Bayes' formula conditional probabilities I am having troubles finding out $P(A)$ and $P(B)$ in the following problem:

The specificity of a test is $99.6\%$ and the sensitivity is
$99.7\%$. $1\%$ is expected to have had an infection
at some point. We are
interested in the probability of having been infected
. Define the events $A = \text{"Have had an infection”}$ and $B = \text{”test is positive”}$.
What are the odds for $A$ given the evidence $B$?

I know that I need to calculate $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ but how do I find $P(A)$ and $P(B)$ from the information given?

In the information it is given that '$1\%$ is expected to have had it at some point', which in this question translates to $\mathbb{P}(A) = 0.01$.
A trick in using Bayes' formula is to rewrite the denominator in terms of conditional probabilities. That is, we can write:
\begin{align*}
\mathbb{P}(B) = \mathbb{P}(B | A^c)\mathbb{P}(A^c) + \mathbb{P}(B | A)\mathbb{P}(A) = 
(1-\mathbb{P}(B^c | A^c))\mathbb{P}(A^c) + \mathbb{P}(B | A)\mathbb{P}(A)
\end{align*}
and all the probabilities in the latter expression are given in the question.
