Question

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

Answer 1

Answer:

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