Given a joint moment generating function, find $P(XLet random variables $X$ and $Y$ have joint MGF:
$$M(t_1,t_2) = 1/2e^{t_1+t_2} + 1/4e^{2t_1+t_2} + 1/12e^{t_2} + 1/6e^{4t_1+3t_2}$$  I now need to find $P(X<Y)$. I know how to find the moments as I had to find $V[X]$ for part of this problem, but I am not quite sure how to figure out probability from this.
 A: Let $Z$ be a discrete random variable taking values $1\dots,n$ with respective probabilities $p_1,\dots,p_n$ (of course $p_1+\dots+p_n =1$). 
Define $X$ and $Y$ as follows: let $x_i,y_i,~i=1,\dots,n$ be a sequence of numbers, and let $X=x_i,Y=y_i$ whenever $Z=i$. 
Note then that $X<Y$ if and only if $Z$ is equal to an $i$ for which $x_i<y_i$.  
The MGF of $(X,Y)$ is easily obtained by conditioning on $Z$: 
$$E[e^{t_1 X+ t_2 Y} ] = \sum_{i=1}^n p_i e^{t_1 x_i + t_2 y_i}.$$
In our case $n=4$, $p_1=\frac 12=\frac{6}{12},p_2=\frac 14=\frac{3}{12},p_3=\frac 1{12},p_4=\frac {1}{6}=\frac{2}{12}$, and $x_1=1,y_1=1$, $x_2=2,y_2=1$, $x_3=0,y_3=1$ and $x_4=4,y_3=3$.  
The only value of $i$ such that $x_i<y_i$ is $i=3$. Since $p_3=\frac{1}{12}$, it follows that $P(X<Y) = p_3= \frac{1}{12}$. 
Can you see what is $P(X=Y)$ and $P(Y<X)$ ? 
A: Too big for the comment.
Not sure about a generalized way. But may be something along this:
Put $t_2 = -t_1$ to get the mgf of $Z = X-Y$ which will be
$$E(e^{tZ}) = 1/2 + 1/4e^{t} + 1/12 e^{-t} + 1/6e^{t}$$
Now,
$$\begin{align} E(e^{tZ}) &= E(e^{tZ}|Z\geq0)P(Z\geq0) + E(e^{tZ}|Z<0)P(Z<0) \\\\ &= E(e^{tZ}|Z\geq0)(1-P(Z<0)) + E(e^{tZ}|Z<0)P(Z<0)\end{align}$$
We know $E(e^{tZ})$ and have to find $P(Z<0)$, by somehow computing $E(e^{tZ}|Z<0)$ and $E(e^{tZ}|Z\geq0)$.
Edit: See me answer here for a generalized procedure.
