A certain disease occurs in 4% of the population. A blood test for this disease shows a false positive 12% of the time. That means a test will show positive given that the person does not have the disease. What is the probability that a person tests positive but does not have the disease?

Question

A certain disease occurs in 4% of the population. A blood test for this disease shows a false positive 12% of the time. That means a test will show positive given that the person does not have the disease. What is the probability that a person tests positive but does not have the disease?

Note: The answer is 0.1152, but I don't know how they get it. Please help me and show the steps please. Thanks.

Answer 1

.12 times .96

false positive times % who don't have the disease.

Answer 2

To solve this problem, we can use Bayes' theorem. Bayes' theorem allows us to calculate the probability of an event, given some prior information.

Let's define the events:
A = person has the disease
B = person tests positive

We are given the following probabilities:
P(A) = 0.04 (probability of having the disease)
P(B|A') = 0.12 (probability of testing positive given that the person does not have the disease)

We want to find the probability of a person testing positive but not having the disease, represented as P(A'|B).

Using Bayes' theorem, we have:
P(A'|B) = (P(B|A') * P(A')) / P(B)

Now let's calculate each term:

1. P(B|A'): This is given as 0.12.
2. P(A'): This is the complement of P(A), which means 1 - P(A). Since P(A) is 0.04, P(A') = 1 - 0.04 = 0.96.
3. P(B): To calculate this, we need to consider the two possible situations: a person can test positive and actually have the disease (A and B), or a person can test positive without having the disease (A' and B).

P(B) = P(A and B) + P(A' and B)

P(A and B): This is the probability of having the disease multiplied by the probability of testing positive given that the person has the disease.
P(A and B) = P(A) * P(B|A) = 0.04 * 1 = 0.04

P(A' and B): This is the probability of not having the disease multiplied by the probability of testing positive given that the person does not have the disease.
P(A' and B) = P(A') * P(B|A') = 0.96 * 0.12 = 0.1152

Now we can calculate P(B):
P(B) = P(A and B) + P(A' and B) = 0.04 + 0.1152 = 0.1552

Finally, we can substitute these values into the formula for P(A'|B):
P(A'|B) = (P(B|A') * P(A')) / P(B) = (0.12 * 0.96) / 0.1552 = 0.1152

Therefore, the probability that a person tests positive but does not have the disease is 0.1152.