Given two events, $A$ and $B$, that will each eventually occur, what's the probability $P_{_{AB}}\ \ $ that $A$ occurs before $B$? What you're given are the two pdf's $a(t),b(t)$ such that, as usual, they're both $\ge0$ and$\int_{t=0}^\infty\ a(t)dt=1=\int_{t=0}^\infty\ b(t)dt$.
What I'm thinking, but can't seem to prove, is the following. Let $A(t)=\int_{0}^t\ a(t')dt'$ and $B(t)=\int_{0}^t\ b(t')dt'$ be the cumulative probabilities. Then $$P_{_{AB}}\hspace{10pt} =\int_0^\infty a(t)\left(1-B(t)\right)dt\ = \hspace{5pt}1-P_{_{BA}}$$ Is that right? If so, how's it proved? Or, if not, what is right?
To be found is $P_{AB}:=P\left(X_{B}>X_{A}\right)$ where $X_A,X_B$ denote the waiting times.
Here $X_A$ and $X_B$ are random variables, $\{X_{B}>X_{A}\}$ is an event (of which we want to find the probability) and $A$ and $B$ are not more than labels or indices that serve by discerning.
If $X_{A}$ and $X_{B}$ are independent (this is not mentioned in your question) then indeed:
$$\begin{aligned}P_{AB} & =P\left(X_{B}>X_{A}\right)\\ & =\int_{0}^{\infty}P\left(X_{B}>X_{A}\mid X_{A}=t\right)a\left(t\right)dt\\ & =\int_{0}^{\infty}P\left(X_{B}>t\mid X_{A}=t\right)a\left(t\right)dt\\ & =\int_{0}^{\infty}P\left(X_{B}>t\right)a\left(t\right)dt\\ & =\int_{0}^{\infty}\left(1-B\left(t\right)\right)a\left(t\right)dt \end{aligned} $$as you expected.
Here $B$ denotes the CDF of $X_{B}$ and the fourth equality is based on independence.
This can also be expressed as $1-P_{BA}$ because in this situation we have $P\left(X_{A}=X_{B}\right)=0$.
This follows from the fact that at least one (that is enough) of the random variables has a PDF.