About 20 women in 80,000 have breast cancer. Suppose the mammogram detecting breast cancer when a...
About 20 women in 80,000 have breast cancer. Suppose the mammogram detecting breast cancer when a woman has breast cancer is 0.90 and the mammogram detecting breast cancer on a woman who does not have breast cancer is 0.01. If a woman has breast cancer, what is the probability that the mammogram will not detect breast cancer? If the mammogram shows breast cancer, what is the probability that she truly has breast cancer? If the mammogram shows a woman has breast cancer, what is the probability that she does not have breast cancer?
Homework Answers
P(breast cancer)= 20/80000
= 0.00025
P(positive | breast cancer)= 0.90
P(positive | no breast cancer)= 0.01
Thus, P(negative | breast cancer)= 1- P(positive | breast cancer)
= 1-0.90
= 0.10
P(breast cancer | positive)= P(positive | breast cancer)* P(breast cancer)/P(positive | breast cancer)* P(breast cancer) + P(positive | no breast cancer)* P(no breast cancer)
= 0.90*0.00025/ (0.90*0.00025 + 0.01*0.99975)
= 0.000225/ (0.000225+0.0099975)
= 0.02201027
P(no breast cancer | positive)= P(positive | no breast cancer)* P(no breast cancer)/P(positive | breast cancer)* P(breast cancer) + P(positive | no breast cancer)* P(no breast cancer)
= 0.01*0.99975/ (0.90*0.00025 + 0.01*0.99975)
= 0.0099975/ (0.000225+0.0099975)
= 0.9779897
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