Two types of jobs arrive at a machine. Type $1$ jobs arrive according to a Poisson process with rate 45 per hour and need an exponential service time with mean $1/2$ minute. Type $2$ jobs arrive according to a Poisson process with rate $15$ per hour and need an exponential service time with mean $1$ minute.
Suppose we want to find the Laplace-Stieltjes transform of the sojourn time of an arbitrary job. If I am not mistaken, the service time is an exponential distribution with mean $1/2$ with probability $3/4$ and an exponential distribution with mean $1$ with probability $1$. So, if $B$ denotes the service time, then
$$\tilde{B}(s) = \frac{3}{4} \frac{2}{2+s} + \frac{1}{4} \frac{1}{1+s}$$
or $\tilde{B}(s) = \frac{8+7s}{(4+4s)(1+s)}$. Since we are dealing with a $M/G/1$-queue, we know
$$\tilde{S}(s) = \frac{(1-\rho)\tilde{B}(s)s}{s - \lambda + \lambda\tilde{B}(s)}$$
with $\lambda = 1$ (since $60$ jobs arrive per hour) and $1-\rho = 1 - \lambda E(B) = 1 - \frac{3}{4}\frac{1}{2} - \frac{1}{4}1 = \frac{3}{8}$.
However, using this expression gives something that seems wrong
$$\tilde{S}(s) = \frac{3}{8}\frac{\frac{8+7s}{(4+4s)(1+s)}s}{s - 1 + \frac{8+7s}{(4+4s)(1+s)}}$$
$$\tilde{S}(s) =\frac{3}{8}\frac{8s +7s^2}{(s-1)(4+4s)(1+s) + 8 + 7s}$$
The constant term in $(s-1)(4+4s)(1+s)$ is $-4$, which does not cancel the $+8$ of $8+7s$, which is what I should get. Where did I make a mistake?
EDIT: I found my mistake, $\tilde{B}(s) = \frac{8+7s}{(4+4s)(1+2s)}$, I messed up the denominator.