Answer

We can model the arrival process like a Poisson process. \(\lambda = 5\) and \(\tau = \frac{1}{2}\)

\[ \begin{aligned} P(\lambda, \tau, k) &= \frac{(\lambda \tau)^{k} e^{-\lambda \tau}}{k!} \newline P(5, \frac{1}{2}, 0) &= \frac{(5 * \frac{1}{2})^{0} e^{-5 \times \frac{1}{2}}}{0!} \newline P(5, \frac{1}{2}, 1) &= \frac{(5 * \frac{1}{2})^{1} e^{-5 \times \frac{1}{2}}}{1!} \end{aligned} \]