Questions: A certain disease has an incidence rate of 0.2%. If the false negative rate is 5% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease.

A certain disease has an incidence rate of 0.2%. If the false negative rate is 5% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease.

Transcript text: A certain disease has an incidence rate of $0.2 \%$. If the false negative rate is $5 \%$ and the false positive rate is $1 \%$, compute the probability that a person who tests positive actually has the disease. $\square$

Solution

Solution Steps

Step 1: Define the Given Values

We are given the following values:

Incidence rate of the disease: $ P(Disease) = 0.002 $
False negative rate: $ P(Negative | Disease) = 0.05 $
False positive rate: $ P(Positive | No Disease) = 0.01 $

Step 2: Calculate the Probability of a Positive Test Given the Disease

The probability of testing positive given that the person has the disease is: \[ P(Positive | Disease) = 1 - P(Negative | Disease) = 1 - 0.05 = 0.95 \]

Step 3: Calculate the Overall Probability of Testing Positive

To find $ P(Positive) $, we use the law of total probability: \[ P(Positive) = P(Positive | Disease) \cdot P(Disease) + P(Positive | No Disease) \cdot P(No Disease) \] Substituting the known values: \[ P(No Disease) = 1 - P(Disease) = 1 - 0.002 = 0.998 \] Thus, \[ P(Positive) = (0.95 \cdot 0.002) + (0.01 \cdot 0.998) = 0.0019 + 0.00998 = 0.01188 \]

Step 4: Apply Bayes' Theorem

Now, we can calculate the probability of having the disease given a positive test result using Bayes' theorem: \[ P(Disease | Positive) = \frac{P(Positive | Disease) \cdot P(Disease)}{P(Positive)} \] Substituting the values: \[ P(Disease | Positive) = \frac{0.95 \cdot 0.002}{0.01188} \approx 0.1599326599326599 \]