Question

A laboratory blood test is 99% effective in diagnosing a certain disease when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy persons tested (i.e., if a healthy person is tested, then with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Hint:

Bayes' Theorem is useful for calculating conditional probabilities, especially in problems involving false positives or negatives.

Answer 1

Step 1: Applying Bayes' Theorem.
We can solve this using Bayes' Theorem, which states: \[ P(\text{Disease} | \text{Positive}) = \frac{P(\text{Positive} | \text{Disease}) \cdot P(\text{Disease})}{P(\text{Positive})} \] Step 2: Calculating the components.
- \( P(\text{Disease}) = 0.001 \) (0.1% of the population has the disease) - \( P(\text{Positive} | \text{Disease}) = 0.99 \) (the test is 99% effective) - \( P(\text{Positive} | \text{Healthy}) = 0.005 \) (false positive rate) - \( P(\text{Healthy}) = 1 - P(\text{Disease}) = 0.999 \) The total probability of a positive test result is given by: \[ P(\text{Positive}) = P(\text{Positive} | \text{Disease}) \cdot P(\text{Disease}) + P(\text{Positive} | \text{Healthy}) \cdot P(\text{Healthy}) \] Substitute the values: \[ P(\text{Positive}) = (0.99 \cdot 0.001) + (0.005 \cdot 0.999) = 0.00099 + 0.004995 = 0.005985 \] Step 3: Substituting into Bayes' Theorem.
Now, we can substitute all the values into Bayes' Theorem: \[ P(\text{Disease} | \text{Positive}) = \frac{0.99 \cdot 0.001}{0.005985} = \frac{0.00099}{0.005985} \approx 0.165 \] Step 4: Conclusion.
Therefore, the probability that a person has the disease given that their test result is positive is approximately 0.165, or 16.5%.