Question

A certain virus infects one in every 2000 people. a test used to detect the virus in a person is positive 96% of the time if the person has the virus and 4% of the time if the person does not have the virus. Let A be the event "that the person is infected" and B be the event "the person tests positive."Find the probability that a person does not have the virus given that they test negative, i.e. find P (not A|not B)

Answer 1

From the given data, the following Table is calculated:

	A=Infected	not A =Not infected	Total
B = Test positive	0.0005 X 0.96 = 0.00048	0.9995 X 0.04 = 0.03998	0.04046
not B = Test negative	0.0005 - 0.00048 = 0.00002	0.9995 - 0.03998 = 0.95952	0.95954
Total	1/2000 = 0.0005	1- 0.0005 = 0.9995	1.00

P(not A/ not B) = P(not A AND not B)/P(not B)

                          = 0.95952/0.95954 = 0.99997916

So,

Answer is:

0.99997916