Suppose I have the following events:
- Event A: Contracted HIV
- Event B: Tested Positive with HIV
Given $1$ in $1000$ people gets contracted with HIV and the HIV Test gives a wrong test result for $1$ in $100$ people. If a person is tested Positive with HIV, I need to find the probability of him really contracted with HIV.
So I need to find $P(A|B)$ and that $P(A|B)=\frac{P(B|A) \cdot P(A)}{P(B)}$.
To find $P(B)$, I need to find all that are tested positive. So...
$P(B)=\frac{1}{1000} \cdot \frac{99}{100}+\frac{999}{1000} \cdot \frac{1}{100}=\frac{1098}{100000}$
$P(B|A)=\frac{1}{1000} \cdot \frac{99}{100}=\frac{99}{100000}$
$P(A)=\frac{1}{1000}$
Therefore, $P(A|B)=\frac{P(B|A) \cdot P(A)}{P(B)}=\frac{\frac{99}{100000}\cdot \frac{1}{1000}}{\frac{1098}{100000}}=\frac{11}{122000}=0.0000901639=0.009016\%$
While this may seem ok to me at first, I suspected my answer to be wrong because $0.009016\%$ as a figure is a little hard to believe. I did a search on the internet and looks like the answer could be just $ \frac{99}{1098} = 9.016\% $.
Is my initial answer wrong? What has gone wrong with my initial answer of $0.009016\%$? I simply followed the Bayes' Conditional formula.