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Answer:
0.0098
Step-by-step explanation:
Given,
- Probability of test is positive if the person has disease, P(T/D)=0.99
Probability of test is negative if the person has disease,
P(T'/D) = 1-0.99 = 0.01
- Probability of test is negative if the person hasn't disease, P(T'/D')= 0.99
Probability of test is positive if the person hasn't disease,
P(T/D') = 1 - 0.99 = 0.01
- Probability of occurrence of disease, P(D) = 0.0001
Probability of not occurrence of disease,
P(D) = 1 - 0.0001 = 0.9999
Probability that test will be positive either disease is present or not,
P(T) = P(T/D).P(D)+P(T/D').P(D')
=0.99 x 0.0001 + 0.01 x 0.9999
= 0.000099 + 0.009999
= 0.010098
So, the probability that the person will have disease if the test is positive,
= 0.0098
So, the required probability will be 0.0098.