b) In a large city, 8% of the inhabitants have contracted a particular disease. A test for this disease is positive in 80% of people who have the disease and is negative in 80% of people who do not have the disease. What is the probability that a person for whom the test result is positive has the disease?

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We shall denote by $A$ and $B$ the following events:

Event $A$ : A person has a disease;

Event $B$ : A test for a person is positive;

The aim is to find a conditional probability $p(A|B)$ . Namely, we have (https://en.wikipedia.org/wiki/Conditional_probability):

$p(A|B)=\frac{p(A\cap B)}{p(B)}$

By direct computations we have: $p(A\cap B)=0.08\cdot0.8=0.064$

In order to compute $p(B)$ we use the law of total probability. Namely, we have:

$p(B)=0.08\cdot0.8+0.92\cdot0.2=0.248$

Thus, we receive:

$p(A|B)=0.2581$ (it is rounded to 4 decimal places)

Answer: 0.2581 (it is rounded to 4 decimal places)