Suppose a medical test has the following characteristics: Pr(Test +ve Patient Di

Question

Suppose a medical test has the following characteristics: Pr(Test +ve Patient Diseased) = 0.98 Pr(Test-ve | Patient Not Diseased) 0.99 (a) Find Pr(Test -ve | Patient Diseased) and Pr(Test +ve | Patient Not Diseased). Suppose that 1 in 20,000 people have this disease so Pr(Patient Diseased) 0.00005 (b) Compute Pr(Test +ve). Hint: Find Pr(Test +ve, Patient Diseased) and Pr(Test +ve, Patient Not Diseased) (c) Use Bayes' rule to find Pr(Patient Diseasedl Test +ve) (d) Give au ive explamation for the disexepancy be r Patient Diseased | Test +ve) and Pr(Test+ve Patient Diseased).

Explanation / Answer

(a) Pr(Test -ve l Pateint Diseased) = 1 - 0.98 = 0.02

Pr(Test + ve l Patient not diseased) = 1 - 0.99 = 0.01

(b) Pr(Patient Diseased) = 1/20000 = 0.00005

Pr(Patient not diseased) = 1 - 0.00005 = 0.99995

Pr(Test +ve ) = Pr(Test +ve l Patient Diseased) * Pr(Diseased) + Pr(Test +ve l Patient not diseased) * Pr(Patient not diseased)

= 0.98 * 0.00005 + 0.01 * 0.99995 = 0.01005

(b) Here by Bayes' Rule

Pr(Pateint Diseased l Test +ve) = Pr(Test + ve l Patient Diseased) * Pr(Patient Diseased) / [Pr(Test + ve l Patient Diseased) * Pr(Patient Diseased) + Pr(Test + ve l Patient not Diseased) * Pr(Patient not Diseased)]

Pr(Patient Diseased l Test +ve) = (0.98 * 0.00005)/ (0.98 * 0.00005 + 0.01 * 0.99995)

= 0.005

(d) Here, as Pr(Patient Diseased l Test +ve) means among who are test positive, it will find who are patient diseased, which are very few in number . So, as there are Pr(Test +ve l Patient Diseased) ,here it is given that patient diseased who are found positive. So there will be very less number of diseased patients, so out of them positive ones will be very high.

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