Solving Bayes Theorem equation -> I can't calculate proper result
Bayes theorem: $P(A|B) = \frac{P(B|A) P(A)}{P(B)}$

In our case: $P(disease|test_+) = \frac{P(test_+|disease) P(disease)}{P(test_+)}$

Given:

$P(test_+|disease) = 0.85 \\
P(test_-|healthy) = 0.9 \\
P(disease) = 0.02$

Since a test can be positive while the patient has a disease and it also can be positive while the patient is healthy, the overall probability for a positive test is calculated as follows:

$P(test_+) = P(test_+|disease) P(disease) + P(test_+|healthy) P(healthy) \\
P(test_+|healthy) = 1 - P(test_-|healthy) \\
P(healthy) = 1 - P(disease)$

So:

$
\begin{align}
P(disease|test_+)
&amp;= \frac{P(test_+|disease) P(disease)}{P(test_+|disease) P(disease) + P(test_+|healthy) P(healthy)} \\
&amp;= \frac{P(test_+|disease) P(disease)}{P(test_+|disease) P(disease) + (1 - P(test_-|healthy)) (1 - P(disease))} \\
&amp;= \frac{0.85 \times 0.02}{0.85 \times 0.02 + (1 - 0.9) (1 - 0.02)} \\
&amp;= \frac{0.017}{0.017 + 0.1 \times 0.98} = \frac{0.017}{0.115} \approx 0.148
\end{align}
$

So, the probability that you have the disease given that you were tested positively is around 14,8% in this case.