The probability that a person has a deadly virus is 5 in one thousand. A test will correctly diagnose this disease 95% of the time and incorrectly on 20% of occasions. (a) find the probability of this test giving a correct diagnosis. (b) given that the test diagnoses the patient as having the disease, what is the probability that the patient does not have the disease? (c) given that the test diagnoses the patient as not having the disease, what is the probability that the patient does have the disease?
Answers
D= diseased
+ve = diagnosed as having disease
q9a
P(correct diagnosis)
= P(D & +ve) + P(D' & -ve)
(5/1000)*.95 + (995/1000)*.8
= .8008
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q9b
P(D' & +ve) = 995/1000*.2 = 199/1000
P(D & +ve) = 5/1000*.95 = 4.75/100
P(+ve) = sum of the above = 203.75/1000
P(D' | +ve) = [199/1000] / [203.75/1000]
= 0.9767
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q9c
P(D & -ve) = 5/1000*.05 = .25/1000
P(D' & -ve) = 995/1000*.8 = 796/1000
P(-ve) = sum of the above = 796.25/1000
P(D | -ve) = [.25/1000] / [796.25/1000]
= .0003
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q10a
P[+ve]
= P(D & +ve) + P(D' & +ve)
= .03*.9 + .97*.02
= .027 + .0194
= .0464
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q10b
P(D | +ve) = .027/.0464
=.5819
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q10c
P(D' & -ve) = .97*.98 =.9506
P(D & -ve) = .03*.1 = .003
P(-ve) = sum of the above = .9536
P(D' | -ve) = .9506/.9536
=.99
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