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Chapter 3: Problem 15

Suppose an influenza epidemic strikes a city. In \(10 \%\) of families the mother has influenza; in \(10 \%\) of families the father has influenza; and in \(2 \%\) of families both the mother and father have influenza...15 Based on Problem 3.12, what is the conditional probability that the father has influenza given that the mother does not have influenza?

Short Answer

Expert verified
The conditional probability that the father has influenza given the mother does not is approximately 0.0889.

Step by step solution

01

Identify Given Data

From the problem, we are provided with the following probabilities:- Probability that the mother has influenza, \( P(M) = 0.10 \).- Probability that the father has influenza, \( P(F) = 0.10 \).- Probability that both the mother and father have influenza, \( P(M \cap F) = 0.02 \).Our task is to find the conditional probability that the father has influenza given that the mother does not have influenza, which is denoted by \( P(F|M^c) \).
02

Use Complementary Probability

We need to find the probability that the mother does not have influenza:- \( P(M^c) = 1 - P(M) = 1 - 0.10 = 0.90 \).
03

Apply the Law of Total Probability

To find \( P(F \cap M^c) \), use the law of total probability:\[ P(F) = P(F \cap M) + P(F \cap M^c) \] Rearrange to find \( P(F \cap M^c) \):\[ P(F \cap M^c) = P(F) - P(F \cap M) = 0.10 - 0.02 = 0.08 \]
04

Calculate Conditional Probability

Utilize the definition of conditional probability:\[ P(F|M^c) = \frac{P(F \cap M^c)}{P(M^c)} \]Substitute the known values:\[ P(F|M^c) = \frac{0.08}{0.90} \approx 0.0889 \]

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Key Concepts

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

Law of Total Probability
The Law of Total Probability is a fundamental concept in probability theory. It helps us compute the likelihood of an event happening by considering all possible ways in which that event can occur. Imagine it as a way of breaking down complex probability problems into simpler parts.
In our exercise, we need to find the probability that the father has influenza, specifically when the mother does not have influenza. We have to consider two possibilities: when both parents have influenza and when only the father does. The Law of Total Probability allows us to think about these separate cases and combine them to achieve the total probability.
For example:
  • The probability that the father has influenza includes cases where both parents have influenza and cases where only the father has influenza.
  • We write this as: \[ P(F) = P(F \cap M) + P(F \cap M^c) \]
This formula tells us to add the probability of the father being sick in scenarios where the mother is also sick, and where she isn't, to achieve the total probability that the father is sick.
Complementary Probability
Complementary Probability is another handy concept that complements our understanding of probability events. Simply put, it's the probability that an event does not happen. If you know the chance of something happening, its complement is just what's left over to make everything add up to 1.
This concept is crucial in our exercise because we're looking to explore what happens when the mother does not have influenza. We denote this as the complement of the mother having influenza, written as \(P(M^c)\). Given that the probability of the mother having the flu is \(0.10\), the complementary probability is:
  • Probability mother does not have flu: \[ P(M^c) = 1 - P(M) = 1 - 0.10 = 0.90 \]
This allows us to specifically focus on outcomes where the mother is not sick, making it easier to apply our formula later on.
Biostatistics
Biostatistics is a crucial field that applies statistical principles to biological experiments and healthcare studies. It plays an essential role in understanding and making decisions in various medical scenarios, such as disease outbreaks like influenza.
In the context of our exercise, biostatistics helps evaluate the spread of influenza within families by using probability concepts to determine the likelihood of certain health outcomes. Through probability calculations, like the conditional probability seen in our exercise, we gain insights into how the disease could progress and influence family members differently.
Biostatistics involves complex models, but at its core, it relies heavily on probability theory, allowing us to estimate and predict health-related outcomes efficiently. It aids in making informed decisions on healthcare policy and patient management, considering various combinations of health events and their probabilities.

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Most popular questions from this chapter

Piriformis syndrome is a pelvic condition that involves malfunction of the piriformis muscle (a deep buttock muscle), which often causes back and buttock pain with sciatica(pain radiating down the leg). An electrophysiologic test to detect piriformis syndrome involves measuring nerveconduction velocity (NCV) at two nerves in the leg (the tibial and peroneal nerves) with the leg flexed in a specific position. Increases in \(\mathrm{NCV}\) in these nerves are often associated with piriformis syndrome. The resulting test, called the flexion abduction and internal rotation (FAIR) test, is positive if the average \(\mathrm{NCV}\) in these nerves is delayed by \(2+\) seconds relative to normal.A small study compared the FAIR test results with patient self-reports of how they feel on a visual analog scale (VAS) of \(0-10,\) with 0 indicating no pain and 10 very severe pain. The results were as shown in Table 3.17 .Suppose physicians consider the FAIR test the gold standard, with a FAIR test result of \(\geq 2\) defined as a true positive and a FAlR test result of \(<2\) defined as a true negative. Suppose a VAS of \(\leq 4\) is considered a good clinical response based on self-report (a test-negative) and a VAS of \(\geq 5\) is considered a bad clinical response (a test-positive).What is the specificity of the VAS?

Consider a family with a mother, father, and two children. Let \(A_{1}=\left\\{\text { mother has influenzal, } A_{2}=\\{ \text { father has influenzal, }\right.\) \(A_{3}=\left\\{\text { first child has influenzal, } A_{4}=\\{ \text { second child has influ- }\right.\) enzal, \(B=\) lat least one child has influenzal, \(C=\) lat least one parent has influenzal, and \(D=\\{\) at least one person in the family has influenzal.Express \(D\) in terms of \(B\) and \(C .\)

A dominantly inherited genetic disease is identified over several generations of a large family. However, about half the families have dominant disease with complete penetrance, whereby if a parent is affected there is a \(50 \%\) probability that any one offspring will be affected. Similarly, about half the families have dominant disease with reduced penetrance, whereby if a parent is affected there is a \(25 \%\) probability that any one offspring will be affected. Suppose in a particular family one parent and two of the two offspring are affected.What is the probability that the mode of transmission for this particular family is dominant with complete penetrance? Is this a prior probability or a posterior probability?

Suppose an influenza epidemic strikes a city. In \(10 \%\) of families the mother has influenza; in \(10 \%\) of families the father has influenza; and in \(2 \%\) of families both the mother and father have influenza.Based on Problem 3.12, what is the conditional probability that the father has influenza given that the mother has influenza?

The familial aggregation of respiratory disease is a wellestablished clinical phenomenon. However, whether this aggregation is due to genetic or environmental factors or both is somewhat controversial. An investigator wishes to study a particular environmental factor, namely the relationship of cigarette-smoking habits in the parents to the presence or absence of asthma in their oldest child age 5 to 9 years living in the household (referred to below as their offspring). Suppose the investigator finds that (1) if both the mother and father are current smokers, then the probability of their offspring having asthma is \(.15 ;(2)\) if the mother is a current smoker and the father is not, then the probability of their offspring having asthma is \(.13 ;(3)\) if the father is a current smoker and the mother is not, then the probability of their offspring having asthma is \(.05 ;\) and (4) if neither parent is a current smoker, then the probability of their offspring having asthma is .04.Suppose the smoking habits of the parents are independent and the probability that the mother is a current smoker is \(.4,\) whereas the probability that the father is a current smoker is .5. What is the probability that both the father and mother are current smokers?
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