Answer to Question #213730 in Statistics and Probability for Kuzi

Question

In a study, physicians were asked what the odds of breast cancer would be in a woman who was initially thought to have a 1% risk of cancer but who ended up with a positive mammogram result (a mammogram accurately classifies about 80% of cancerous tumors and 90% of benign tumors). Ninety-five (95) out of a hundred physicians estimated the probability of cancer to be about 75%. Do you agree?

Answer 1

Events

p = mammogram result is positive

B = tumor is benign,

M = tumor is malignant

$P(M) = 1 \; \%=0.01 \\ P(B) = 1 -P(M) \\ = 1 -0.01 \\ = 0.99 \\ P(p|M) = 0.8 \\ P(\bar{p}|B)=0.9 \\ P(p|B) = 1 -0.9=0.1$

The probability that a patient with a positive mammogram actually has a tumor (Bayes' formula in this case) is

$P(M|p) = \frac{P(p|M)P(M)}{P(p|M)P(M) + P(p|B)P(B)} \\ = \frac{0.8 \times 0.01}{(0.8 \times 0.01) + (0.1 \times 0.99)} \\ = 0.0748 \\ = 7.48 \; \%$

I do not agree.