Respuesta :
Answer:
0.8974 = 89.74% probability that the person actually has the disease.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Positive test.
Event B: Has the disease.
Probability of a positive test:
0.7 of 0.2(has the disease).
0.02 of 100 - 20 = 80%(does not have the disease). Thus:
[tex]P(A) = 0.7*0.2 + 0.02*0.8 = 0.156[/tex]
Probability of a positive test and having the disease:
0.7 of 0.2. So
[tex]P(A \cap B) = 0.7*0.2 = 0.14[/tex]
What is the probability that the person actually has the disease?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.14}{0.156} = 0.8974[/tex]
0.8974 = 89.74% probability that the person actually has the disease.
