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For example

Assume that a test for a disease gives a positive result for 2.5% of people who do not have the disease, but does not test negative if the person has the disease.

What is the probability that a person who tested positive has the disease if 3% of people have the disease?

With Bayes theorem, we could technics listed here http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_Probability/BS704_Probability6.html.

But how would you solve the noted problem only using a probability tree? I want to know how because it will help me visualize how Bayes Theorem is working in a different abstraction.

As I have heard ALL Probabilities can be solved with probabilities trees. I would like to see the computation done in Bayes Theorem solved/expanded into a probability tree.

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A tree diagram still needs use of the formula to determine a conditional probability. My attempt here identifies the different combinations of outcomes the same as a table.

enter image description here

The diagram below is a more formal presentation, but again, without a formula or explanation of how to arrive at the values, I don't think it helps with conceptual understanding any more than a table.

enter image description here

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  • $\begingroup$ Thank you for your post. It helps illustrate the first half of what I was asking. Although I have heard that ALL Probabilities can be solved with probabilities trees. I would like to see the computation done in Bayes Theorem solved in a probability tree. Does my question make sense? $\endgroup$ – user1787331 Oct 20 '18 at 18:15
  • $\begingroup$ Yes, it's a way of displaying the results but I don't think a computation is any different. See my edit above. $\endgroup$ – Phil H Oct 20 '18 at 19:30
  • $\begingroup$ I can see everything happening, except I have two questions. 1. Are you getting the value in the second tree; first branch; top leaf; .05425 from the fact all probabilities should sum to 100%? E.g since we know .94575 then the other leaf is .94575 + x = 100? Question 2. I am not seeing the correlation between the .05425 and .5330. Where it says "Positive" could we insert your top tree? I am not seeing how those numbers are connected. Thanks again for your help. $\endgroup$ – user1787331 Oct 22 '18 at 16:38
  • $\begingroup$ 1. Yes, probabilities should sum to $1$ ($100\%$) where the branches split. 2. Sorry, I had a typo in there, it should have been $.5530$ and not $.5330$ as in the calculation in the $1$ st diagram. $\frac{.03}{.05425} = .5530$ $\endgroup$ – Phil H Oct 22 '18 at 17:05
  • $\begingroup$ One more comment, a notion that helps with understanding many conditional probability problems, especially in false positive and false negative scenarios, is the probability of having the disease if testing positive is the proportion of true positives to true plus false positives. And the probability of not having the disease is the proportion of false positives to true plus false positives. $\endgroup$ – Phil H Oct 22 '18 at 17:16

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