If $f_h$ is the pdf of channel $h$, where $h \sim exp(\mu)$. And $I_i$ is the total interference and $I_i \forall i\in \{1,2,...n\}$ is also a random variable. The pdf of $h$ is given as:
$$f_h = \frac{P_mhr^{-\alpha}}{\sum_{i=0}^nI_i} = SINR$$
Where $P_m$ is the Transmitter (Tx) power, $r$ is the distance between Tx and Rx, and $\alpha$ is the pathloss exponent.
For a given threshold, the probability of success for successful communication is then :
$$p = \mathbb{P} \bigg[ SINR > T\bigg] = \mathbb{P} \bigg[ \frac{P_mhr^{-\alpha}}{I_r} > T\bigg]$$
$$ p = 1 - \mathbb{P}\bigg[ \frac{P_mhr^{-\alpha}}{\sum_{i=0}^nI_i} \leq T\bigg]$$
$$ p = 1 - \mathbb{P}\bigg[ h \leq \frac{TI_rr^\alpha}{P_m}\bigg]$$
Now, my question is do I calculate $\mathbb{P}\bigg[ h \leq \frac{T\sum_{i=0}^nI_ir^\alpha}{P_m}\bigg]$ as follows ?
$$\mathbb{P}\bigg[ h \leq \frac{T\sum_{i=0}^nI_ir^\alpha}{P_m}\bigg] = \mathbb{E}_{\sum_{i=0}^nI_i}\Bigg[ \mathbb{P}\bigg[ h \leq \frac{T\sum_{i=0}^nI_ir^\alpha}{P_m}\bigg]\Bigg]$$
i.e. the CDF of $h$ depending on expected value of random variable $\sum_{i=0}^nI_i$
Furthermore, how can I get the solution for it? Note that $I_i = Ag$, where $A$ is a constant and $g \sim exp(1)$
$$\mathbb{P}\bigg[ h \leq \frac{T\sum_{i=0}^nI_ir^\alpha}{P_m}\bigg] = E_{I_r}\Bigg[\exp\Bigg(-\frac{T\sum I_rr^a}{Pm}\Bigg)\Bigg]$$
Now, we know that there is a relation between the Expectation of random variable $X$ and its Laplace transform ie.e. $\mathcal{L}_X(s) = E[e^{-sX}]$. So we can write above equation as (considering s = $Tr^a/\text{P_m}$):
$E_{I_r}[exp(...)] = \mathcal{L}_I(s) = \int_v^\infty\frac{1}{1 + v^a/s} dv$ (according to a paper titled "Tractible Approach to Coverage and Rate in Cellular Networks". Following this paper, rest of the steps can be solved.