Can I break down $P(h \geq (A + B)$, given all $ A,B,h$ are all random variables. Will the following rule works?
$$P[h \geq (A + B)] = P(h\geq A) + P(h\geq B)$$
Actually, in one of my mathematical analysis, I end up with a complex expression which can be simplified to $P[h \geq (A + B)]$. I believe I can move forward if I can break it down somehow. Further explanations of variables are as below.
$h \sim \exp(\lambda')$ and $g \sim \exp(\lambda'')$
$A = a(1 + e^{sh})$, $B = bg(1+e^{sh})$
$a,b,s, \lambda', \lambda''$ are constants.
As has already been stated, your approach does not work. Here is an idea: Since $h,g$ are independent (this has not directly been stated in the question but I assume it's true), the following should hold: \begin{align*}\textbf{P}(h\geq A+B)&=\textbf{P}(h\geq(1+e^{sh})(a+bg))=\textbf{P}(h\geq(1+e^{sh})(a+bg))\\&=\textbf{P}(h/(1+e^{sh})\geq(a+bg))=\textbf{E}[\textbf{P}(h/(1+e^{sh})\geq(a+bg)|h)]\\&=\textbf{E}[F_g((h/(1-e^{sh})-a)/b)]\end{align*} where $F_g$ is the CDF of $g$. This is the case if $b>0$. Cases $b=0$ and $b<0$ follow analogously.