4. In a certain region of the country it is known from past experi- ence that th

Question

4. In a certain region of the country it is known from past experi- ence that the probability of selecting an adult over 40 years of age with cancer is .05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is .78 and the probability of incorrectly diagnosing a person without cancer as having the disease is .06, what is the probability that a person diagnosed as What is the probability that a person does not have the disease if they are diagnosed as not having the disease? In both cases assume that the person is having cancer actually has the disease? over 40

Explanation / Answer

a)probability of diagnosed as cancer =P(have cancer and diagnosed +does not have cancer and diagnosed)

=0.05*0.78+(1-0.05)*0.06=0.096

therefore probability of having disease given diagnosed =P(have cancer and diagnosed)/P(diagnosed)

=0.05*0.78/0.096=0.40625

b)

P(not diagnosed) =1-P(diagnosed) =1-0.0904 =0.904

therefore probability of not having disease given diagnosed as not having=

=P(not have and not diagnosd)/P(not diagnosed) =(1-0.05)*(1-0.06)/0.904=0.987832

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