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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?

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1 Answer 1

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+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 $s_c = q_c.$

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  • $\begingroup$ 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. $\endgroup$
    – rorance_
    Dec 9, 2020 at 3:08

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