Answer:
The proportion of the population which is infected is =0.07692
Step-by-step explanation:
Let the events be,
A: a person is infected.
B: the test result is positve
According to the question we have,
[tex]P(B|A)=90/100=0.9,[/tex]
[tex]P(B^c|A^c)=95/100=0.95[/tex]
[tex]P(A|B)=0.6[/tex]
So
[tex]P(B|A)=0.9 \Rightarrow \frac{P(A\cap B)}{P(A)}=0.9 \Rightarrow P(A\cap B)=0.9P(A)[/tex]
[tex]P(A|B)=0.6 \Rightarrow \frac{P(A\cap B)}{P(B)}=0.6 \Rightarrow P(A\cap B)=0.6P(B)[/tex]
Hence,
[tex]0.6P(B)=0.9P(A) \Rightarrow P(B)=1.5P(A)[/tex]
Now
[tex]P(B^c|A^c)=0.95 \Rightarrow \frac{P(A^c\cap B^c)}{P(A^c)}=0.95 \Rightarrow \frac{P(A\cup B)^c}{1-P(A)}=0.95[/tex]
[tex]\Rightarrow \frac{1-P(A\cup B)}{1-P(A)}=0.95[/tex]
[tex]\Rightarrow 1-[P(A)+P(B)-P(A\cap B)]=0.95-0.95P(A)[/tex]
[tex]\Rightarrow 1-P(A)-1.5P(A)+0.9(A)+0.95P(A)=0.95 \Rightarrow 0.65P(A)=0.05[/tex]
[tex]\Rightarrow P(A)=\frac{1}{13}\Rightarrow P(A)=0.07692[/tex]
The probability that a person is infected is [tex]\frac{1}{13}=0.07692.[/tex]
So the probability that a person is infected is =0.07692.
The proportion of the population which is infected is [tex]\frac{1}{13}=0.07692[/tex]
Hence the proportion of the population which is infected is =0.07692
