Let's split each period into $n$ intervals. There's a continuum $u$ of unemployed and $v$ of vacancies. During each interval, there is a total of $X/n$ job offers. That means that each unemployed gets a job offer with probability $X/(nu)$ (assuming that $n$ is eventually so small that the chance of multiple job offers to the same person will go to zero).
Single individual Number of job offers - for a single individual - in the interval is distributed $Binomial(k, n, \frac{X}{un})$. Letting $n\to\infty$, the distribution in the continuous-time analog converges to $Poisson(k, \frac{X}{u})$.
I'm interested in the probability of $x$ individuals getting at least one job interval during a whole interval. That is $(1 - Binomial(0, n, \frac{X}{nu}))^x$:
$$ (1 - (1 - \frac{X}{un})^n)^x $$
If I'm not mistaken, the continuous-time analog is
$$ (1 - e^\frac{-X}{u})^x$$
Is that correct? The combination of exponential and power function is making me quite uncomfortable.