Let and be two discrete random variables with the joint moment generating function
$$_{,}(t_{1},t_{2})=(\frac{1}{3} e^{t_{1}} + \frac{2}{3})^{2} (\frac{2}{3} e^{t_{2}} + \frac{1}{3})^{3} ,t_{1},t_{2} \in R.$$ Then $P$ $( 2 + 3 > 1)$ equals
My Approch:
$ _{} = _{,}(t_{1},0) = (\frac{1}{3} e^{t_{1}} + \frac{2}{3})^{2} $ similar to MGF of $Binomial(2,\frac{1}{3})$
$ _{Y} = _{,}(0,t_{2}) = (\frac{2}{3} e^{t_{2}} + \frac{1}{3})^{3} $ similar to MGF of $Binomial(3,\frac{2}{3})$
$P$ $( 2 + 3 > 1) = 1 - P(X=0) P(Y=0) = 1 - (\frac{2}{3})^{2}(\frac{1}{3})^{3} \simeq 0.98 $
Is this a right approach to solve for this kind type of question?
What if in case, I don't know the MGF of known distribution?
Is there any general way to solve this kind of problems?