Conditional Probability - Testing for Disease

Tutorial on how to calculate conditional probability (Bayes Theorem) for two events P(A), P(B), P(B|A) with two examples using

Probability: The Addition Rule & Conditional Probability

Simple probability question that introduces both the addition rule and the concept of conditional probability.

I hope it helps :)

Harold Walden

G'day guys of got a fairly popular probability question today specifically it's going to be a conditional probability question to the question is asking us to suppose everyone takes a test for whether or not they have a particular disease which can produce either a positive or negative result the probability of an individual being infected with this particular disease is 0.02 it also says that regardless of whether an individual has disease the probability that the test will be correct is 0.95 so what we have to do guys is in a group 100 people were gonna have to estimate approximately

so we have to do guys is the first time a is saying in a group for hundred people approximately how many would return a positive result from the test and part B says that the person receives a positive result from the test find the probability that although the result is positive they do not have the disease

Okay so let's get started the first of me to do is draw a tree diagram so I can identify what all of the different outcome is gonna be and what the results are conditional on the first step in drawing a tree diagram is we can either have the disease or not have the disease and from those two branches we can either return a positive result or negative result so from here we are going to import our probabilities

so the question says the probability that we have the disease is zero. 02 and the probability then must be that we don't have the disease is the rest of 0.98 now from this the probability that the test is correct is 0.95 or 95% to the correct results are if we have the disease and we returned a positive if we don’t have the disease and we returned a negative result to each of those Australians are going to have a 0.95 probability attached to them

so from there the other ones didn't are wrong that is the test comes back wrong is going to be the rest of the probability so we're going to have 0.05 for this one and 0.05 for this one cool so from here we are going to work out what our final probabilities after our four different outcomes from a tree diagram

[syamkriz]

This is really good - From India. I was searching for this

[oleholgerson]

Great explanation, very well done. Thank you

[drkmenon]

Your calculation says that after doing a test with specificity of 95%, here we have 72.1% probability that you don't have the disease even if test is positive. That is not correct. P(D`|T+) = P(T+|D`) × P(D)/P(T+), your formula seems to assume that P(D`T+) is synonymous with P(T+|D`). The correct answer is 1.44%. Correct me if I am wrong.