Respuesta :
Answer:
0.2177 = 21.77% conditional probability that she does, in fact, have 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: Test positive
Event B: Has the disease
Probability of a positive test:
90% of 3%(has the disease).
1 - 0.9 = 0.1 = 10% of 97%(does not have the disease). So
[tex]P(A) = 0.90*0.03 + 0.1*0.97 = 0.124[/tex]
Intersection of A and B:
Positive test and has the disease, so 90% of 3%
[tex]P(A \cap B) = 0.9*0.03 = 0.027[/tex]
What is the conditional probability that she does, in fact, have the disease
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.027}{0.124} = 0.2177[/tex]
0.2177 = 21.77% conditional probability that she does, in fact, have the disease
