We can solve both parts of this problem using Bayes' Theorem and the law of total probability.

Given information:

• The probability a person is infected:
  $P(A) = \frac{1}{200} = 0.005$

• The probability a person is not infected:
  $P(A') = 1 - P(A) = 0.995$

• The probability of testing positive given the person has the virus (true positive):
  $P(B|A) = 0.90$

• The probability of testing positive given the person does not have the virus (false positive):
  $P(B|A') = 0.08$

a) Find $P(A|B)$:

We want the probability that a person has the virus given that they tested positive. This is written as $P(A|B)$, and it can be calculated using Bayes' Theorem:

$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$

Now, we need to calculate $P(B)$, the total probability that the test is positive, which can be found using the law of total probability:

$P(B) = P(B|A)P(A) + P(B|A')P(A')$

Substitute the known values:

$P(B) = (0.90)(0.005) + (0.08)(0.995)$

$P(B) = 0.0045 + 0.0796 = 0.0841$

Now, using Bayes' Theorem:

$P(A|B) = \frac{(0.90)(0.005)}{0.0841}$

$P(A|B) = \frac{0.0045}{0.0841} \approx 0.0535$

Therefore, the probability that a person has the virus given that they tested positive is approximately:

$P(A|B) \approx 0.0535$

b) Find $P(A'|B')$:

We want the probability that a person does not have the virus given that they tested negative, i.e. $P(A'|B')$. Again, we use Bayes' Theorem:

$P(A'|B') = \frac{P(B'|A')P(A')}{P(B')}$

First, we calculate $P(B')$, the probability of testing negative. Using the law of total probability:

$P(B') = P(B'|A)P(A) + P(B'|A')P(A')$

Where:

• $P(B'|A) = 1 - P(B|A) = 1 - 0.90 = 0.10$
• $P(B'|A') = 1 - P(B|A') = 1 - 0.08 = 0.92$

Substitute the known values:

$P(B') = (0.10)(0.005) + (0.92)(0.995)$

$P(B') = 0.0005 + 0.9144 = 0.9149$

Now, using Bayes' Theorem:

$P(A'|B') = \frac{(0.92)(0.995)}{0.9149}$

$P(A'|B') = \frac{0.914}{0.9149} \approx 0.9990$

Therefore, the probability that a person does not have the virus given that they tested negative is approximately:

$P(A'|B') \approx 0.9990$

Final answers:

• $P(A|B) \approx 5.35 \%$
• $P(A'|B') \approx 99.90 \%$

Would you like further clarification on any steps or details in the calculations?

5 Relative Questions:

1. How would the probabilities change if the accuracy of the test (true positive rate) increased?
2. What happens to $P(A|B)$ if the false positive rate decreases?
3. Can you calculate $P(B|A')$ for different levels of infection rates?
4. How does Bayes' Theorem apply to real-world diagnostics in other fields, like cancer testing?
5. What would be the effect of a different infection rate on the result?

Tip:

Bayes' Theorem is powerful for medical diagnostics, helping assess how reliable a positive or negative test result really is based on test accuracy and overall infection rates.