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

In which
P(B|A) is the probability of event B happening, given that A happened.
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

Intersection of A and B:
Positive test and has the disease, so 90% of 3%

What is the conditional probability that she does, in fact, have the disease

0.2177 = 21.77% conditional probability that she does, in fact, have the disease