# Calculating the probability of someone having zero to n diseases given overlapping prevalence

I'm trying to calculate the proportion of people in a population who have a combination of disease $$A$$, $$B$$ and $$C$$. This can range from $$0,0,0$$ (the set of people with no disease) to $$1,1,1$$ (the set of people with all three diseases).

I would like to calculate the probability each of these 8 outcomes (from $$0,0,0$$ to $$1,1,1$$) given three inputs, which are the prevalence of each disease (e.g. 80% of people have disease $$A$$, 70% of people have disease $$B$$ and 50% of people have disease $$C$$). I would also like to generalise this to $$n$$ diseases in the future. It is important to be able to distinguish between each outcome, because they have different effects on people (e.g. $$0,1,1$$ has a different chance to kill someone than $$1,0,1$$).

Given that we know nothing about the relationship between these diseases and that the prevalences do not sum to 100%, I do not think this has a single solution. However, if we expect the diseases to be independent of each other, I assume this is solvable, but I do not know how to do it. If I reduce the diseases to two, this becomes easier. For example:

If the prevalence of Disease $$A$$ is $$80\%$$ and the prevalence of Disease $$B$$ is $$70\%$$:

• $$1,1$$: The proportion of people with $$A \& B$$ = $$0.8 * 0.7 = 0.56$$
• $$1,0$$: People with $$A$$ only = $$0.8 - 0.56 = 0.24$$
• $$0,1$$: People with $$B$$ only = $$0.7 - 0.56 = 0.14$$
• $$0,0$$: People with no disease = $$1 - (0.56+0.24+0.14) = 0.06$$

But in the case of three diseases, I get stuck. For example:

Prevalence of Disease $$A$$ is $$80\%$$, Prevalence of Disease $$B$$ is $$70\%$$, Prevalence of Disease $$C$$ is $$50\%$$

• $$1,1,1$$ = $$0.8 * 0.7 * 0.5 = 0.28$$

From here, i'm not sure what to do. Perhaps I need to simultaneously solve the next set of equations e.g. $$1,0,1$$ and $$1,1,0$$, but i'm not sure.

Is this on the right track? Or does this problem become unsolvable with $$>2$$ diseases?

+1 for the query, very good analysis so far.

To carry it into three people, using:

80% of people have disease A, 70% of people have disease B and 50% of people have disease C

and

However, if we expect the diseases to be independent of each other, I assume this is solvable, but I do not know how to do it.

Let $$p_a = 0.8 =$$ chance of person having disease A.
Let $$p_b = 0.7 =$$ chance of person having disease B.
Let $$p_c = 0.5 =$$ chance of person having disease C.

Let $$q_a = 1 - p_a.$$
Let $$q_b = 1 - p_b.$$
Let $$q_c = 1 - p_c.$$

You define 8 possibilities, where each person either does or doesn't have each of the 3 diseases.

Then, each of the 8 possibilities will compute to $$s_a \times s_b \times s_c$$, where if the possibility refers to:

the person having disease A, set $$s_a = p_a,$$ else $$s_a = q_a.$$
the person having disease B, set $$s_b = p_b,$$ else $$s_b = q_b.$$
the person having disease C, set $$s_c = p_c,$$ else $$s_c = q_c.$$

• Thank you for the answer. Just to supplement this, I thought i'd provide an example of the computation steps to show how it's done, just in case anyone finds that useful. So, for case 0,0,0 - where no one has any disease, Sa = 1-0.8, Sb = 1-0.7, and Sc = 1-0.5. For case 1,1,1 - where someone has every disease, Sa = 0.8, Sb = 0.7, Sc = 0.5. 1,0,0 = 0.8 x 0.3 x 0.5 and so on. Dec 9, 2020 at 3:08