The transforms associated with two independent discrete random variables X and Y are $M_x(s) = (\frac{1}{2} {e^2}^s + \frac{1}{2} {e^4}^s)^7, M_Y(s) = e^8(e^s - 1)$ Find P(X + Y = 15).
This is what I have so far: $(\frac{1}{2} {e^2}^s + \frac{1}{2} {e^4}^s)^7 = {e^{14}}^s(\frac{1}{2}+\frac{1}{2}{e^2}^s)^7 \quad$ let W~Binomial(7,$\frac{1}{2})\quad$ then X=2W+14 $\quad$ Also Y~Poisson(8)
Now P(X + Y = 15) = P(2W + Y = 1) , 2W~Binomial(14 , $\frac{1}{2}$)$\quad$ Let Z = 2W + Y $\quad P_z(k) = \sum_{k = 0}^{14} P_x(k)P_y(z-k) = \sum_{k = 0}^{14} \binom{14}{k}{2}^{-14}\frac{{e}^{-8}{8}^{z-k}}{8!}$
I'm stuck here. I used the summation calculator and the answer doesn't seem right. Please let me know where do I get it wrong or perhaps there's a better way to approach?