As the emails arrive in accordance with a Poisson process the arrival time for the first email, $T$, is an exponentially distributed random variable with parameter $\lambda$
\[
P(T \le t ) = 1 - e^{-\lambda t} \qquad \rightarrow \qquad P(T = t ) =
\frac{\textrm{d}P(T \le t)}{\textrm{d}t} = \lambda e^{-\lambda t}
\]
We can thus calculate the expectated time to complete the job by conditioning on the arrival time of first email. We have:
\[
\mathbb{E}(D) = (c+d)\int_0^c \lambda e^{-\lambda t} \textrm{d}t + c \int_c^\infty \lambda e^{-\lambda t} \textrm{d}t
\]
This integral is solved as follows:
\[
\begin{aligned}
\mathbb{E}(D) & = (c+d)\lambda \int_0^c e^{-\lambda t} \textrm{d}t + c\lambda \int_c^\infty e^{-\lambda t} \textrm{d}t \\
& = (c+d)\lambda \left[ -\frac{1}{\lambda}e^{-\lambda t} \right]_0^c + c\lambda \left[ -\frac{1}{\lambda} e^{-\lambda t} \right]_c^\infty \\
& = (c+d)[1 - e^{-\lambda c}] + ce^{-\lambda c} \\
& = (c+d) - de^{-\lambda c}
\end{aligned}
\]
It will take the engineers either $c$ hours or $c+d$ hours to complete the job. The time they take will never be equal to the expectation. Hence you
should allocate $c+d$ hours of their time as this is the longer of the two times.
