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5 July, 09:35

3% of women at age forty who participate in routine mammography screening have breast cancer. 85% of women with breast cancer will get positive mammographies. 9.5% of women without breast cancer will also get positive mammographies. Suppose that a woman in this age group had a positive mammography result in a routine screening. Using the information given, what is the probability that she actually has breast cancer

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  1. 5 July, 09:50
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    P (B/T) = 0.217

    Step-by-step explanation:

    denoting the event T = a woman has a positive mammography, then

    P (T) = probability a woman has breast cancer * probability that a woman has positive mammography given that she has breast cancer + probability a woman has not breast cancer * probability that a woman has positive mammography given that she has not breast cancer = 0.03 * 0.85 + 0.97 * 0.095 = 0.117

    then for conditional probability we can use the theorem of bayes. Denoting the event B = a woman selected at random has breast cancer then

    P (B/T) = P (B∩T) / P (T) = 0.03 * 0.85 / 0.117 = 0.217

    where

    P (B/T) = probability that a woman selected at random has breast cancer given that she had a positive mammography

    P (B∩T) = probability that a woman selected at random has breast cancer and a positive mammography
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