In a city, 5% of people have a certain disease (D) and 95% do not have the disease (D'). A company develops a test that correctly identifies the disease in 98% of cases (T) and incorrectly identifies the disease in 2% of cases (T'). If a person tests positive, what is the probability that they actually have the disease?

Step 1 ：Step 1: Define the events. Let's denote D as the event 'person has the disease' and T as the event 'test is positive'. We are asked to find P(D|T), the probability that a person has the disease given that they tested positive.

Step 2 ：Step 2: Apply Bayes' Theorem: $$P(D|T) = \frac{P(T|D) \cdot P(D)}{P(T)}$$

Step 3 ：Step 3: Calculate P(T|D) and P(D). We know that P(T|D), the probability that the test is positive given that the person has the disease, is 0.98. And P(D), the probability that a random person has the disease, is 0.05.

Step 4 ：Step 4: Calculate P(T). This is the total probability that the test is positive, and it's given by $$P(T) = P(T|D) \cdot P(D) + P(T|D') \cdot P(D')$$ where P(T|D') is the probability that the test is positive given that the person does not have the disease (false positive), and P(D') is the probability that a person does not have the disease.

Step 5 ：Step 5: Substitute the given values into the equation for P(T). We know that P(T|D') is 0.02 and P(D') is 0.95. So, $$P(T) = 0.98 \cdot 0.05 + 0.02 \cdot 0.95 = 0.049$$

Step 6 ：Step 6: Substitute the calculated values of P(T|D), P(D) and P(T) into the Bayes' Theorem equation to find P(D|T). $$P(D|T) = \frac{0.98 \cdot 0.05}{0.049} = 1$$