Answer:
0.1356
Explanation:
This is a probability exercise. Let's define some probability concepts.
Given two events A and B :
(A∩B) = (A,B)
(A,B) is the intersection event where A and B occur both at the same time.
We define [tex]P(A/B)[/tex] as the conditional probability '' The probability of the event A given that we know that the event B occurred'' as :
[tex]P(A/B)=\frac{P(A,B)}{P(B)}[/tex]
Where [tex]P(B)>0[/tex]
Now, if A is an event and [tex]A'[/tex] is its complement ⇒
[tex]P(A)=1-P(A')[/tex]
Finally we define the probability of the union between two events A and B :
P(A∪B) = P(A) + P(B) - P(A,B)
If the events A and B are independent between them ⇒ P(A,B) = 0 ⇒
P(A∪B) = P(A) + P(B)
Let's define the following events for this exercise :
D : ''People taking this test that have the disease''
[tex]P(D)=0.89[/tex]
[tex]P(D')=1-P(D)=1-0.89=0.11[/tex]
[tex]P(D')=0.11[/tex]
P : ''The test is positive''
[tex]P(P/D)=0.97[/tex]
[tex]P(P'/D)=1-P(P/D)=1-0.97=0.03[/tex]
[tex]P(P'/D)=0.03[/tex]
And [tex]P(P'/D')=0.99[/tex]
We are looking the probability of [tex]P(P')[/tex]
P(P') = P [(P'∩D) ∪ (P'∩D')]
Given that this events are independent between them :
[tex]P(P')=P(P',D)+P(P',D')[/tex] (I)
Let's write the conditionals for this problem :
[tex]P(P'/D)=\frac{P(P',D)}{P(D)}[/tex] ⇒
[tex]P(P',D)=P(P'/D).P(D)[/tex]
[tex]P(P',D)=(0.03).(0.89)[/tex] (II)
And the another conditional :
[tex]P(P'/D')=\frac{P(P',D')}{P(D')}[/tex] ⇒
[tex]P(P',D')=P(P'/D').P(D')[/tex]
[tex]P(P',D')=(0.99).(0.11)[/tex] (III)
Replacing (II) and (III) in (I) :
[tex]P(P')=(0.03).(0.89)+(0.99).(0.11)[/tex]
[tex]P(P')=0.1356[/tex]
We find that the probability of the test indicating that the person does not have the disease ( P(P') ) is 0.1356
