Probability that event A occurs when event A competes with event B I have two processes, A and B, which compete with each other. When event A happens, B cannot happen. When event B occurs, A cannot occur.
The “likelihood” of event A happening at a given time (in the absence of competition) is given by the equation:
$$1 - e^{-kt}$$
where t means time and k is the parameter describing the shape that this curve takes.
The “likelihood” of event B happening in the absence of competition is given by a similar equation,
$$1 - e^{-qt}$$
where t is time and q describes the parameter of the curve. Furthermore, the curve for event B starts after a time delay T.

Intuitively, I know that the probability of event A winning the competition at a given time t is related to the value of $1 - e^{-kt}$ at time t as well as the probability that event A has not happened before time t, and the probability that event B has not happened before time t.
My question is: what is a proper mathematical way of describing the probability that event A wins the competition and happens before event B?
 A: The probability that $A$ happens in time interval $[0,t]$ is $P(A)=1-\exp(-kt)$ for $t>0$ and zero otherwise. The probability that $B$ happens in time interval $[0,t]$ is $P(B)=1-\exp(-q(t-T))$ for $t>T$ and zero otherwise. Therefore if $A$ happens at time $t\in [0,T)$, whose probability is $1-\exp(-kT)$, then $A$ wins with certainty. 
Now consider the case where $A$ happens at time $t\geq T$. The probability that $A$ happens in time interval $(t,t+dt)$ is $k\exp(-kt)dt$. For $A$ to win, $B$ must happen during $(t,\infty)$ and whose probability is $\exp(-q(t-T))$. Hence the probability that $A$ wins in this case is $\exp(-q(t-T))\times k\exp(-kt)dt$. Accounting for all possible times during which $A$ can occur gives:
\begin{align}
P(\textrm{$A$ wins})&=1-e^{-kT}+\int_{T}^\infty dt~e^{-q(t-T)}~k~e^{-kt}\\
&=1-\frac{q}{k+q}e^{-kT}
\end{align}
See that as $T\to\infty$ then $P(\textrm{$A$ wins})\to 1$ as it should.
A: I suspect that the question can be rephrased as: what is $\mathsf P(T+Y>X)$ if $X$ and $Y$ are independent random variables haveing exponential distribution with parameters $k$ and $q$ respectively?
$$\mathsf P(T+Y>X)=\int_0^{\infty}\int_0^{\infty}[T+y>x]\mathsf P_Y(dy)\mathsf P_X(dx)\tag1$$
Here $[T+y>x]$ denotes the function that takes value $1$ if $T+y>x$ and takes value $0$ otherwise.
For a fixed positive $x$ we have $\int_0^{\infty}[T+y>x]\mathsf P_Y(dy)=\mathsf P(Y>x-T)$ so that $(1)$ can be rewritten as:$$\mathsf P(T+Y>X)=\int_0^{\infty}\mathsf P(Y>x-T)\mathsf P_X(dx)\tag2$$
Exponential distribution with parameter $k$ has PDF $ke^{-kx}$ so $(2)$ can be rewritten as:$$\mathsf P(T+Y>X)=k\int_0^{\infty}\mathsf P(Y>x-T)e^{-kx}dx\tag3$$
The RHS of $(3)$ can be split up:$$k\int_0^{\infty}\mathsf P(Y>x-T)e^{-kx}dx=k\int_{0}^{T}\mathsf{P}\left(Y>x-T\right)e^{-kx}dx+k\int_{T}^{\infty}\mathsf{P}\left(Y>x-T\right)e^{-kx}dx$$$$=k\int_{0}^{T}e^{-kx}dx+k\int_{T}^{\infty}e^{-\left(q+k\right)x}dx=1-e^{-kT}+\frac{k}{q+k}e^{-\left(q+k\right)T}$$
