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carlosleblanc26
16.03.2020 •
Mathematics
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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Ответ:
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
Ответ: