A certain disease has an incidence rate of 0.1%. If the false negative rate is 5% and the false positive rate is 5%, compute the probability that a person who tests positive actually has the disease.

1.87Incorrect  

Give your answer accurate to at least 3 decimal places


To solve this problem, we can use Bayes' theorem to find the probability that a person actually has the disease given that they tested positive. Let's define the information provided:

- **P(Disease)** = 0.1% = 0.001 (incidence rate of the disease)
- **False negative rate** = 5%, meaning that if a person has the disease, there is a 5% chance the test will incorrectly indicate they don't have it.
- **False positive rate** = 5%, meaning that if a person does not have the disease, there is a 5% chance the test will incorrectly indicate they do have it.

Let:
- \( P(D) \): Probability of having the disease = 0.001
- \( P(\text{No Disease}) \): Probability of not having the disease = 1 - 0.001 = 0.999
- \( P(\text{Positive} | \text{Disease}) \): Probability of testing positive given having the disease = \( 1 - \text{False Negative Rate} = 1 - 0.05 = 0.95 \)
- \( P(\text{Positive} | \text{No Disease}) \): Probability of testing positive given not having the disease = False Positive Rate = 0.05

We want to find \( P(\text{Disease} | \text{Positive}) \), the probability that a person actually has the disease given that they tested positive.

Using Bayes' theorem:

\[
P(\text{Disease} | \text{Positive}) = \frac{P(\text{Positive} | \text{Disease}) \times P(\text{Disease})}{P(\text{Positive})}
\]

where \( P(\text{Positive}) \) is the total probability of testing positive, given by:

\[
P(\text{Positive}) = P(\text{Positive} | \text{Disease}) \times P(\text{Disease}) + P(\text{Positive} | \text{No Disease}) \times P(\text{No Disease})
\]

Substituting the values:

1. Calculate \( P(\text{Positive}) \):

   \[
   P(\text{Positive}) = (0.95 \times 0.001) + (0.05 \times 0.999)
   \]
   \[
   = 0.00095 + 0.04995 = 0.0509
   \]

2. Substitute into Bayes' theorem:

   \[
   P(\text{Disease} | \text{Positive}) = \frac{0.95 \times 0.001}{0.0509}
   \]
   \[
   = \frac{0.00095}{0.0509} \approx 0.0187
   \]

So, the probability that a person who tests positive actually has the disease is approximately **0.019** (or 1.87%).

Would you like a deeper explanation of the steps?

### Related Questions

1. How does changing the incidence rate affect the probability of actually having the disease when testing positive?
2. What happens if we lower the false negative rate to 2%?
3. How would increasing the false positive rate impact the result?
4. How does Bayes' theorem help in medical testing?
5. How can we interpret a low probability of having the disease despite a positive test result?

**Tip:** Small changes in the false positive rate can greatly affect probabilities in low-incidence conditions.

A certain disease has an incidence rate of 0.1%. If the false negative rate is 5% and the false positive rate is 5%, compute the probability that a person who tests positive actually has the disease.

Discover how to compute the probability that a person who tests positive for a disease actually has it, given a 0.1% incidence rate, a 5% false negative rate, and a 5% false positive rate. This problem involves Bayes' theorem and conditional probability, ideal for college-level probability studies.

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