The moment generating functions of two independent variables $X$ and $Y$ are $M_X(t)=\exp(2e^t-2)$ and $M_Y(t)=\left(\frac34e^t+\frac14\right)^{10}$. What are (a) $P(X+Y=2)$; (b) $P(XY=0)$; (c) $E[XY]$?
For (a), I did it in two ways that yield different answers.
First method: $M_{X+Y}(t)=\exp(2e^t-2)\left(\frac34e^t+\frac14\right)^{10}$, so $P(X+Y=2)=\frac{d^2}{d(e^t)^2}M_{X+Y}(t)\big|_{e^t=0}=\frac{467}{524288e^2}.$
Second method: $X$ is Poisson with parameter $2$ and $Y$ is Binomial with parameters $(10,\frac34)$. So $P(X+Y=2)=\sum_{i=0}^2e^{-2}\frac{2^i}{i!}{10\choose 2-i}\left(\frac34\right)^{2-i}\left(\frac14\right)^{8+i}=\frac{467}{1048576e^2}.$
Where did I go wrong?
For (b) and (c), can you check my solutions?
(b) $P(XY=0)=P(X=0)+P(Y=0)-P(X=0,Y=0)=e^{-2}+\frac1{4^{10}}-\frac{e^{-2}}{4^{10}}$
(c) $E[XY]=E[X]E[Y]=2\left(10\cdot\frac34\right)=15$
Thanks in advance.