/*! 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 62 About 8 women in 100,000 have ce... [FREE SOLUTION] | 91Ó°ÊÓ

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About 8 women in 100,000 have cervical cancer (C), so \(\mathrm{P}(\mathrm{C})=0.00008\) and \(\mathrm{P}(\mathrm{no} \mathrm{C})=0.99992 .\) The chance that a Pap smear will incorrectly indicate that a woman without cervical cancer has cervical cancer is \(0.03 .\) Therefore, $$ \mathrm{P}(\text { test } \operatorname{pos} \mid \text { no } \mathrm{C})=0.03 $$ What is the probability that a randomly chosen women who has this test will both be free of cervical cancer and test positive for cervical cancer (a false positive)?

Short Answer

Expert verified
The probability that a randomly chosen woman who has this test will be free of cervical cancer and test positive for cervical cancer (a false positive) is \(2.998\% \)

Step by step solution

01

Understand the given probabilities

Given probabilities are \n1) The probability of a randomly chosen woman having cervical cancer, denoted as \(\mathrm{P}(\mathrm{C})=0.00008\) \n2) The probability of her not having cervical cancer, denoted as \(\mathrm{P}(\mathrm{noC})=0.99992\) \n3) The probability that the test incorrectly shows a woman without cervical cancer as having the disease, denoted as \(\mathrm{P}(\text { test pos } \mid \text { no } \mathrm{C})=0.03\)
02

Apply the rule of joint probability

The rule of joint probability states that the probability of two events, A and B, both occurring is given by the probability of A times the probability of B given A. In this context, the two events are: a woman being free of cervical cancer and the test being positive. This can be expressed as \(\mathrm{P}(\text{noC and test pos}) = \mathrm{P}(\text{noC}) \cdot \mathrm{P}(\text{test pos} \mid \text{noC})\)
03

Calculate the probability

Substituting the given values into the joint probability formula gives \(\mathrm{P}(\text{noC and test pos}) = 0.99992 \cdot 0.03 = 0.02998\). Hence, the probability that a randomly chosen woman who has this test will be free of cervical cancer and test positive for cervical cancer is \(2.998\% \)

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Key Concepts

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

Probability Theory
Probability theory lies at the heart of making predictions about events when uncertainty is involved. It provides us with the mathematical framework to describe the likelihood of occurrence of different outcomes. In this context, probability is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

There are different types of probabilities, such as marginal, joint, and conditional probabilities that help us understand various aspects of chance. Joint probability, a crucial concept in our exercise, is specifically the probability of two or more events occurring simultaneously. Calculating joint probabilities allows us to determine how likely two events are to occur together, which is particularly useful in complex situations where events may not be independent of one another.
Conditional Probability
Conditional probability is a measure of the probability of one event occurring with some relationship to one or more other events. For example, the probability of a test showing a false positive is greatly dependent on whether or not the condition - in our case, cervical cancer - is actually present.

Formally, the conditional probability of event A given event B is denoted as \(P(A \mid B)\) and is calculated by dividing the probability of both events A and B occurring together (the joint probability) by the probability of event B. In medical testing, understanding conditional probability is essential because it aids healthcare professionals in interpreting test results accurately and in weighing the likelihood of various diagnoses.
False Positive Rate
The false positive rate is a critical measure in diagnostic testing, especially for conditions like cervical cancer. It refers to the proportion of individuals who do not have a condition but test positive for it. A high false positive rate means that many people who do not have the condition are being told they do, which can lead to unnecessary stress and further testing.

In our exercise, the false positive rate is denoted by \(P(\text{test pos} \mid \text{no C}) = 0.03\), or 3%. This means that for every 100 women without cervical cancer who undergo the Pap smear test, about 3 will receive a false positive result. It's important for individuals to understand this concept as it directly impacts the interpretation of their test results and the subsequent medical advice they may receive.
Pap Smear Test Statistics
Pap smear tests are a preventative measure used to detect cervical cancer in women. The statistics concerning the Pap smear test, such as its sensitivity, specificity, and the rates of false positives, are key in evaluating its effectiveness as a diagnostic tool. In probability terms, these statistics reflect the performance of the test in distinguishing between those who have cervical cancer and those who do not.

With the figures provided in the exercise, the joint probability calculation is used to determine how frequently a false positive occurs in the general population – a necessary piece of information for understanding the real-world impact of test statistics. These statistics help health professionals and researchers to improve testing protocols and ensure that patients receive the most appropriate care based on the risks and benefits implied by the test outcomes.

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Most popular questions from this chapter

a. On a true/false quiz in which you are guessing, what is the probability of guessing correctly on one question? b. What is the probability that a guess on one true/false question will be incorrect?

A multiple-choice test has 10 questions. Each question has four choices, but only one choice is correct. Which of the following methods is a valid simulation of a student who guesses randomly on each question. Explain. (Note: there might be more than one valid method.) a. Ten digits are selected using a random number table. Each digit represents one question on the test. If the digit is even, the answer is correct. If the digit is odd, the answer is incorrect. b. The digits 1, 2, 3, 4 represent the students attempt on one question. All other digits are ignored. The 1 represents a correct choice. The digits 2 , 3 , and 4 represent an incorrect choice. c. The digits \(1,2,3,4,5,6,7,8\) represent the student's attempt on one question. The digits 0 and 9 are ignored. The digits 1 and 2 represent a correct choice and the digits \(3,4,5,6,7,8\) represent an incorrect choice.

a. Explain how you could use a random number table (or the random numbers generated by software or a calculator) to simulate rolling a fair four-sided die 20 times. Assume you are interested in the probability of rolling a 1 . Then report a line or two of the random number table (or numbers generated by a computer or calculator) and the values that were obtained from it. b. Report the empirical probability of rolling a 1 on the four-sided die from part (a), and compare it with the theoretical probability of rolling a 1 .

a. How many outcomes are in the sample space? b. Assuming all of the outcomes in the sample space are equally likely, find each of the probabilities: i. all tails in 4 tosses ii. only 1 tail in 4 tosses iii. at most 1 tail in 4 tossesThe sample space given here shows all possible sequences for tossing a fair coin 4 times. The sequences have been organized by the number of tails in the sequence.

Suppose a person is selected at random from a large population. a. Label each pair of events as mutually exclusive or not mutually exclusive. i. The person has traveled to Mexico; the person has traveled to Canada. ii. The person is single; the person is married. b. Give an example of two events that are mutually exclusive when a person is selected at random from a large population.

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