It is given that only 0.1 % of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows :

For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive.

Based on the above informations, answer the question :

The probability that the person is actually having COVID given that he is tested as COVID positive is:

$\frac{10}{121}$

Let,

A = Person selected has covid.

B = Person don't have covid.

C = Person have judge report positive.

$P(A/C)=\frac{0.001×0.9}{0.001×0.9+0.999×0.01}=\frac{9×10^{-4}}{9×10^{-4}+99.9×10^{-4}}$

$=\frac{9×10^{-4}}{10^{-4}(9+99.9)}=\frac{9}{108.9}=0.083$ (approx) or $\frac{90}{1089}$