a lab blood test is 99%effective in detecting acertain disease. however test also yield a false positive result for .5% of the healthy person tested. if .1 % of the population actually has the disease what is the probability that a person has the disease given that his result is positive
Answers
Let E
1
and E
2
be the respective events that a person has a disease and a person has no disease
.
Since E
1
and E
2
are events complimentary to each other.
∴P(E
1
)+P(E
2
)=1
⇒P(E
2
)=1−P(E
1
)=1−0.001=0.999
Let A be the event that the blood test result is positive.
P(E
1
)=0.1% =
100
0.1
=0.001
P(A∣E
1
)=P(result is positive given the person has disease)=99% =0.99
P(A∣E
2
)=P(result is positive given that the person has no disease)=0.5% =0.005
Probability that a person has a disease, given that his test result is positive, is given by P(E
1
∣A).
By using Baye's theorem, we obtain
P(E
1
∣A)=
P(E
1
)⋅P(A∣E
1
)+P(E
2
)⋅(A∣E
2
)
P(E
1
)⋅P(A∣E
1
)
=
0.001×0.99+0.999×0.005
0.001×0.99
=
0.00099+0.004995
0.00099
=
0.005985
0.00099
=
5985
990
=
665
110
=
133
22
=0.165
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