Answer:
0.9378 = 93.78% probability that a randomly selected person has diabetes, given that his test is positive.
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: Person has diabetes.
Probability of a positive test:
0.95 out of 0.1(person has diabetes).
0.007 out of 1 - 0.1 = 0.9(person does not has diabetes). So
[tex]P(A) = 0.95*0.1 + 0.007*0.9 = 0.1013[/tex]
Probability of a positive test and having diabetes:
0.95 out of 0.1. So
[tex]P(A \cap B) = 0.95*0.1 = 0.095[/tex]
What is the probability that a randomly selected person has diabetes, given that his test is positive?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.095}{0.1013} = 0.9378[/tex]
0.9378 = 93.78% probability that a randomly selected person has diabetes, given that his test is positive.
