What is the probability that you have more than 3 deaths in an infected family of 6 if each pair of people have a different probability of death? If say 2 people individually  had a .05 chance of survival, 2 individually had a .10 chance of survival, and 2 individually had a .20 chance of survival (pulled these numbers out of a hat)
What is the chance that the family has more than 3 deaths.
The method I have is quite tedious and I was wondering if there was a better way than just going case by case of the combinations and summing the results. I'm not even sure if that gives me the right answer, and I will check later using monte carlo.
Does this have a well known probability distribution? I know some small parts are binomial.
I wanted to know the probability distribution of death in our family, which has people in different age groups, if everyone in the family got covid. Other than by using simulation, I don't know a reasonable way to scale up what I did to include more age groups and more people. So I was curious to know if there is a way to do that, or at least approximate the probability.
 A: Consider the polynomial
$$f(x) = (.05 + .95 x)^2 (.10 + .90 x)^2 (.20 + .80 x)^2$$
When expanded, we have
$$f(x) = 0.000001 +0.000064 x+0.00159 x^2+0.01948
   x^3+0.123865 x^4+0.387144 x^5+0.467856 x^6$$
The relation of this expanded polynomial to the original problem is that the coefficient of $x^n$ is the probability that there will be exactly $n$ deaths, so the probability that there will be zero deaths is $0.000001$, the probability of exactly one death is $0.000064$, the probability of exactly two deaths is $0.00159$, etc.
$f(x)$ is an example of a generating function.  If you find this approach interesting and would like to learn more about generating functions, there are many resources in the answer to this question: How can I learn about generating functions?
A: I slightly modified Steve's answer to specifically fit what I was asking for
by changing this line
p_1, p_2, p_3, p_4, p_5, p_6 = symbols('p_1 p_2 p_3 p_4 p_5 p_6')

to this
p_1, p_2, p_3, p_4, p_5, p_6 = symbols('p_1 p_1 p_2 p_2 p_3 p_3')

you will get:
$$2 p_{1}^{2} p_{2} \left(1 - p_{2}\right) \left(1 - p_{3}\right)^{2} + 2 p_{1}^{2} p_{3} \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right) + p_{1}^{2} \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right)^{2} + 2 p_{1} p_{2}^{2} \left(1 - p_{1}\right) \left(1 - p_{3}\right)^{2} + 8 p_{1} p_{2} p_{3} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) + 4 p_{1} p_{2} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right)^{2} + 2 p_{1} p_{3}^{2} \left(1 - p_{1}\right) \left(1 - p_{2}\right)^{2} + 4 p_{1} p_{3} \left(1 - p_{1}\right) \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right) + 2 p_{1} \left(1 - p_{1}\right) \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right)^{2} + 2 p_{2}^{2} p_{3} \left(1 - p_{1}\right)^{2} \left(1 - p_{3}\right) + p_{2}^{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{3}\right)^{2} + 2 p_{2} p_{3}^{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right) + 4 p_{2} p_{3} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right) \left(1 - p_{3}\right) + 2 p_{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right) \left(1 - p_{3}\right)^{2} + p_{3}^{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right)^{2} + 2 p_{3} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right)$$
which suits the three groups of two in the question.
