Question

A medical test gives a correct result 90% of the time for infected individuals and 95% of the time for non-infected individuals. All members of the population are tested. The probability that a person is infected given they have a positive test result is 0.6. What proportion of the population is infected?

Answer 1

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,

$P(B|A)=90/100=0.9,$

$P(B^c|A^c)=95/100=0.95$

$P(A|B)=0.6$

So

$P(B|A)=0.9 \Rightarrow \frac{P(A\cap B)}{P(A)}=0.9 \Rightarrow P(A\cap B)=0.9P(A)$

$P(A|B)=0.6 \Rightarrow \frac{P(A\cap B)}{P(B)}=0.6 \Rightarrow P(A\cap B)=0.6P(B)$

Hence,

$0.6P(B)=0.9P(A) \Rightarrow P(B)=1.5P(A)$

Now

$P(B^c|A^c)=0.95 \Rightarrow \frac{P(A^c\cap B^c)}{P(A^c)}=0.95 \Rightarrow \frac{P(A\cup B)^c}{1-P(A)}=0.95$

$\Rightarrow \frac{1-P(A\cup B)}{1-P(A)}=0.95$

$\Rightarrow 1-[P(A)+P(B)-P(A\cap B)]=0.95-0.95P(A)$

$\Rightarrow 1-P(A)-1.5P(A)+0.9(A)+0.95P(A)=0.95 \Rightarrow 0.65P(A)=0.05$

$\Rightarrow P(A)=\frac{1}{13}\Rightarrow P(A)=0.07692$

The probability that a person is infected is   $\frac{1}{13}=0.07692.$

So the probability that a person is infected is =0.07692.

The proportion of the population which is infected is   $\frac{1}{13}=0.07692$

Hence the proportion of the population which is infected is   =0.07692