Answer:
0.00002 = 0.002% probability of actually having 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: Having the disease
Probability of having a positive test:
0.05 of 1 - 0.000001(false positive)
0.99 of 0.000001 positive. So
[tex]P(A) = 0.05*(1 - 0.000001) + 0.99*0.000001 = 0.05000094[/tex]
Probability of a positive test and having the disease:
0.99 of 0.000001. So
[tex]P(A \cap B) = 0.99*0.000001 = 9.9 \times 10^{-7}[/tex]
What is the probability of actually having the disease
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{9.9 \times 10^{-7}}{0.05000094} = 0.00002[/tex]
0.00002 = 0.002% probability of actually having the disease
