Math Problem Statement

Considere um teste de laboratório para detectar uma determinada doença. Definem-se os seguintes eventos: A = pessoa tem a doença. B = teste teve resultado positivo para a doença. Sabe-se que: P(B | A) = 0.95 P(B | A^c) = 0.001 e que a proporção de doentes da população é 0.0017. Qual a probabilidade de que uma pessoa tenha de fato a doença dado que o teste foi positivo?

Solution

The problem is asking for the probability that a person actually has a disease given that they tested positive for it. This is a classic example of applying Bayes' Theorem. Let's define the events:

  • AA: The person has the disease.
  • BB: The test result is positive for the disease.

We are given:

  • P(BA)=0.95P(B | A) = 0.95: The probability of testing positive given that a person has the disease.
  • P(BAc)=0.001P(B | A^c) = 0.001: The probability of testing positive given that a person does not have the disease.
  • P(A)=0.0017P(A) = 0.0017: The proportion of people in the population who have the disease.

We want to find P(AB)P(A | B), the probability that a person has the disease given that they tested positive.

Applying Bayes' Theorem

Bayes' Theorem is given by:

P(AB)=P(BA)P(A)P(B)P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)}

Where P(B)P(B) is the total probability of a positive test result, which can be found using the law of total probability:

P(B)=P(BA)P(A)+P(BAc)P(Ac)P(B) = P(B | A) \cdot P(A) + P(B | A^c) \cdot P(A^c)

Now let's compute the values:

P(Ac)=1P(A)=10.0017=0.9983P(A^c) = 1 - P(A) = 1 - 0.0017 = 0.9983

P(B)=(0.95×0.0017)+(0.001×0.9983)P(B) = (0.95 \times 0.0017) + (0.001 \times 0.9983)

Calculating P(B)P(B)

Let's compute this step-by-step.The total probability of a positive test result, P(B)P(B), is approximately 0.00261330.0026133.

Calculating P(AB)P(A | B)

Now we use Bayes' Theorem to calculate P(AB)P(A | B):

P(AB)=P(BA)P(A)P(B)P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)}

Substitute the values:

P(AB)=0.95×0.00170.0026133P(A | B) = \frac{0.95 \times 0.0017}{0.0026133}

Let's compute this.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can guide you through the final calculation manually:

Manual Calculation

To calculate P(AB)P(A | B):

P(AB)=0.95×0.00170.0026133P(A | B) = \frac{0.95 \times 0.0017}{0.0026133}

  1. Multiply 0.95×0.0017=0.0016150.95 \times 0.0017 = 0.001615.
  2. Then divide by 0.00261330.0026133.

So, P(AB)0.0016150.00261330.618P(A | B) \approx \frac{0.001615}{0.0026133} \approx 0.618.

This means the probability that a person actually has the disease given a positive test result is approximately 61.8%.

If you have more questions or want further details, feel free to ask!

Related Questions

  1. What is the significance of Bayes' Theorem in medical testing?
  2. How does the accuracy of a test impact the post-test probability?
  3. What are false positives and false negatives in the context of medical testing?
  4. How can the sensitivity and specificity of a test be interpreted?
  5. How does the prevalence of a disease affect the probability of actually having the disease after a positive test?

Tip

Understanding the base rate fallacy can help in interpreting probabilities correctly, especially in medical diagnostics where rare conditions are involved.

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Math Problem Analysis

Mathematical Concepts

Probability
Bayes' Theorem

Formulas

P(A | B) = (P(B | A) * P(A)) / P(B)
P(B) = P(B | A) * P(A) + P(B | A^c) * P(A^c)

Theorems

Bayes' Theorem

Suitable Grade Level

College/Advanced High School