Math Problem Statement

  1. A disease afflicts 1 person in 10 in a population. A test for this disease results in 5% false positives. Assume that a random person is tested for the disease (this person is not selected for the test because there are other symptoms indicating the presence of the disease nor are there any special signs of being disease free). Assume that there are no false negatives from this test. A person tests positive. What is the probability that the person truly has the disease? B. Now assume that false negatives are 15%. What is the probability that the person truly has the disease?

Solution

Part A: No False Negatives

Let’s define the given data first:

  • The disease afflicts 1 person in 10, i.e., the probability that a person has the disease (prior probability) is: P(D)=110=0.1P(D) = \frac{1}{10} = 0.1
  • The probability that a person does not have the disease is: P(¬D)=1P(D)=10.1=0.9P(\neg D) = 1 - P(D) = 1 - 0.1 = 0.9
  • The test results in 5% false positives, i.e., the probability of testing positive given the person does not have the disease is: P(T+¬D)=0.05P(T^+ | \neg D) = 0.05
  • There are no false negatives, meaning if a person has the disease, the test will always come back positive: P(T+D)=1P(T^+ | D) = 1

We need to find the probability that a person has the disease given that they test positive, i.e., P(DT+)P(D | T^+).

Using Bayes’ Theorem: P(DT+)=P(T+D)P(D)P(T+)P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} Where P(T+)P(T^+) (the total probability of testing positive) is given by: P(T+)=P(T+D)P(D)+P(T+¬D)P(¬D)P(T^+) = P(T^+ | D)P(D) + P(T^+ | \neg D)P(\neg D)

Substitute the values: P(T+)=(1)(0.1)+(0.05)(0.9)=0.1+0.045=0.145P(T^+) = (1)(0.1) + (0.05)(0.9) = 0.1 + 0.045 = 0.145

Now, calculate P(DT+)P(D | T^+): P(DT+)=(1)(0.1)0.145=0.10.1450.6897P(D | T^+) = \frac{(1)(0.1)}{0.145} = \frac{0.1}{0.145} \approx 0.6897

Thus, the probability that the person truly has the disease given that they tested positive is approximately 68.97%.


Part B: With 15% False Negatives

Now, we are given that there are 15% false negatives, meaning the probability that a person tests negative given that they have the disease is: P(TD)=0.15P(T^- | D) = 0.15 Thus, the probability that a person tests positive given that they have the disease is: P(T+D)=10.15=0.85P(T^+ | D) = 1 - 0.15 = 0.85

Using Bayes’ Theorem again: P(DT+)=P(T+D)P(D)P(T+)P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)}

We need to recalculate P(T+)P(T^+) with the new information: P(T+)=P(T+D)P(D)+P(T+¬D)P(¬D)P(T^+) = P(T^+ | D)P(D) + P(T^+ | \neg D)P(\neg D) Substitute the values: P(T+)=(0.85)(0.1)+(0.05)(0.9)=0.085+0.045=0.13P(T^+) = (0.85)(0.1) + (0.05)(0.9) = 0.085 + 0.045 = 0.13

Now, calculate P(DT+)P(D | T^+): P(DT+)=(0.85)(0.1)0.13=0.0850.130.6538P(D | T^+) = \frac{(0.85)(0.1)}{0.13} = \frac{0.085}{0.13} \approx 0.6538

Thus, the probability that the person truly has the disease given that they tested positive with 15% false negatives is approximately 65.38%.


Would you like more details on any step, or do you have any questions?

Here are some related questions to consider:

  1. How does the sensitivity of a test impact the probability of disease?
  2. What happens to the probability if the false positive rate is reduced?
  3. How would the prior probability of the disease (prevalence) affect the results?
  4. Can Bayes' Theorem be applied to other real-world scenarios like drug testing?
  5. How does the test accuracy affect public health policies?

Tip: False negatives are critical in disease testing because missing a case could lead to further spread of the disease or delayed treatment!

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Math Problem Analysis

Mathematical Concepts

Probability
Bayes' Theorem
False Positives
False Negatives

Formulas

Bayes' Theorem: P(A | B) = (P(B | A) * P(A)) / P(B)
Total Probability of Testing Positive: P(T^+) = P(T^+ | D)P(D) + P(T^+ | ¬D)P(¬D)

Theorems

Bayes' Theorem

Suitable Grade Level

Undergraduate / Advanced High School