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About 1% of women aged between 40-50 contracts breast cancer. A woman with cancer has a 90% chance of positive test from a mammogram, while a woman who doesn't have cancer has a 10% chance of a false positive test. What is the probability that a woman has breast cancer given she just got a positive test.
Using Bayes rule we can easily calculate the result to be 8.3%. So this means that the chance that a woman is tested positive and actually has cancer is ~8 out of 100. So does this mean that there is a 91.7% chance that a woman who doesn't have cancer but is tested positive? Or isn't this the false positive case given as 10%?
What am I missing?

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  • $\begingroup$ Your interpretation is right.. You calculated the probability that a woman has breast cancer given that she just got a positive test. The complement-91.7%- is the probability that she does not have breast cancer. 10% is the probability of a woman getting a positive test, not the probability of a woman not having cancer given a positive test. $\endgroup$ – Goldname Feb 9 '17 at 6:10
  • $\begingroup$ 91.7% is P(doesn't have cancer| positive test). 10% is P(positive test | doesn't have cancer) $\endgroup$ – true blue anil Feb 9 '17 at 6:17
  • $\begingroup$ The 'predictive power of a positive test' PVP = P(Cancer|Pos Mam), which is what you have calculated. You want is P(No Cancer|Pos Mam), which is 1 - P(Cancer|Pos Mam). But be careful not to mix this up with PVN = P(No Cancer|Neg Mam), which requiires a different application of Bayes' Theorem. $\endgroup$ – BruceET Feb 9 '17 at 7:27

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