Suppose that 3% of the population ha a certain disease. We have a test for the disease. 97% of those who actually have the disease will test positive. 12% of those who do not have the disease will also test positive. What is the probability that a randomly selected person has the disease and tests positive?

Probability of both/events occurring is found by multiplying the probabilities of the individual events.

.03 * .97 = ?

To solve this problem, we can use Bayes' theorem. Bayes' theorem helps us calculate conditional probabilities based on prior probabilities and test results.

Let's break down the probabilities given in the problem:

- P(D) = 0.03: The probability of a randomly selected person having the disease is 3%.
- P(Pos|D) = 0.97: The probability of testing positive given that the person has the disease is 97%.
- P(Pos|¬D) = 0.12: The probability of testing positive given that the person does not have the disease is 12%.

Now, we want to calculate the probability of a randomly selected person having the disease and testing positive, denoted as P(D|Pos).

Using Bayes' theorem, we have:

P(D|Pos) = (P(Pos|D) * P(D)) / P(Pos)

To calculate P(Pos), we need to consider both true positives (those who have the disease and test positive) and false positives (those who do not have the disease but test positive).

P(Pos) = [P(Pos|D) * P(D)] + [P(Pos|¬D) * P(¬D)]

To find P(¬D) (the probability of not having the disease), we can subtract P(D) from 1 since it's mutually exclusive (a person either has the disease or does not have it).

P(¬D) = 1 - P(D)

Now, let's plug in the values and calculate the probabilities:

P(¬D) = 1 - 0.03 = 0.97

P(Pos) = [(0.97 * 0.03) + (0.12 * 0.97)]

P(D|Pos) = (0.97 * 0.03) / P(Pos)

Simplifying the equation:

P(D|Pos) = (0.0291) / P(Pos)

Now, we can substitute the value of P(Pos) and calculate P(D|Pos).