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I want to ask something about a common medical situation:

  • Suppose the probability of a medical symptom X (let's say blue tongue:-) ), to occur in a person at a specific time moment is 1/1000. Pretty rare.
  • Suppose also the probability of a disease A to occur in a person at a specific time moment is 1/200. Not that rare.
  • We know also that disease A causes the symptom X in 1/10 cases.
  • Suppose also the probability of another disease B to occur in a person at a specific time moment is 1/2000. Somewhat rare.
  • We know also that disease B causes the symptom X in 1/40 cases.

Obviously the probability of:
- Someone getting the symptom X is 1/1000.
- Someone getting the symptom X because of disease A is (1/200)(1/10) = 1/2000
- Someone getting the symptom X because of disease B is (1/2000)(1/40) = 1/80000

Now suppose that someone is diagnosed with the disease B AND then develops the symptom X. We also don't know anything about whether he has or not the disease A or any other disease that can cause symptom X. We just know he has disease B and the symptom X.

My question is what is the probability the symptom X has occurred because of disease B? Is it 1/40 or something much higher?

Because any doctor for example would say that this can't be a coincidence (that this person had this symptom X after he was diagnosed with disease B, that causes in some cases the symptom X) and that the symptom X was caused by disease B with very big probability, but I want to quantify this probability. How much is it?

For the general case, i.e with general numbers, with known $P(A)$, $P(B)$, $P(X)$, $P(X|A)$, $P(X|B)$, I want to find probability of X being caused by B and not A or anything else, knowing that X and B are true.

Are there any references books about these kind of problems?

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The way I would look at this is as follows:

P(B caused X) = P(X given B) / sum of this and other ways to get X

P(B caused X) = P(X given B) / ( P(X given B) + P(X given A)P(A) )

P(B caused X) = 1/40 / ( 1/40 + 1/200 x 1/10 )

P(B caused X) = 50 / 51 or about 98% chance that B caused X not an undiagnosed A condition

Depending upon your assumptions, this could change a bit. For example, if you believe that 1/1000 people get X (even with B) that is not caused by B. That is, this small possibility is included in the 1/40. If you go the other way the equation becomes:

P(B caused X) = P(X given B) / ( P(X given B) + P(X given A)P(A) + P(X not from A or B) )

P(B caused X) = 1/40 / ( 1/40 + 1/200 x 1/10 + 1/1000 )

P(B caused X) = 50 / 53 or about 94% chance that B caused X not an undiagnosed A condition or just among the whole population

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  • $\begingroup$ << The way I would look at this is as follows: P(B caused X) = P(X given B) / sum of this and other ways to get X >> The problem with this is that: a) I don't get the reason the following equality is true. How it is being derived logically? b) Since we know someone is diagnosed with B and has the symptom X, then the probability of both is 1. I.e P(B)=P(X)=1 So for example P(X|B) is not 1/40 but 1. $\endgroup$
    – Fliers
    Apr 11 '15 at 20:14
  • $\begingroup$ Furthermore following another thinking procedure i come up with a different result: Let's say that all the diseases that create the symptom X are B, A, A2, A3, ..., An. And the symptom X has probability 1/1000 to occur generally from any of these diseases. P(B causes X) = 1 - P(B hasn't caused X) = 1 - P(X was not caused by B and was caused by A or A2 or A3 or... An) = 1 - P(X was not caused by B)·P(X was caused by A or A2 or A3 or... An) = 1 - (39/40)·(1/1000 - 1/80000) Since P(X was caused by A or A2 or A3 or... An) is the probability of X to occur(1/1000) minus P(to occur from B). $\endgroup$
    – Fliers
    Apr 11 '15 at 20:23
  • $\begingroup$ Formatting in comments is terrible and i can't enter line breaks so sorry for this hideous formatting. $\endgroup$
    – Fliers
    Apr 11 '15 at 20:27
  • $\begingroup$ So one definition of probability is the number of simple events that support the event of interest divided by the total number of simple events. Unless the simple events are not equally probable, in which case the numbers are simply scaled. There are only so many ways to end up with X, so we take all these ways and sum their probabilities to get the denominator. The numerator (simple events that satisfy or probability of them) is given in your assumptions - it's the odds of getting X from B in general not for the specific individual. $\endgroup$
    – MikeP
    Apr 12 '15 at 21:25

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