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A screening test has a 90% chance of registering breast cancer if it
exists, as well as a 20% chance of falsely registering cancer when it
does not exist. About one in one hundred women requesting the
screening test end up diagnosed with breast cancer. Ms. X has just been
told that her screening test was positive. What is the probability that 
she has breast cancer?

What is the probability that Ms. X has breast cancer?
– A. About 1%.
– B. About 90%.
– C. About 85%.
– D. About 80%.
– E. About 4%.

P(BC) = .9 --> has breast cancer

P(BC') = .1 --> doesn't have breast cancer

P(FC) = .8 --> falsely registering breast cancer

p(FC') = .2 --> not falsely registering cancer)

My answer (below) is dead wrong, I would like to know what I did wrong. Thanks for taking the time to read.

(.9)(.8)/(.9)(.1) + (.8)(.2) = 2.88

Daniel Fischer
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obsolutemal
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  • Possibly helpful: https://math.stackexchange.com/questions/2279851/applied-probability-bayes-theorem/2279888#2279888 – Ethan Bolker Nov 01 '17 at 22:41

1 Answers1

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She was tested and registered positive.   What is the probability that she has it given that the test is positive?

Bayes' Rule and the Law of Total Probability say:

$$\mathsf P(\operatorname{Has}\mid \operatorname{Positive})~{=\dfrac{\mathsf P(\operatorname{Has})~\mathsf P(\operatorname{Positive}\mid \operatorname{Has})}{\mathsf P(\operatorname{Has})~\mathsf P(\operatorname{Positive}\mid \operatorname{Has})+\mathsf P(\operatorname{HasNot})~\mathsf P(\operatorname{Positive}\mid \operatorname{HasNot})} \\ =\dfrac{(0.01)(0.90)}{(0.01)(0.90)+(0.99)(0.20)}}$$

You had the right form: you used the wrong data.   That one in one hundred have it means that ninety nine in one hundred do not have it.   $90\%$ of the former test positive; and $20\%$ of the later do too.

Graham Kemp
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