Math Problem Statement

In a hospital, tests are used to determine whether a patient has a disease or not. Suppose 1% of the population has a certain disease D. A medical test is used to determine whether one has the disease. It is not 100% accurate in the following ways: 95% of people with the disease test positive (true positive), 2% of people without the disease test positive (false positive). If a person tests positive, what is the probability they actually have the disease?

Solution

This is a classic case of conditional probability, and we can use Bayes' Theorem to solve it.

Given:

  • 1% of the population has the disease DD (i.e., P(D)=0.01P(D) = 0.01).
  • The probability that the test is positive given the person has the disease is P(PositiveD)=0.95P(\text{Positive} | D) = 0.95 (true positive rate).
  • The probability of a false negative (test is negative given the person has the disease) is 5%, which is P(NegativeD)=0.05P(\text{Negative} | D) = 0.05.
  • The probability that the test is negative given the person does not have the disease is P(Negative¬D)=0.98P(\text{Negative} | \neg D) = 0.98.
  • The probability of a false positive (test is positive given the person does not have the disease) is 2%, which is P(Positive¬D)=0.02P(\text{Positive} | \neg D) = 0.02.

We are asked to find:

P(DPositive)P(D | \text{Positive}), the probability that a person has the disease given that the test result is positive.

Bayes' Theorem:

P(DPositive)=P(PositiveD)P(D)P(Positive)P(D | \text{Positive}) = \frac{P(\text{Positive} | D) \cdot P(D)}{P(\text{Positive})}

Where: P(Positive)=P(PositiveD)P(D)+P(Positive¬D)P(¬D)P(\text{Positive}) = P(\text{Positive} | D) \cdot P(D) + P(\text{Positive} | \neg D) \cdot P(\neg D) P(¬D)=1P(D)=0.99P(\neg D) = 1 - P(D) = 0.99.

Now, let's calculate:

  1. P(Positive)=(0.950.01)+(0.020.99)P(\text{Positive}) = (0.95 \cdot 0.01) + (0.02 \cdot 0.99) P(Positive)=0.0095+0.0198=0.0293P(\text{Positive}) = 0.0095 + 0.0198 = 0.0293

  2. Now, apply Bayes' Theorem: P(DPositive)=0.950.010.0293=0.00950.02930.324P(D | \text{Positive}) = \frac{0.95 \cdot 0.01}{0.0293} = \frac{0.0095}{0.0293} \approx 0.324

Answer:

The probability that the person actually has the disease given a positive test result is approximately 0.324 (or 32.4%).

Would you like any further details or explanations?

Here are 5 related questions you might be interested in:

  1. How does increasing the disease prevalence affect the result?
  2. What happens to the probability if the test's false positive rate decreases?
  3. How does Bayes' Theorem apply in other medical contexts?
  4. Can we calculate the probability of not having the disease given a negative test?
  5. How does sample size affect confidence in medical tests?

Tip: Always be cautious with test results, especially when the disease prevalence is low, as false positives can be common!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Conditional Probability
Bayes' Theorem

Formulas

P(D | Positive) = (P(Positive | D) * P(D)) / P(Positive)
P(Positive) = P(Positive | D) * P(D) + P(Positive | ¬D) * P(¬D)

Theorems

Bayes' Theorem

Suitable Grade Level

College level or Advanced High School (Grades 11-12)