Accurate Disease Test Results

Solution & Explanation

To solve the problem of determining the probability that Bob does not have the disease despite receiving a positive test result, we can use Bayes' Theorem. Bayes' Theorem allows us to update our beliefs about the probability of an event based on new evidence. In this context, the event of interest is whether Bob actually has the disease given a positive test result.

Given Data:

• False Positive Rate (FPR): 1% or 0.01
• False Negative Rate (FNR): 15% or 0.15
• Bob's Test Result: Positive
• Friends' Test Results: Six friends tested negative

Assumptions:

• The prevalence of the disease in the population is not given, but we'll assume a low prevalence as is common for many diseases.
• The tests are independent.
• We assume that Bob and his friends share the same health condition.

Steps to Solution:

1. Define the Events:

   • Let $D$ be the event that Bob has the disease.
   • Let $\neg D$ be the event that Bob does not have the disease.
   • Let $T^+$ be the event that Bob tests positive.

2. Calculate the Probability of Testing Positive Given Bob has the Disease (True Positive Rate):

   • $P(T^+|D) = 1 - \text{FNR} = 0.85$

3. Calculate the Probability of Testing Positive Given Bob does not have the Disease (False Positive Rate):

   • $P(T^+|\neg D) = \text{FPR} = 0.01$

4. Assume a Prior Probability of Having the Disease:

   • Without specific prevalence data, assume a low prior probability, say $P(D) = 0.001$ (1 in 1000), as diseases are generally rare.

5. Calculate the Probability of Bob Testing Positive:

   Using the law of total probability:

   $P(T^+) = P(T^+|D) \cdot P(D) + P(T^+|\neg D) \cdot P(\neg D)$

   $P(T^+) = 0.85 \cdot 0.001 + 0.01 \cdot 0.999 = 0.00085 + 0.00999 = 0.01084$

6. Apply Bayes' Theorem to Find the Probability that Bob does not have the Disease Given a Positive Test Result:

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

   $P(\neg D|T^+) = \frac{0.01 \cdot 0.999}{0.01084} = \frac{0.00999}{0.01084} \approx 0.921$

Conclusion:

Given the assumptions and calculations, the probability that Bob does not have the disease despite a positive test result is approximately 92.1%. This highlights the importance of considering false positives, especially when the prevalence of the disease is low.