Collection of binary random variables A portfolio of 13 different companies have loans. Company $i$ defaults on their loan with probability $p_i$ and survives with probability $q_i=1-p_i$. Let $Y_i=1$ denote default.
Question: How could I get to a reasonable guess, preferably using probability theory, on what is the probability of default for the whole portfolio?
I am pondering about the risk on a loan to the entire portfolio. I am free to make assumptions and be creative, for example not find an exact solution but bound the probability by some interval. I made three scenarios, each with different assumptions:


*

*I) If one defaults, all other defaults. 

*II) The firms are independent

*III) They are positively/negatively correlated in the sense that if one falls, it is a higher/lower probability of the other falling. 


I)
Pr(portfolio.defaults) = Pr(someone defaults) = 1 - Pr(no one defaults). Here my thinking stops, can I 
II)
Here it is like a binomial with $n=13$ although the probabilities $p$ are not the same but are changing. I could make an upper bound and say it is equal to $Bin(n=13, p=\max\{p_i\})$ but surely there is a distribution for this? Maybe I could use the PDF of a Bin directly via ${\displaystyle \Pr(Y_i=k)={\binom {n}{k}}p_i^{k}(1-p_i)^{n-k}}$
I) and II)
We have that Pr(portfolio.default) = by assummption I = Pr(someone defaults) = 1 - Pr(no one defaults). Since they are independent by assumption II we have $$Pr(\text{portfolio.default}) = 1 - (q_1 \times ... \times q_{13})$$
III)
I have not started thinking about this scenario. 
 A: First, to ease notation let $N$ be the total number of defaults, i.e.
$$
N = \sum_{i=1}^{13} Y_i.
$$
Concerning the first scenario: in order to be able to compute $\mathbb P(N=0)$ you need to make some assumptions about the joint distribution of the vector $(Y_1, Y_2, \dots, Y_{13})$. In general you can only say that
$$
\mathbb P(N=0)=\mathbb P(Y_1=0, \dots, Y_{13}=0)
$$
and to get an expression for the right hand side you need the joint distribution.
For II:
This formula is only useful in theory but not really applicable in practice (especially if your portfolio consists of more than 13 companies)
$$
\mathbb{P}(N=n) = \sum_{I \subseteq \{1,\dots,13\}\\ \quad  |I|=n} \prod_{i \in I}p_i \cdot \prod_{j \in \{1,\dots,13\}\setminus I}(1-p_j).
$$
So you consider a sum for all subsets of size $n$ and use the independence to get the product of the $p_i$.
One classic approach is to use the Poisson approximation in this case. There it is assumed $N$ has a Poisson distribution with parameter $\lambda=p_1+p_2+\dots+p_{13}$, which simplifies the calculation of $\mathbb P (N=n)$ and then you bound the error.
For the third scenario it is difficult to say anything other than 
$$
\mathbb P(N=0)=\mathbb P(Y_1=0, \dots, Y_{13}=0),
$$
since knowing all pairwise correlation does not describe the joint distribution (this is only the case for joint normal distributions).
So you would have to be more specific.
