Talk:Confusion of the inverse

== "physicians were asked what the chances of malignancy with a 1% prior probability of occurring and a positive test result from a diagnostic known to be 80% accurate with a 10% false positive rate for that type of test" == This sentence is so complex that is almost impossible to understand, unless you read it several times. Some rewording might be useful.--85.69.200.81 (talk) 16:50, 14 July 2010 (UTC)

http://rationalwiki.org/index.php/Conditional_Probability_Fallacy link is broken. — Preceding unsigned comment added by 66.68.25.227 (talk) 03:39, 17 July 2011 (UTC)

Revision of Example 2?
(I have updated the following slightly--Nø (talk) 09:33, 30 June 2014 (UTC))

The article [http://www.dcscience.net/?p=6473 On the hazards of significance testing. Part 1: the screening problem] on the site DC's Improbable Science has a closely matching example. Below I adapt our example (that I wrote years ago for the page Conditional probability, from where it was later merged into this article) with the values from Improbable Science. Perhaps we should use this version instead - and perhaps some of the text I improvised should be replaced by some of the text from Improbable Science, properly sourced. I won't go ahead with this at this time, but feel free...!

Suppose 100 individuals in a group of 10000 suffer from the disease, and the rest are well. Choosing an individual at random,
 * $$P(\text{ill}) = 1\% = 0.01\text{ and }P(\text{well})=99\%=0.99.$$

Suppose that the screening test has a specificity of 95%, i.e., when it is applied to a person not having the disease, there is a 95% chance of getting a negative result, but a 5% chance of getting a false positive result:
 * $$P(\text{negative}|\text{well})=95\%,\text{ and } P(\text{positive}|\text{well})=5\%.$$

Finally, suppose that the test has a sensitivity of 80%, i.e., when it is applied to a person having the disease, there is an 80% chance of getting a positive result, but also a 20% chance of a false negative result:
 * $$P(\text{positive}|\text{ill})=80\%,\text{ and }P(\text{negative}|\text{ill})=20\%.$$

Calculations

The fraction of individuals in the whole group who are well and test negative:


 * $$P(\text{well}\cap\text{negative})=P(\text{well})\times P(\text{negative}|\text{well})=99\%\times95\%=94.05\%.$$

In absolute numbers, this would be 9405 true negatives out of the entire group of 10000.

The fraction of individuals in the whole group who are ill and test positive (true positives):


 * $$P(\text{ill}\cap\text{positive}) = P(\text{ill})\times P(\text{positive}|\text{ill}) = 1\%\times80\% = 0.80\%.$$

The fraction of individuals in the whole group who have false positive results:


 * $$P(\text{well}\cap\text{positive})=P(\text{well})\times P(\text{positive}|\text{well})=99\%\times5\%=4.95\%.$$

The fraction of individuals in the whole group who have false negative results:


 * $$P(\text{ill}\cap\text{negative})=P(\text{ill})\times P(\text{negative}|\text{ill}) = 1\%\times20\% = 0.20\%.$$

Furthermore, the fraction of individuals in the whole group who test positive:



\begin{align} P(\text{positive}) & {} =P(\text{well }\cap\text{ positive}) + P(\text{ill} \cap \text{positive}) \\ & {} = 4.95\%+0.80\%=5.75\%. \end{align} $$

Finally, the probability that an individual who tests positive is actually well:


 * $$P(\text{well}|\text{positive})=\frac{P(\text{well}\cap\text{positive})} {P(\text{positive})} = \frac{4.95\%}{5.75\%}= 86\%.$$

In other words, only 80 out of the 575 individuals who test positive are actually ill. — Preceding unsigned comment added by Nø (talk • contribs) 13:24, 27 June 2014 (UTC)