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Diseases \(D_{1}, D_{2}\), and \(D_{3}\) cause symptom A with probabilities \(0.5,0.7\), and \(0.8\), respectively. If \(5 \%\) of a population have disease \(D_{1}, 2 \%\) have disease \(D_{2}\), and \(3.5 \%\) have disease \(D_{3}\), what percent of the population have symptom A? Assume that the only possible causes of symptom A are \(D_{1}, D_{2}\), and \(D_{3}\) and that no one carries more than one of these three diseases.

Short Answer

Expert verified
Therefore, 6.7% of the population have symptom A.

Step by step solution

01

Calculate percentage with disease \(D_{1}\) and symptom A

Multiply the population percentage with disease \(D_{1}\) (5%) by the probability of \(D_{1}\) causing symptom A (0.5). This will give us \(5 \% \times 0.5 = 2.5 \% \)
02

Calculate percentage with disease \(D_{2}\) and symptom A

Similarly, for disease \(D_{2}\), multiply its population percentage (2%) by the symptom A causing probability (0.7). This equals to \(2 \% \times 0.7 = 1.4 \% \)
03

Calculate percentage with disease \(D_{3}\) and symptom A

Do the same for disease \(D_{3}\) to find its contribution to symptom A. This will be \(3.5 \% \times 0.8 = 2.8 \% \)
04

Sum percentages

Sum up these percentage values to find the total percentage of the population with symptom A. This will be \(2.5 \% + 1.4 \% + 2.8 \% = 6.7 \% \)

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Key Concepts

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

Disease Probability
Understanding disease probability is an essential aspect of health statistics. To determine how many people might exhibit symptoms linked to diseases, we start with the probability of a symptom appearing due to a particular disease. If a disease is present in a population, the probability of it causing a specific symptom is key to predicting broader health outcomes.

In the context of our exercise, the given probabilities indicate how likely it is for symptoms (like symptom A) to present when diseases such as \(D_1, D_2,\) or \(D_3\) are present:
  • Disease \(D_1\) can cause symptom A 50% of the time,
  • Disease \(D_2\) can cause symptom A 70% of the time,
  • Disease \(D_3\) can cause symptom A 80% of the time.
Calculating these disease probabilities allows us to infer the prevalence of certain symptoms in a population, which is an important task in public health.
Symptom Analysis
Symptom analysis involves assessing which symptoms are likely to occur due to certain diseases. This process helps determine how symptoms distribute across a population. It's important to evaluate each disease in relationship to its symptoms.

In our exercise, for instance, we calculate how much each disease contributes to symptom A across the general population:
  • The occurrence rate of disease \(D_1\) in the population is 5%. Multiplying this by the probability that \(D_1\) causes symptom A (0.5) yields a 2.5% rate of symptom A attributable to \(D_1\).
  • For disease \(D_2\), with a population rate of 2% and a symptom causing probability of 0.7, symptom A occurs in 1.4% of the population.
  • Similarly, for disease \(D_3\), its 3.5% prevalence and 0.8 probability of causing symptom A result in a 2.8% occurrence of the symptom in the population.
Analyzing symptoms in this systematic way enables healthcare professionals to more effectively diagnose and treat conditions.
Population Health Statistics
Population health statistics give a broad view of how health issues affect a given group of people. These statistics are crucial in informing public health initiatives and resource allocation.

By adding together the probabilities of each disease leading to symptom A, we learn what portion of the population experiences this symptom. In our solved problem, the total is 6.7%, which indicates that 6.7% of the population exhibits symptom A due to the presence of diseases \(D_1, D_2,\) and \(D_3\).
  • This overall percentage is the sum total of those with different diseases showing the same symptom.
  • Such statistical analysis helps in anticipating healthcare needs and in finding ways to prevent the spread of diseases.
Understanding these statistics helps public health officials prioritize health interventions and make decisions grounded in data.

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Most popular questions from this chapter

Adam and three of his friends are playing bridge. (a) If, in holding certain hand, Adam announces that he has a king, what is the probability that he has at least one more king? (b) If, for some other hand, Adam announces that he has the king of diamonds, what is the probability that he has at least one more king? Compare parts (a) and (b) and explain why the answers are not the same.

Two persons arrive at a train station, independently of each other, at random times between 1:00 P.M. and 1:30 P.M. What is the probability that one will arrive between 1:00 P.M. and 1:12 P.M., and the other between 1:17 P.M. and \(1: 30\) P.M.?

An experiment consists of first tossing an unbiased coin and then rolling a fair die. If we perform this experiment successively, what is the probability of obtaining a heads on the coin before a 1 or 2 on the die?

Solve the following problem, from the "Ask Marilyn" column of Parade Magazine, August 9, 1992 . Three of us couples are going to Lava Hot Springs next weekend. We're staying two nights, and we've rented two studios, because each holds a maximum of only four people. One couple will get their own studio on Friday, a different couple on Saturday, and one couple will be out of luck. We'll draw straws to see which are the two lucky couples. I told my wife we should just draw once, and the loser would be the couple out of luck both nights. I figure we'll have a two-out-of-three \(\left(66 \frac{2}{3} \%\right)\) chance of winning one of the nights to ourselves. But she contends that we should draw straws twice- first on Friday and then, for the remaining two couples only, on Saturdayreasoning that a one-in-three \(\left(33 \frac{1}{3} \%\right)\) chance for Friday and a one-in-two \((50 \%)\) chance for Saturday will give us better odds. Which way should we go?

A fair die is thrown twice. If the second outcome is 6 , what is the probability that the first one is 6 as well?

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