Using Bayes’ Theorem to Find the Probability of Disease Given a Positive Test

Q22. In a certain city, 1% of the population has a rare disease. A diagnostic test for the disease gives a true positive rate of 95% (the probability that the test is positive if the person has the disease) and a false positive rate of 5% (the probability that the test is positive if the person does not have the disease). If a person tests positive, what is the probability that they actually have the disease? (Hint: Use Bayes’ Theorem to solve.)

Background

Topic: Conditional Probability & Bayes’ Theorem

This question tests your understanding of how to use Bayes’ Theorem to update probabilities based on new evidence—in this case, the probability that a person actually has a disease given a positive test result.

Key Terms and Formulas

• Prevalence (Prior Probability): $P(D)$ = Probability a person has the disease = 0.01

• True Positive Rate (Sensitivity): $P(+|D)$ = Probability test is positive given disease = 0.95

• False Positive Rate: $P(+|\neg D)$ = Probability test is positive given no disease = 0.05

• Bayes’ Theorem:

$P(D|+) = \frac{P(+|D) \cdot P(D)}{P(+|D) \cdot P(D) + P(+|\neg D) \cdot P(\neg D)}$

• $P(D|+)$ = Probability of disease given a positive test

• $P(\neg D)$ = Probability a person does not have the disease = 1 - P(D)$

Step-by-Step Guidance

1. Identify and write down all the given probabilities:

   • $P(D) = 0.01$

   • $P(\neg D) = 1 - 0.01 = 0.99$

   • $P(+|D) = 0.95$

   • $P(+|\neg D) = 0.05$

2. Write out Bayes’ Theorem for this context:

   $P(D|+) = \frac{P(+|D) \cdot P(D)}{P(+|D) \cdot P(D) + P(+|\neg D) \cdot P(\neg D)}$

3. Substitute the known values into the formula:

   $P(D|+) = \frac{0.95 \times 0.01}{0.95 \times 0.01 + 0.05 \times 0.99}$

4. Calculate the numerator: $0.95 \times 0.01$

5. Calculate the denominator: $0.95 \times 0.01 + 0.05 \times 0.99$

Try solving on your own before revealing the answer!