|cc} & & \text { Diagnosis } \\ & \begin{array}{c} \text { Diagnosis } \\ \text { Delayed } \end{array} & \begin{array}{c} \text { Not } \\ \text { Delayed } \end{array} \\ \hline \begin{array}{l} \text { Mammogram Report Benign } \\ \text { Mammogram Report } \\ \text { Suspicious } \end{array} & 32 & 89 \\ & 8 & 304 \\ \hline \end{array} $$ Consider the following events: \(B=\) the event that the mammogram report says benign \(S=\) event that the mammogram report says suspicious \(D=\) event that diagnosis is delayed a. Assume that these data are representative of the larger group of all women with breast cancer. Use the data in the table to find and interpret the following probabilities: i. \(\quad P(B)\) ii. \(P(S)\) iii. \(P(D \mid B)\) iv. \(P(D \mid S)\) b. Remember that all of the 433 women in this study actually had breast cancer, so benign mammogram reports were, by definition, in error. Write a few sentences explaining whether this type of error in the reading of mammograms is related to delayed diagnosis of breast cancer.

The paper "Good for Women, Good for Men, Bad for People: Simpson's Paradox and the Importance of Sex-Spedfic Analysis in Observational Studies" (Journal of Women's Health and Gender-Based Medicine [2001]: \(867-872\) ) described the results of a medical study in which one treatment was shown to be better for men and better for women than a competing treatment. However, if the data for men and women are combined, it appears as though the competing treatment is better. To see how this can happen, consider the accompanying data tables constructed from information in the paper. Subjects in the study were given either Treatment \(\mathrm{A}\) or Treatment \(\mathrm{B}\), and survival was noted. Let \(S\) be the event that a patient selected at random survives, \(A\) be the event that a patient selected at random received Treatment \(\mathrm{A}\), and \(B\) be the event that a patient selected at random received Treatment \(\mathrm{B}\). a. The following table summarizes data for men and women combined: $$ \begin{array}{l|ccc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 215 & 85 & \mathbf{3 0 0} \\ \text { Treatment B } & 241 & 59 & \mathbf{3 0 0} \\ \text { Total } & \mathbf{4 5 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? b. Now consider the summary data for the men who participated in the study: $$ \begin{array}{l|rrr} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 120 & 80 & \mathbf{2 0 0} \\ \text { Treatment B } & 20 & 20 & 40 \\ \text { Total } & \mathbf{1 4 0} & \mathbf{1 0 0} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? c. Now consider the summary data for the women who participated in the study: $$ \begin{array}{l|rrc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 95 & 5 & \mathbf{1 0 0} \\ \text { Treatment B } & 221 & 39 & \mathbf{2 6 0} \\ \text { Total } & \mathbf{3 1 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? d. You should have noticed from Parts (b) and (c) that for both men and women, Treatment \(A\) appears to be better. But in Part (a), when the data for men and women are combined, it looks like Treatment \(\mathrm{B}\) is better. This is an example of what is called Simpson's paradox. Write a brief explanation of why this apparent inconsistency occurs for this data set. (Hint: Do men and women respond similarly to the two treatments?)

Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cap B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

The article "Chances Are You Know Someone with a Tattoo, and He's Not a Sailor" (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50 . The accompanying data are consistent with summary values given in the article. $$ \begin{array}{l|cc} & \text { At Least One Tattoo } & \text { No Tattoo } \\ \hline \text { Age 18-29 } & 18 & 32 \\ \text { Age 30-50 } & 6 & 44 \\ \hline \end{array} $$ Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities. a. \(P(\) tattoo \()\) b. \(P(\) tattoo \(\mid\) age \(18-29)\) c. \(P(\) tattoo \(\mid\) age \(30-50\) ) d. \(P(\) age \(18-29 \mid\) tattoo \()\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

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.