1% of women at age forty who participate in routine screening have breast
cancer. 80% of women with breast cancer will get positive
mammographies. 9.6% of women without breast cancer will also get positive
mammographies. A woman in this age group had a positive mammography in
a routine screening. What is the probability that she actually has breast
cancer?
麻煩列式,唔該!!
Probability 問題 5分
2007-05-20 10:25 pm
回答 (2)
2007-05-20 10:40 pm
✔ 最佳答案
我的想法如下:Let
event A = "she has breast cancer "
event B = "she get +ve mammographies"
Probability that she REALLY has breast cancer
= P(A|B)
= P(A and B) / P(B)
= 1%x80% / (80%+9.6%)
= 0.008/0.896
= 8/896
= 0.00893 (to 3 sig. fig.)
2007-05-21 6:59 pm
Formula of conditional probability: P(A|B) = P(A and B) / P(B)
And so, possibility that a woman has breast cancer (A) given the condition that she has a positive mammography (B)
= P (has breast cancer and has positive result) / P (has positive result)
= P (has breast cancer and has positive result) / [P (has breast cancer and has positive result) + P(no breast cancer but has positive result)]
= 1% * 80% / (1% * 80% + 99% * 9.6%)
= 7.7640%
收錄日期: 2021-04-18 22:16:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070520000051KK02484