Answer:
The probability is 0.00019989
Step-by-step explanation:
We are going to define the conditional probability as :
[tex]P(A/B)=\frac{P(A,B)}{P(B)} \\P(B)> 0[/tex]
Where P(A,B) is the probability of the event (A∩B).
Let's define the following events :
B : ''The HIV test is positive''
A : ''The adult is infected''
Also we define :
[tex]P(A)=1-P(A^{c} )[/tex]
Where [tex]A^{c}[/tex] is the event where A does not occur.
In the exercise :
[tex]P(A)=0.0001\\P(B/A)=0.999\\P(B^{c} /A^{c} )=0.9999[/tex]
Where
[tex]P(B/A)\\P(B^{c}/A^{c})[/tex]
represents the test accuracy
We are ask about P(B).
We write :
[tex]P(A^{c})=1-P(A)=1-0.0001=0.9999\\ P(A^{c})=0.9999[/tex]
[tex]P(B/A)=\frac{P(B,A)}{P(A)} \\P(B,A)=P(B/A).P(A)[/tex]
[tex]P(B/A^{c})=1-P(B^{c}/A^{c})=1-0.9999=0.0001\\ P(B/A^{c})=0.0001[/tex]
[tex]P(B/A^{c})=\frac{P(B,A^{c})}{P(A^{c})} \\[/tex]
[tex]P(B,A^{c})=P(B/A^{c}).P(A^{c})[/tex]
P(B) = P{ [(B∩A)∪[B∩(A^{c})]}
P(B) = P(B∩A) +P[B∩(A^{c})]
Using the conditional equations
[tex]P(B)=P(B/A).P(A)+P(B/A^{c} ).P(A^{c})\\ P(B)=(0.999).(0.0001)+(0.0001).(0.9999)=(1.9989).10^{-4} =0.00019989\\P(B)=0.00019989[/tex]
