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Probability (http://www.cancer-forums.net/probability-
Mooney Posts: 54 Joined: Wed Jan 08, 2014 11:53 pm
t186089.html#p769076) by Mooney » Wed Mar 30, 2016 11:22 pm
Hi, professor. Here is the question. A certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that a person is diagnosed as having cancer. I hope you'll assist me. The answer for this question is 0.096 but I don't know the solution. Thanks for your help. T
Probability (http://www.cancer-forums.net/probability-
Burhtun Posts: 40 Joined: Sun Mar 16, 2014 12:36 am
t186089.html#p769145) by Burhtun » Thu Mar 31, 2016 10:19 am
Hello Orev, Use Bayes Rule!!! P(diagnosed w/cancer)
USEFUL LINKS:
=P(having cancer AND diagnosed as such) + P(no cancer AND diagnosed as having cancer)
American Cancer Society National Cancer Institute Cancer Definition
=P(have cancer)*P(diagnosed with cancer|have cancer) + P(no cancer)*P(diagnosed with cancer|no cancer)
Pablo Horteg
=(0.05)(0.78)+(0.95)*(0.06)
Cancer
=0.039 + 0.057 =0.096 OK? Abe T
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Hern Posts: 48 Joined: Wed Apr 09, 2014 7:12 pm
t186089.html#p769161) by Hern » Thu Mar 31, 2016 12:49 pm
Hi, professor. Here is the question. A certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that a person is diagnosed as having cancer. I hope you'll assist me. The answer for this question is 0.096 but I don't know the solution. Thanks for your help. T
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